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Varieties of Integration

c� 2015 byThe Mathematical Association of America (Incorporated)

Library of Congress Catalog Card Number 2015944846

Print Edition ISBN 978-0-88385-359-7

Electronic Edition ISBN 978-1-61444-217-2

Printed in the United States of America

Current Printing (last digit):10 9 8 7 6 5 4 3 2 1

The Dolciani Mathematical Expositions

NUMBER FIFTY-ONE

Varieties of Integration

C. Ray RosentraterWestmont College

Published and Distributed by

The Mathematical Association of America

DOLCIANI MATHEMATICAL EXPOSITIONS

Council on Publications and Communications

Jennifer J. Quinn, Chair

Committee on Books

Fernando Gouvea, Chair

Dolciani Mathematical Expositions Editorial Board

Harriet S. Pollatsek, Editor

Elizabeth Denne

Ricardo L. Diaz

Emily H. Moore

Michael J. Mossinghoff

Margaret M. Robinson

Ayse A. Sahin

Dan E. Steffy

Robert W. Vallin

Joseph F. Wagner

The DOLCIANI MATHEMATICAL EXPOSITIONS series of the Mathematical As-sociation of America was established through a generous gift to the Association fromMary P. Dolciani, Professor of Mathematics at Hunter College of the City Universityof New York. In making the gift, Professor Dolciani, herself an exceptionally talentedand successful expositor of mathematics, had the purpose of furthering the ideal ofexcellence in mathematical exposition.

The Association, for its part, was delighted to accept the gracious gesture initiat-ing the revolving fund for this series from one who has served the Association withdistinction, both as a member of the Committee on Publications and as a member ofthe Board of Governors. It was with genuine pleasure that the Board chose to namethe series in her honor.

The books in the series are selected for their lucid expository style and stimulatingmathematical content. Typically, they contain an ample supply of exercises, manywith accompanying solutions. They are intended to be sufficiently elementary for theundergraduate and even the mathematically inclined high-school student to under-stand and enjoy, but also to be interesting and sometimes challenging to the moreadvanced mathematician.

1. Mathematical Gems, Ross Honsberger

2. Mathematical Gems II, Ross Honsberger

3. Mathematical Morsels, Ross Honsberger

4. Mathematical Plums, Ross Honsberger (ed.)

5. Great Moments in Mathematics (Before 1650), Howard Eves

6. Maxima and Minima without Calculus, Ivan Niven

7. Great Moments in Mathematics (After 1650), Howard Eves

8. Map Coloring, Polyhedra, and the Four-Color Problem, David Barnette

9. Mathematical Gems III, Ross Honsberger

10. More Mathematical Morsels, Ross Honsberger

11. Old and New Unsolved Problems in Plane Geometry and Number Theory,Victor Klee and Stan Wagon

12. Problems for Mathematicians, Young and Old, Paul R. Halmos

13. Excursions in Calculus: An Interplay of the Continuous and the Discrete, RobertM. Young

14. The Wohascum County Problem Book, George T. Gilbert, Mark Krusemeyer, andLoren C. Larson

15. Lion Hunting and Other Mathematical Pursuits: A Collection of Mathematics,Verse, and Stories by Ralph P. Boas, Jr., edited by Gerald L. Alexanderson andDale H. Mugler

16. Linear Algebra Problem Book, Paul R. Halmos

17. From Erdos to Kiev: Problems of Olympiad Caliber, Ross Honsberger

18. Which Way Did the Bicycle Go? . . . and Other Intriguing Mathematical Myster-ies, Joseph D. E. Konhauser, Dan Velleman, and Stan Wagon

19. In Polya’s Footsteps: Miscellaneous Problems and Essays, Ross Honsberger

20. Diophantus and Diophantine Equations, I. G. Bashmakova (Updated by JosephSilverman and translated by Abe Shenitzer)

21. Logic as Algebra, Paul Halmos and Steven Givant

22. Euler: The Master of Us All, William Dunham

23. The Beginnings and Evolution of Algebra, I. G. Bashmakova and G. S. Smirnova(Translated by Abe Shenitzer)

24. Mathematical Chestnuts from Around the World, Ross Honsberger

25. Counting on Frameworks: Mathematics to Aid the Design of Rigid Structures,Jack E. Graver

26. Mathematical Diamonds, Ross Honsberger

27. Proofs that Really Count: The Art of Combinatorial Proof, Arthur T. Benjaminand Jennifer J. Quinn

28. Mathematical Delights, Ross Honsberger

29. Conics, Keith Kendig

30. Hesiod’s Anvil: falling and spinning through heaven and earth, Andrew J.Simoson

31. A Garden of Integrals, Frank E. Burk

32. A Guide to Complex Variables (MAA Guides #1), Steven G. Krantz

33. Sink or Float? Thought Problems in Math and Physics, Keith Kendig

34. Biscuits of Number Theory, Arthur T. Benjamin and Ezra Brown

35. Uncommon Mathematical Excursions: Polynomia and Related Realms, DanKalman

36. When Less is More: Visualizing Basic Inequalities, Claudi Alsina and Roger B.Nelsen

37. A Guide to Advanced Real Analysis (MAA Guides #2), Gerald B. Folland

38. A Guide to Real Variables (MAA Guides #3), Steven G. Krantz

39. Voltaire’s Riddle: Micromegas and the measure of all things, Andrew J. Simoson

40. A Guide to Topology, (MAA Guides #4), Steven G. Krantz

41. A Guide to Elementary Number Theory, (MAA Guides #5), Underwood Dudley

42. Charming Proofs: A Journey into Elegant Mathematics, Claudi Alsina and RogerB. Nelsen

43. Mathematics and Sports, edited by Joseph A. Gallian

44. A Guide to Advanced Linear Algebra, (MAA Guides #6), Steven H. Weintraub

45. Icons of Mathematics: An Exploration of Twenty Key Images, Claudi Alsina andRoger B. Nelsen

46. A Guide to Plane Algebraic Curves, (MAA Guides #7), Keith Kendig

47. New Horizons in Geometry, Tom M. Apostol and Mamikon A. Mnatsakanian

48. A Guide to Groups, Rings, and Fields, (MAA Guides #8), Fernando Q. Gouvea

49. A Guide to Functional Analysis, (MAA Guides #9), Steven G. Krantz

50. A Mathematical Space Odyssey: Solid Geometry in the 21st Century, ClaudiAlsina and Roger B. Nelsen

51. Varieties of Integration, C. Ray Rosentrater

MAA Service CenterP.O. Box 91112

Washington, DC 20090-11121-800-331-1MAA FAX: 1-301-206-9789

Preface

While the primary audience for this book is an advanced undergraduatemathematics student, the contents will appeal to any mathematician whohas wondered how the integrals introduced in elementary calculus and inreal analysis courses fit together. By the time a young mathematician hascompleted the first year of graduate school, she will have encountered threeversions of the integral: Riemann, introduced in elementary calculus; Dar-boux, studied in a first real analysis course and often still called a Riemannintegral; and Lebesgue, developed in an advanced analysis course. Most of-ten, these integrals are studied in isolation and with very little connectionor comparison made between the different definitions. This book provides acomparative study of four approaches to integration over an interval Œa; b�:Riemann, Darboux, Lebesgue, and gauge.

In addition to serving as a reference, this book can serve as a text for asecond course in real analysis. Indeed, this manuscript is written with suchusers particularly in mind. The prerequisite first course should include thestandard topics of supremum, infimum, compactness, the mean value the-orem, and sequences of functions. The reader should also be familiar withusing the formal "-ı definitions of limit and continuity in proofs. A series ofappendices containing statements of the most relevant definitions and resultsfrom a first real analysis course is provided for readers who have encoun-tered the requisite ideas but may need to refresh their memories. In addition,readers may find the notational index found at the beginning of the indexhelpful.

While the most celebrated milestone in the development of calculus comesfrom the late 17th century work of Newton and Leibniz, questions and ideasthat lie at the heart of integral calculus were introduced by Eudoxus (4thcentury BCE) and Archimedes (3rd century BCE). The ideas of the differ-ential and integral calculus (brought together by Newton and Leibniz) werepowerful forces in the advancement of science. But cracks in the founda-tions of the subject (identified early on by Bishop Berkeley) became increas-ingly apparent toward the end of the 18th century. By the end of the 19th

ix

x Preface

century, Cauchy, Riemann, and Darboux had addressed these foundationalissues and had provided solid foundations for the integral calculus. But thenewly grounded integral calculus was challenged again by a new set of is-sues arising from applications in differential equations. Lebesgue, Henstock,Kurzweil, and others developed new approaches to meet these challenges.Subsequently, Lebesgue’s ideas have been extended into other domains inwhich both the regions of integration and the types of values produced bythe integrals are greatly generalized.

This book explores the critical contributions by Riemann, Darboux,Lebesgue, Henstock, and Kurzweil and provides a glimpse of more recentvariations of the integral. Though historical background is useful in moti-vating and framing the questions to be addressed, the primary focus of thisbook is not historical. The primary goals are (1) to provide you with an un-derstanding of and an appreciation for the work done in formalizing andextending the ideas of the integral calculus, (2) to help you to think like amathematician, and (3) to provide you with an opportunity to develop into areader of professional mathematics. While the first goal is self-explanatory,the other two deserve some explanation.

Most often mathematical texts present material in a final, polished form.This is good. However, the budding mathematician is often left with the im-pression that ideas spring full-grown from the ground and that they couldnever have been otherwise. Since you are reading this book, you alreadyknow that mathematical knowledge is hard-won, but you may not have thesense for how many different paths have been tried and abandoned in the pro-cess of developing the theories that we have today. The general structure ofthis book, as it moves through the development of various integrals, encour-ages you to think about how mathematics develops and the implications ofdifferent approaches. This book also addresses the issue in a more localizedfashion. The exercises will continually prod you to think about why certainchoices in a proof were made or why an alternate, apparently simpler or moreobvious approach was not used. The intended effect is that you, the reader,will come away with a greater flexibility in the way you see mathematics.

This text also seeks to improve your mathematics reading ability. Proof-oriented mathematics texts try to include all the details and backgroundused in a proof. Student writing tends to reflect this practice by includinglong sequences of algebraic derivations. This happens even (particularly?)when a student’s solution or proof fails to address a critical logical step ortwo. Mathematics found in professional journals takes a different approach.The writing is terse. The reader is expected to be familiar with a significant

Preface xi

breadth of background information and to understand (or at least appreciate)the motivation for the work. The professional proof then provides naviga-tional landmarks that the reader is expected to use to construct a path fromthe hypothesis to the conclusion. In particular, the reader is expected to fillin much of the algebra. To use another metaphor, the version of a proof thatappears in a professional journal serves as a skeleton and the reader mustfill in many details to “flesh out” the proof. This type of writing requiressignificant engagement on the part of its readers since, before understandinga proof, the reader must identify the places where details must be filled inand must construct the bridging argument or computation. Consequently, itis not uncommon to spend an hour or even a couple of days working to digesta proof.

You may well ask why this difference in writing styles exists. This is anexcellent question to which I offer only tentative, partial answers. One factoris the mathematical culture. This is the way mathematicians have written forquite some time and the practice is not likely to change soon. Of course, thisanswer does nothing to explain how things came to be this way. One sig-nificant pressure in this direction is the cost of publishing or, moving furtherinto the past, the effort required to make papyrus sheets. Until fairly recently,it was customary to charge authors or their institutions a fee for preparing apaper for publication. These charges could be as high as several hundreddollars per page. In addition, most journals have limits on the length of thepapers they will publish. Both of these practices exert pressure on authors tocompress their writing.

With the widespread use of TEX, papers no longer need to be retyped intoa form suitable for publication so the assessment of page charges has largelydisappeared. Online publishing has the potential to eliminate the constraintson article length. In this new environment, perhaps you will help create acultural shift so that professional mathematics papers will include more ofthe details. In the meantime, it is important for developing mathematiciansto learn to read journals as they are written rather than as one might wishthey were written.

This book attempts to help you become more adept at this task. The proofsat the beginning of the book will call attention to the places where detailsmust be filled in. Usually this takes the form of a reference to an exercisethat often (but not always) includes suggestions about how to approach theproblem. These suggestions should be taken for what they are: suggestions.You should feel free to approach the problem a different way or have an-other type of insight. Some exercises will ask you to fill in minor steps in

xii Preface

a proof and will not require a great deal of work. Other exercises will askyou to modify a proof to fill in the missing details signaled by the phrase“similarly, . . . ” that appears so frequently in mathematical writing. These ex-ercises are generally grouped under the classification of “Filling the Gaps.”Other exercises ask you to reflect on the structure of definitions and proofs.Still other exercises are independent of the proofs and are designed to helpyou gain a deeper understanding of the material or greater facility workingwith the ideas. The latter types of exercise are labeled as “Deeper Reflection”and may demand more significant effort. Occasionally, a separate section la-beled “Related Ideas: Deeper Reflections” is included at the end of a chapter.The problems in this section build on but are not directly related to the ideasin the chapter. Since they play such an important role in this text, you shouldat least read all of the “Filling the Gaps” exercises even if you do not intendto complete any of them. It is highly recommended that you read the “DeeperReflections” exercises as well.

As the book progresses, the exercises will remain, but they will be refer-enced less frequently in the main body. At this point, you will be well servedby keeping a finger or bookmark in the exercises when reading a proof. Notethose places where you have identified details that should be verified. Thenread the exercises to see if there are any significant details that you havemissed.

While the proofs use a style comparable to that used in a professional jour-nal, the transitional material is more conversational, informal, and reflective.The transitions provide a time to discuss the historical reasons for the trackof investigation, to suggest motivational questions, to outline the general arcof the work, and to compare the ways that the different approaches to inte-gration affect the way that proofs are constructed.

Thank you for engaging this book. Please address any error notices, com-ments, and suggestions to [email protected].

I would like to thank Professor Russell Howell and students Tyler Bran-nan and Daniel Ray for their careful reading of the book and for their manycorrections and suggestions. I also owe a huge debt of gratitude to the re-viewers and editors of the Dolciani series whose subsequent suggestions foradditions and modifications greatly strengthened the book and whose carefulreading identified many needed corrections.

Contents

Preface ix

1 Historical Introduction 11.1 Greek foreshadowing . . . . . . . . . . . . . . . . . . . . . 11.2 Newton and Leibniz . . . . . . . . . . . . . . . . . . . . . 41.3 Cauchy, Riemann, and Darboux . . . . . . . . . . . . . . . 71.4 Lebesgue . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5 Henstock and Kurzweil . . . . . . . . . . . . . . . . . . . . 131.6 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 The Riemann Integral 252.1 Riemann integrability . . . . . . . . . . . . . . . . . . . . . 272.2 Subintervals . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3 The fundamental theorems . . . . . . . . . . . . . . . . . . 392.4 Convergence theorems . . . . . . . . . . . . . . . . . . . . 412.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3 The Darboux integral 513.1 Darboux integrability . . . . . . . . . . . . . . . . . . . . . 533.2 Comparing Riemann and Darboux integration . . . . . . . . 583.3 Additional integrability results . . . . . . . . . . . . . . . . 613.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4 A Functional zoo 714.1 Dirichlet and friends . . . . . . . . . . . . . . . . . . . . . 714.2 Trigonometric series . . . . . . . . . . . . . . . . . . . . . 724.3 Friends of Cantor . . . . . . . . . . . . . . . . . . . . . . . 774.4 Volterra’s example . . . . . . . . . . . . . . . . . . . . . . 81

xiii

xiv Contents

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5 Another Approach: Measure Theory 915.1 Measurable sets I . . . . . . . . . . . . . . . . . . . . . . 935.2 Outer measure . . . . . . . . . . . . . . . . . . . . . . . . 955.3 Measurable sets II . . . . . . . . . . . . . . . . . . . . . . 975.4 Sigma algebras . . . . . . . . . . . . . . . . . . . . . . . . 1005.5 Measurable sets III . . . . . . . . . . . . . . . . . . . . . . 1035.6 Measurable functions . . . . . . . . . . . . . . . . . . . . 1055.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 1095.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6 The Lebesgue Integral 1196.1 Variations . . . . . . . . . . . . . . . . . . . . . . . . . . 1196.2 Reconciling the approaches . . . . . . . . . . . . . . . . . 1246.3 Convergence theorems . . . . . . . . . . . . . . . . . . . . 1346.4 The fundamental theorems . . . . . . . . . . . . . . . . . . 1416.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 1496.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 159

7 The Gauge Integral 1617.1 Definition and basic examples . . . . . . . . . . . . . . . . 1627.2 The art of constructing gauges . . . . . . . . . . . . . . . . 1677.3 Basic integrability results . . . . . . . . . . . . . . . . . . 1727.4 Absolute integrability . . . . . . . . . . . . . . . . . . . . 1777.5 Convergence theorems . . . . . . . . . . . . . . . . . . . . 1867.6 The fundamental theorems . . . . . . . . . . . . . . . . . . 1907.7 Integral relationships . . . . . . . . . . . . . . . . . . . . . 1997.8 Loose ends and Dini derivatives . . . . . . . . . . . . . . . 2017.9 Some reflections on the gauge integral . . . . . . . . . . . 2087.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 2097.11 References . . . . . . . . . . . . . . . . . . . . . . . . . . 226

8 Stieltjes-type Integrals and Extensions 2278.1 Examples and counterexamples . . . . . . . . . . . . . . . 2288.2 Basic integrability theorems . . . . . . . . . . . . . . . . . 2308.3 Evaluation theorems . . . . . . . . . . . . . . . . . . . . . 2328.4 Convergence theorems . . . . . . . . . . . . . . . . . . . . 2378.5 Connecting to measure theory . . . . . . . . . . . . . . . . 2418.6 Integration with measures . . . . . . . . . . . . . . . . . . 247

Contents xv

8.7 Extensions to other types of measures . . . . . . . . . . . . 2518.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 2568.9 References . . . . . . . . . . . . . . . . . . . . . . . . . . 269

9 A Look Back 2719.1 Basic approaches . . . . . . . . . . . . . . . . . . . . . . . 2719.2 Integrable functions . . . . . . . . . . . . . . . . . . . . . 2739.3 Convergence theorems . . . . . . . . . . . . . . . . . . . . 2749.4 The fundamental theorems . . . . . . . . . . . . . . . . . . 2769.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . 277

10 Afterword: L2 Spaces and Fourier Series 27910.1 L2 spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 28110.2 Completeness . . . . . . . . . . . . . . . . . . . . . . . . 28810.3 Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29110.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . 29910.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 30010.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 308

Appendices: A Compendium of Definitions and Results 309A.1 Sets of real numbers . . . . . . . . . . . . . . . . . . . . . 309A.2 Infimums and supremums . . . . . . . . . . . . . . . . . . 311A.3 Sequences of real numbers . . . . . . . . . . . . . . . . . 312A.4 Real-valued functions . . . . . . . . . . . . . . . . . . . . 314A.5 Sequences of functions . . . . . . . . . . . . . . . . . . . 315A.6 Complex numbers . . . . . . . . . . . . . . . . . . . . . . 316A.7 Inner product spaces and projections . . . . . . . . . . . . 317

Index 319

About the Author 325

CHAPTER 1

Historical Introduction

The Riemann integral is usually introduced in elementary calculus classesvia a problem with roots in early Greek mathematics—the problem of find-ing the area of a region R. Modern readers of mathematics expect the areato be expressed as a number. But what does this number mean? One way ofinterpreting the number is to think of it as representing the length of one sideof a rectangle whose other side has unit length. Then the areas of the regionR and the rectangle are the same.

1.1 Greek foreshadowingFollowing this line of thought, a Greek solution to an area problem consistedof finding a square (or equivalently a rectangle or triangle) having the samearea as the region in question.

The lunes of Hippocrates of Chios (c. 470–410 BCE) provide an earlyexample of this type of problem. Hippocrates’ work was done in an attemptto find a square or triangle with the same area as a given circle. Referring toFigure 1.1, Hippocrates was able to prove that the area of a portion of thecircle with diameter AD has the same area as a triangle. Specifically, thecrescent AEDF has the same area as the triangle ACD. (See exercise 1.)Hippocrates’ result reduced the task of finding a square to represent the areaof a circle to the problem of constructing a square to represent the area of theregion AGDE. About two millennia later, Ferdinand von Lindemann wouldshow that this task is impossible by demonstrating that � is a transcendentalnumber. (See [Baker] and [Lindemann].)

Archimedes of Syracuse (287–212 BCE) proved that the area of a circleis equal to the area of a triangle whose base has the same length as the cir-cumference of the circle and whose height is the radius of the circle. (Seeexercises 7 and 8.) In modern notation, the area of the triangle, and so of thecircle, is 1

2� .2�r/ � r or �r2.

1

2 CHAPTER 1. Historical Introduction

CA B

D

E

F

G

Figure 1.1. Lune of Hippocrates

While Archimedes’ formulation of the result may seem odd to us, we needto appreciate that the first glimmers of algebraic notation did not appearuntil the 3rd-century CE work of Diophantus of Alexandria and that alge-braic notation did not reach a state of relative maturity until the 17th centuryin the work of Descartes and Fermat. Consequently, the statement that thearea of a circle is �r2 would have been far more foreign to the Greeks thanArchimedes’ statement relating the area of the circle and a triangle is to us.

While the development of algebra is critical to the development of mod-ern calculus, Archimedes’ work on the area of a circle engaged in signif-icant calculus-like thinking without algebra. Here we review a result ofArchimedes that has an even stronger calculus flavor: the quadrature1 of theparabola.

Suppose that points A and B are the endpoints of a section of a parabolaas illustrated in Figure 1.2. Let C be the intersection between the parabolaand the line l constructed through the midpointM of AB and parallel to theaxis of the parabola. (In terms of coordinate geometry, C is the point on theparabola whose x-coordinate is midway between x-coordinates of the twoendpoints.) Then the area of the parabolic section between A and B is equalto 4

3the area of the inscribed triangle ABC .

Notice that the section of the parabola determined by A and B consists ofthree parts: the triangle ABC , the parabolic section determined by A and C ,and the parabolic section determined by C and B . Expand the triangle ABCto become an inscribed polygon P2 by attaching the inscribed triangle foreach smaller parabolic section. In this first stage, the added triangles use thepoints C1 and C2 located at the intersections of the parabola with the lines

1 The problem of quadratures is the problem of finding the area of an object. The word quadra-ture refers to finding a square of the same area.

1.1. Greek foreshadowing 3

Section of a parabola Inscribed triangle

Inscribed polygon

B

A

B

A

M

Cl

B

A

M

Cl

M1

M2

C1

C2

l1

l2

Figure 1.2. Quadrature of the Parabola

l1 and l2 parallel to the axis of the parabola and through the midpoints ofAC and CB respectively. (A similar triangles argument shows that the linesl1 and l2 could also be defined as passing through the midpoints of AMand MB .) This process can be repeated indefinitely introducing more, butsmaller, triangles and parabolic sections. Each succeeding stage producesa polygon that better fits the original parabolic section. At each stage, thecombined area of the added triangles is 1

4the area of triangles on which they

are based. Exercise 2 outlines a proof of this fact.Let T be the area of ABC , let P be the area of the parabolic section

between A and B , and let Pn be the area of the polygon that results fromn iterations of the process described in the preceding paragraph. Using theresults of the exercises,

P0 D T

P1 D T C1

4T D

1

3

�4�

1

4

�T

P2 D T C1

4T C

1

42T D

1

3

�4�

1

42

�T

P3 D T C1

4T C

1

42T C

1

43T D

1

3

�4�

1

43

�T

� � �

Pn D T C1

4T C

1

42T C � � � C

1

4nT D

1

3

�4�

1

4n

�T:

A modern student with some background in calculus would conclude that

P D

1XkD0

T

4kD

T

1 � 14

D4

3T

or

P D limn!1

1

3T

�4 �

1

4n

�D4

3T .

Not only did Archimedes lack the tools of calculus, but the Greek mindwould have recoiled at the very idea of the completed infinity implicit inP1kD0

T

4k. Instead, Archimedes argued that P could not be greater than 4

3T

nor could it be less than 43T . Hence P must be 4

3T .

Suppose that P were less than 43T . Then for a sufficiently large value of

n, we would find that P < 43T � 1

3�4nT D 1

3T�4 � 1

4n

�D Pn. But this is

impossible since the polygon is contained in the section of the parabola. Onthe other hand, ifP were greater than 4

3T , we could keep inscribing triangles

until Pn > 43T . But Pn D 1

3T�4 � 1

4n

�is manifestly less than 4

3T: Thus by

a double reductio ad absurdum argument, P D 43T .

1.2 Newton and LeibnizAlthough many of the foundational ideas were previously known in a vagueform, the 17th and 18th centuries witnessed the birth of what we now callcalculus. Isaac Newton (1642–1727) and Gottfried Leibniz (1646–1716)realized the power of the antiderivative and introduced its use. For Newtonand Leibniz the integral was the antiderivative. Using modern notation, givena function f on Œa; b�,

NZ b

a

f .x/ dx D F .b/ � F .a/ , where F 0 D f .

(Here the N reflects the fact that we are referring to Newton’s integral.) Inthis context, the fundamental theorem of calculus and the definition of theintegral are one and the same. There is nothing to prove.

Of course the preceding statement is a gross oversimplification of the sit-uation. For Newton, the fluxion and the fluent were inverse operations inthat the fluxion represented the ratio of the change in a quantity over an in-finitely small time interval and a fluent captured the movement of an objectthat had a given ratio of change to time. Similarly, for Leibniz the integralaccumulated the results of an infinite number of infinitesimals in the form of

1.2. Newton and Leibniz 5

differentials—mathematical objects smaller than any real number but hav-ing some of the properties of real numbers. So instead of starting with adefinition of the integral based on the area under the curve and showing thatit can be computed using an antiderivative, Newton and Leibniz started withan integral defined by an antiderivative and needed to show that the integralrepresents the area under a curve.

Figure 1.3 is a drawing used by Leibniz to support a geometric proof ofthe connection between the antiderivative and the area under a curve.

I shall now show that the general problem of quadratures can be reducedto the finding of a [curve] that has a given law of tangency.

—Leibniz [Struik, pg 282]

In other words, the problem of finding the area under a given curve(AH.H/ in the diagram) is to be solved by finding a second curve (C.C/in the diagram) whose tangent line at any point has a slope that is equal tothe distance of the curve AH.H/ from the line AF.F /. In more modernterminology, the problem of finding the area under the curve y D f .x/ is tobe solved by finding another function F whose derivative is f .

Figure 1.3. Liebniz diagram

In the diagram, the line AF.F / is essentially what we would call the x-axis and the curves AH.H/ and C.C/ are viewed from the perspective ofthis axis. However, C.C/ does not represent a negative function. Instead,the viewer should concentrate on the distance between the curve and the“x-axis”. Similarly, while the slope2 of the tangent line TC appears to benegative according to modern conventions, the distance betweenAF.F / and

2 Note that Leibniz never mentions slope.


TC is increasing near C so the slope is actually positive. The segments GL,LC , CE, and the curve between C and .C / are all infinitesimals and haveno values—unassignable in Leibniz’s terminology. However, the segmentscan be used in ratios. Leibniz used the similarity of TBC , GLC , and CEC 0

(where C 0 is the unmarked point where TC intersects E.C/) together withthe fact that E.C/ and EC 0 are essentially equal to draw his conclusion.3

Leibniz assumed, without verification, that the desired curve C.C/ exists.The Newton-Leibniz integral proved to be surprisingly useful for solving

both physical and mathematical problems. As is illustrated in the followingexample, the previous development of solid algebraic notation and proce-dures was critical to the integral’s usefulness.

Independently of his work on the calculus, Newton developed a general-ized binomial formula. By solving for leading coefficients in special cases,Newton concluded that

.1C x/˛ D

1XkD1

˛

k

!xk

where ˛

k

!D˛ .˛ � 1/ .˛ � 2/ � � � .˛ � k C 1/

kŠ.

When ˛ is a positive integer,�˛k

�agrees with the usual definition of the

binomial coefficient for 0 � k � ˛ and, since ˛� .˛C 1/C 1 D 0,�˛k

�D 0

for k � ˛ C 1. This means that Newton’s formula agrees with the standardbinomial theorem for positive integer values of ˛.

Newton did not prove his formula for general values of ˛. That would haveto wait until Lagrange and Cauchy provided error bounds for Taylor series.Instead, Newton convinced himself of the correctness of his expression for.1C x/˛ by considering special cases. Using ˛ D 1

2for example, Newton

truncated the series after a manageable number of terms and verified that,when squared, the truncated series matched 1C x through the same numberof terms. (See exercise 11.)

y = 1 – x2

Figure 1.4. Area of �=4

3 For a detailed explanation, see [Nauenberg].

1.3. Cauchy, Riemann, and Darboux 7

In Newton’s capable hands, the generalized binomial expansion provideda tool for approximating � by computing the area under y D

p1 � x2 from

x D 0 to x D 1.

�

4D N

Z 1

0

p1 � x2 dx

D NZ 1

0

1XkD0

1=2

k

!.�1/k x2k dx

D

1XkD0

1=2

k

!.�1/k N

Z 1

0

x2k dx

D

1XkD0

.�1/k

1=2

k

!x2kC1

2k C 1

ˇˇ1

0

D

1XkD0

.�1/k

1=2

k

!1

2k C 1

D 1 �1

6�

1XkD2

1 � 3 � � � .2k � 3/

2kkŠ .2k C 1/.

This expression for � is not particularly elegant or efficient; quite a fewterms are needed to obtain much accuracy. But there are more basic concernshere. No justification is provided for interchanging the infinite sum with theintegral. It happens to work in this case, but the interchange can produceerroneous results. In fact, Newton did not even prove that the sum converged.The ratio test, typically used to show that

P1kD2

1�3��.2k�3/

2kkŠ.2kC1/converges (see

exercise 12), was developed by Jean-Baptiste le Rond d’Alembert who wasonly nine years old when Newton died.

1.3 Cauchy, Riemann, and DarbouxWhile the Newton-Leibniz integral was useful, its foundations were notsolid. Philosopher Bishop George Berkeley strongly criticized mathemati-cians for using techniques with such flimsy groundings.

It must, indeed, be acknowledged, that [Newton] used Fluxions, likethe Scaffold of a building, as things to be laid aside or got rid of, assoon as finite Lines were found proportional to them. But then thesefinite Exponents are found by the help of Fluxions. Whatever thereforeis got by such Exponents and Proportions is to be ascribed to Fluxions:

which must therefore be previously understood. And what are theseFluxions? The Velocities of evanescent Increments? And what arethese same evanescent Increments? They are neither finite Quantitiesnor Quantities infinitely small, nor yet nothing. May we not call themthe Ghosts of departed Quantities?—Bishop Berkeley in The Analyst

Foundational questions about fluxions, differentials, and infinitesimalsmight be ignored, but difficulties related to convergence of infinite serieswere even more apparent and more difficult to disregard. Newton, Leib-niz, and their contemporaries made frequent use of infinite series. But whatdoes an infinite series mean? How can one contemplate adding an infinitudeof terms? Particular series such as

P1nD0 .�1/

n were recognized as beingproblematic quite early on. By regrouping, the series seems to converge to1 or 0.

1C .�1C 1/C .�1C 1/C .�1C 1/C � � � D 1;

but

.1 � 1/C .1 � 1/C .1 � 1/C .1 � 1/C � � � D 0:

Concern over the legitimate handling of the infinitely large and infinitelysmall rose to a level where, in 1784, the Berlin Academy offered a prize toany mathematician who could provide a “clear and precise theory of what iscalled the infinite.” Even though the prize was awarded in 1786, the mathe-matical community still was not satisfied that it had a satisfactory account-ing of the infinite. That would have to wait another 35 years until Cauchypublished Cours d’Analyse in 1821.4 In this text, Cauchy introduced the "-ıdefinition of convergence and eliminated the need to deal with infinitesimals.

Cauchy used the "-ı paradigm not only to define the limit of a functionbut also to provide a clear statement of the meaning of the integral. Cauchy’sdefinition takes a left-endpoint approach.

Definition 1. Let f be a function defined on Œa; b�. Then f is integrableover Œa; b� if there is a value A such that for every " > 0 one can find a valueı > 0 such that ˇ

ˇnXiD1

f .xi�1/ .xi � xi�1/ � A

ˇˇ < "

4 See [Bradley and Sandifer] for an annotated English translation.

for any choice of values a D x0 < x1 < x2 < � � � < xn D b satisfyingxi � xi�1 < ı for 1 � i � n. The value of the integral of f over Œa; b�,denoted by C

R ba f , is A.5

While Cauchy’s definition can be applied more generally, he only ap-plied the definition to continuous (or at least piecewise continuous) func-tions. In the context of continuous functions, the question of integrability ismoot. Cauchy provided an argument that all continuous functions are inte-grable (i.e., the value A required by his definition always exists). In addition,Cauchy proved that if F has a continuous derivative on Œa; b�, then

CZ b

a

F 0 D F .b/ � F .a/ .

You will immediately recognize this statement as a form of the fundamen-tal theorem of calculus. Spend a few minutes comparing Cauchy’s definitionand theorem with those of Newton and Leibniz. Notice the shift from geo-metric conceptualization to a more algebraic presentation.

Even before Cauchy published Cours d’Analyse, mathematical develop-ments were occurring for which Cauchy’s definition proved to be awkwardand inadequate. In 1807, Joseph Fourier submitted a memoir entitled Theoryof the propagation of heat in solid bodies. In his paper, Fourier proposed us-ing series of trigonometric polynomials (what we now call Fourier series) tosolve the differential equation that models the steady-state temperature in alamina or thin plate. Fourier’s paper was rejected. Powerful mathematiciansof the day, including Laplace, Lagrange, and Poisson, objected to Fourier’sapproach in large part because it violated their sense of what a functionshould be. But while Fourier’s methods were viewed as problematic, his ap-proach seemed to produce solutions that agreed with experimental results.

In 1810, the Institut de France announced that the next Grand Prize inMathematics would be on the subject of “the propagation of heat in solidbodies.” Fourier rewrote his paper and submitted it to the competition. In1812, Fourier was awarded the substantial prize associated with the compe-tition, but his prize-winning paper was not immediately published. Whiledisciplinary politics certainly played a role, the delay was also a resultof the view that, while Fourier had the right equations, his methods wereproblematic.

5 The symbol CR

is used to denote the Cauchy integral. It does not refer to a multiple ofan integral. Similar notation will be used throughout the text to identify particular types ofintegrals.

In addition to requiring a modified view of a function, Fourier’s meth-ods raised questions related to convergence of series and the interactionsbetween integration and series.6 Ten years after winning the prize, Fourier’swork was published as The analytical theory of heat. Later (after Fourier waselected secretary of the Institut de France), his original paper was publishedin two parts in the Memoirs of the Academie des Sciences. The second partappeared in 1826, the year of Bernhard Riemann’s birth.

Riemann’s first university lectures, given in 1854, introduced what wouldlater become Riemannian geometry and opened up the idea of using higherdimensions (beyond three or four) to describe physical reality. In this sameyear, Riemann submitted a paper entitled On the representability of a func-tion by a trigonometric series in support of his certification as lecturer at theUniversity of Gottingen. This paper, not published until 1868, introduced ageneral theory of trigonometric series that extended the ideas behind Fourierseries. Riemann’s paper also included a major milestone in the developmentof integral calculus. Section 4: On the concept of a definite integral and theextent of its validity modified Cauchy’s definition of the integral to the onenow found in elementary calculus courses. Specifically, Riemann’s definitionrequired that ˇ

ˇnXiD1

f .ti / .xi � xi�1/ � A

ˇˇ < "

not just when ti is the left endpoint of Œxi�1; xi � but for any choice ofti 2 Œxi�1; xi �. While this change has the appearance (and reality) of com-plicating the definition, it makes many of the proofs involving the integralsignificantly simpler.

Since the functions produced by trigonometric series are not necessarilycontinuous, Riemann was interested in when the integral could be appliedto discontinuous functions. Although the result was not proved until wellafter Riemann’s death, any bounded function that is continuous almost ev-erywhere (a concept to be defined in later chapters) is Riemann integrable.Any function with a finite or even a countably infinite number of points ofdiscontinuity is included in the set of Riemann integrable functions. As aresult, the fundamental theorem of calculus can be extended to conclude that

RZ b

a

F 0 D F .b/ � F .a/

as long as F 0 is bounded and continuous almost everywhere on Œa; b�.

6 See Section 2 on trigonometric series in Chapter 4 for more information on Fourier’s methodand its problems.


The last major player of this era was Gaston Darboux (1842–1917). Dar-boux edited and republished Fourier’s The analytical theory of heat in 1888.Darboux also reformulated Riemann’s definition of the integral by replacingthe value of the function at a single point, f .ti /, with the infimum and supre-mum of the values of f over the interval Œxi�1; xi �. Intuitively, a function isDarboux integrable over Œa; b� if the sums

nXiD1

infŒxi�1;xi �

f � .xi � xi�1/ andnXiD1

supŒxi�1;xi �

f � .xi � xi�1/

are essentially the same. (Here infŒxi�1;xi � f D inf ff .x/ W x 2 Œxi�1; xi �gis the infimum of the set of values of f over the input set Œxi�1; xi �. Thesupremum is to be understood similarly.) If one considers the integral geo-metrically for positive functions, Darboux used inscribed and circumscribedrectangles instead of using general approximating rectangles. This formula-tion of the integral, while again appearing to be more complicated, had theeffect of greatly simplifying the proofs associated with the Darboux integral.

Figure 1.5. Newton’s diagram

Interestingly, Darboux’s definition comes full circle to one of Newton’sresults. In Book I, Section I of The mathematical principles of natural phi-losophy, Newton’s lemma II states that, for decreasing functions, the sumsof the areas of the inscribed rectangles and the sums of the areas of the cir-cumscribed rectangles converge to the area under the original curve. (SeeFigure 1.5.)

. . . Then if the breadth of those parallelograms be supposed to be di-minished and their number augmented in infinitum: I say that the ulti-mate ratios which the inscribed figure AKbLcMdD, the circumscribed

figure AalbmcndoE, and the curvilinear figure AabcdE, will have to oneanother, are ratios of equality.7

1.4 LebesgueWhile Riemann and Darboux provided precise and workable definitions ofthe integral and developed criteria that guaranteed that a function is inte-grable, the state of the theory remained unsatisfactory from the perspectiveof Fourier series (a particular type of trigonometric series which will be in-troduced in Chapter 4). Niels Henrik Abel (1802–1829) and Johann PeterGustav Lejeune Dirichlet (1805–1859) developed theories and techniquesthat could be used to prove that Fourier series converged, but the interactionbetween Fourier series and integrals remained murky. One major difficultywas that, while the convergence of a Fourier series could be established, theresulting function might not be Riemann integrable. Cauchy had proved thatwhen a sequence (or series) of continuous functions converges uniformly,the limit is continuous. In a more general context, the uniform limit of a se-quence (or series) of integrable functions is again integrable. Unfortunately,some of the most interesting problems (including one of Fourier’s originalexamples) involve discontinuous functions. In such cases, the Fourier seriescannot converge uniformly.

By approaching integration from a different perspective, Henri Lebesgue(1875–1941) found a way to address this problem. Instead of partitioningŒa; b�, the domain of f , Lebesgue partitioned f ’s range. In some sense, theLebesgue approach slices the region under a function horizontally insteadof vertically. Whereas a term in a Riemann sum is f .ti / .xi � xi�1/ whereŒxi�1; xi � is a subinterval of Œa; b� and ti 2 Œxi�1; xi �, a Lebesgue term isyi� .Ai / where Œyi�1; yi � is a subinterval of the range of f , Ai is the set ofall x 2 Œa; b� for which yi�1 < f .x/ � yi , and � .Ai / is the “length” ofAi . (See Figure 1.6.) The sums used to define the Lebesgue integral havethe form

nXiD1

yi� .Ai / .

A function is Lebesgue integrable if these sums converge in an appropriatesense.

7 You can read an English translation of Newton’s lemma and proof on Google books atbooks.google.com/books?id=Tm0FAAAAQAAJ. See page 42 and the unnumbered page after44.

1.5. Henstock and Kurzweil 13

Riemann term Lebesgue term

Figure 1.6. Comparison of Riemann and Lebesgue terms

In addition to being visually more complicated, Lebesgue’s approach re-quires us to make sense of the “length” of rather arbitrary sets. This is notan easy task, but the effort to generalize length to a broader class of sets isrichly rewarded. The Fourier series of a bounded function will always con-verge and it will do so in a way that allows termwise integration. Lebesgue’sresults, published in a 1902 paper, finally resolved the issues generated byFourier’s work over 90 years earlier.

1.5 Henstock and Kurzweil

Lebesgue’s formulation of the integral is not the only way to address thequestions flowing from Fourier’s work. The gauge integral, developed pri-marily by Ralph Henstock (1923–2007) and Jaroslav Kurzweil (1926–)around 1960, retains the Riemann integral’s focus on the domain of the func-tion to be integrated but works more locally than does the Riemann integral.

The Riemann integral requires thatˇˇnXiD1

f .ti / .xi � xi�1/ � A

ˇˇ < "

for any choice of a D x0 < x1 < x2 < � � � < xn D b satisfying xi �xi�1 < ı and ti 2 Œxi�1; xi � for 1 � i � n. The gauge integral exertsa more localized control on the size of the subintervals by specifying thatxi � xi�1 < ı .ti / for 1 � i � n. In this formulation, ı .�/ is a positivefunction on Œa; b� called a gauge.

For example, if

f .x/ D

(1px; x > 0

0; x � 0

then f is unbounded and so cannot be Riemann integrable over Œ0; 1�. How-ever if we choose the gauge

ı .t/ D

�t=2; t > 0

1; t � 0

then any choice of points satisfying 0 D x0 < x1 < x2 < � � � < xn D 1,ti 2 Œxi�1; xi �, and xi � xi�1 < ı .ti / for 1 � i � n must take t1 D 0.No other choice for t1 allows the left endpoint of the first interval to be 0.This has the effect of ensuring that the corresponding gauge sum will remainbounded. (See exercise 17.)

Of course, the boundedness of the gauge sums does not guarantee that fis gauge integrable. However, given a positive " it is possible to construct agauge that will ensure that

PniD1 f .ti / .xi � xi�1/ is within " of 2. Thus

gR 10f D 2. Since the process of constructing such a gauge is too involved

for our current purposes, we will not do so here. But we will return to thisexample in Chapter 7.

As is the case for Lebesgue integrable functions, the limit of any sequenceof gauge integrable functions will itself be gauge integrable. This means thatall the manipulations used by Fourier to construct trigonometric series tosolve differential equations are justified in the context of the gauge integral.

Of all the standard integrals on R, the gauge integral has the strongestversion of the fundamental theorem of calculus. As long as F is continuouson Œa; b� and the derivative of F exists at all but a countable number of pointsin Œa; b�,

gZ b

a

F 0 D F .b/ � F .a/ :

1.6 Extensions

Not all modifications of the Riemann integral were made for the purposeof solving the difficulties associated with Fourier’s work or extending thefundamental theorem of calculus. In 1894, Thomas Joannes Stieltjes (1856–1894) published the definition of a new type of integral that did not treatall intervals of the same length equally. Given functions f and g, Stieltjes

1.6. Extensions 15

considered sums of the form

nXiD1

f .ti / Œg .xi / � g .xi�1/�

where a D x0 < x1 < x2 < � � � < xn D b and ti 2 Œxi�1; xi �. When thesums converge to a finite value in an appropriate sense, the value, denoted byR-SR ba f dg, is called the Riemann-Stieltjes integral of f over Œa; b� with re-

spect to g. The Riemann-Stieltjes integral is a generalization of the Riemannintegral since Stieltjes’ integral is Riemann’s integral when g .x/ D x.

While Stieltjes’ work was motivated by the problem of calculating themoment of inertia of an object with varying density, the integral has muchbroader application. In the moment of inertia problem, g .xi / � g .xi�1/reflects the mass between xi�1 and xi . Alternatively, if you wish to modeltravel over terrain of unequal difficulty, the value of g .xi /�g .xi�1/ can beinterpreted as the effort required to travel from xi�1 to xi .

From a mathematical point of view, probably the greatest value of theRiemann-Stieltjes integral is that it provides a single treatment of all typesof probability on the real line. There are two principal types of probabil-ity spaces: discrete and continuous. Discrete spaces model situations wherethe set of possible outcomes forms a finite or countable set. Counting thenumber of phone calls arriving in an hour or counting the number of flipsof a coin required to see the first heads are typical examples. In such cases,computations related to probability and expected value involve sums. Onthe other hand, continuous distributions used to model phenomena such asthe height of an object or the time until the next arrival of an airplane use theRiemann integral for calculation. As a result, the typical probability or statis-tics text written for upper-division mathematics students states many if notmost theorems twice: once for the discrete case and once for the continuouscase. And while texts typically present only one type of proof and indicatethat the other proof is similar, most theorems require two different proofs forcomplete justification.8

The Riemann-Stieltjes integral captures both types of distributions as wellas distributions that are mixtures of the two types. Consequently only onestatement and one proof is needed when probability and statistics is doneusing the Riemann-Stieltjes integral. Moreover, the proof is generally just aseasy or hard as the proof of one of the individual cases in the more common

8 Actually, complete proofs would be even more complicated since a distribution could mixtypes or otherwise not fit neatly into one of the two principle types of distributions.

approach. Why then is the Riemann-Stieltjes integral not used in probabil-ity and statistics courses? Among other reasons is the fact that text writerscannot assume that students have prior knowledge of the Riemann-Stieltjesintegral.

The notion that different intervals should have varying weights has a nat-ural affinity with Lebesgue’s idea of working with the “lengths” of vari-ous sets. Instead of thinking of the length of a set, we should consider its“measure.” The idea of measure encompasses many applications such as theweight of an object over the set A, the total difficulty in traveling over theset A, or the probability of A. To integrate a function relative to the measure�, we use Lebesgue’s approach and partition the range of f . The partitionsgenerate sums of the form

nXiD1

yi� .Ai /

where Œyi�1; yi � is a subinterval of the range of f , Ai is the set of all x 2Œa; b� for which yi�1 < f .x/ � yi , and � .Ai / is the “measure” of Ai . Thefunction is Lebesgue-Stieltjes integrable with respect to � if the sums con-verge appropriately. In this case, the integral is written as M

R ba f d�.9 The

Riemann-Stieltjes and Lebesgue-Stieltjes integrals are connected by building� from the foundational assumption that � ..xi�1; xi �/ D g .xi /� g .xi�1/.

When we step back from particular applications and consider what couldhappen with arbitrary choices of g, things can appear rather odd. Supposethat g is decreasing on the interval Œxi�1; xi �. Then g .xi / � g .xi�1/ willbe negative. This makes no sense if the associated measure represents prob-ability. But other types of interpretations might make sense. If g reflectsthe resistance along a path, then a negative value of g .xi / � g .xi�1/ canbe understood as a section of the path that puts energy back. Perhaps thesection is downhill. A negative value in a moment of inertia applicationmight reflect a certain type of anti-matter or a repelling rather than attractingforce.

Once such possibilities are contemplated, a whole different type of math-ematical universe opens up. Measures can be put on sets much more generalthan R or Rn. Moreover, measures need not be restricted to taking on onlypositive, real, or even complex values. In particular, projection-valued mea-sures defined on subsets of the complex numbers are used in operator theory,a mathematical subject that plays a central role in quantum physics.

9 M is for measure.

1.7. Exercises 17

B

A

M

Cl

M1

M2

C1

C2

l1l2

B

A

M

Cl

M1

M2

C1

C2

l2l3l4

E

FD

Figure 1.7. Quadrature of the parabola

Mathematicians and physicists have extended the concept of the integral ina myriad of other ways that we will not explore in this text. In the remainingchapters, you will be introduced to the ideas and proofs that undergird theideas sketched out in this historical overview, and you should be prepared tostudy an even broader range of interpretations of integration.

1.7 Exercises

1.1 Greek foreshadowing: filling the gaps1. In Figure 1.1, C is the midpoint of AB and thus the center of the large

circle. Prove that the triangle ACD and the crescent AEDF have thesame areas by providing explanations of the following. (Your write-upwill be stronger if you blend the individual explanations into a singleproof rather than answering each part separately.)

(a) How are the areas of the circles with diametersAB and AD related?Why?

(b) How are the areas of the semicircle GDFA and the quarter circleCDEA related?

(c) Why do the crescent AEDF and the triangle ACD have the samearea?

2. Give a geometric proof of the claim in Archimedes’ proof that the com-bined area of triangles ACC1 and BCC2 is 1

4that of ABC . (Use the

following outline to construct a proof.)

(a) Add lines l3 and l4 parallel to AB through C2 and C respectivelyand use D, E, and F to label the intersection points as indicated inthe parabola on the right of Figure 1.7.

(b) Explain why triangles EC2B and EC2C have the same area.


(c) How is the area of M2BE related to the area of MBC ?

(d) Archimedes knew from Euclid’s Elements that CDCMD

MM22

MB2. Use

this fact to prove that M2E is twice as long as C2E.(e) Explain why triangles M2BE and CBC2 have the same area.(f) Argue that the combined area of triangles ACC1 and BCC2 is 1

4

that of ABC .

1.1 Greek foreshadowing: deeper reflections3. (Quadrature of two squares) Given two squares, explain how to construct

a single square with the same area as the combined areas of the twooriginal squares.

4. Given a square, how can one construct a second square with half itsarea?

5. (Quadrature of a rectangle) Given a rectangle, explain how to constructa square of the same area. (Work with a circle whose diameter is the sumof the lengths of the two sides of the rectangle.)

6. (Quadrature of a triangle) Given a triangle, how can one construct asquare of the same area? (Use the previous exercises.)

Exercises 7 and 8 outline Archimedes’ proof that the area of a circle isequal to the area of a triangle with height equal to the radius of the circleand with base length equal to its circumference. Let C be a circle and letT be a triangle with height equal to the radius of C and with base lengthequal to the circumference of C . Let AC and AT be the correspondingareas.

7. Prove that AC 6> AT . (Combine the steps into a flowing exposition.)

(a) Suppose that D D AC � AT > 0.

P2 P3

polygons Pn of 2n sides

1.7. Exercises 19

(b) Construct polygons Pn of 2n sides contained in C is the followingmanner.i. P2 is an inscribed square.

ii. PnC1 is constructed from Pn by adding an isosceles triangle oneach edge whose third vertex is on C .

(c) Explain why more than half of the area between C and Pn is re-moved to make PnC1.

(d) Give a simple description of the area of Pn in terms of its circum-ference.

(e) Explain how this produces a contradiction.

8. Prove that AC 6< AT using circumscribed polygons in an argument sim-ilar to that of the previous exercise.

9. Give a calculus-based proof of the claim that the combined area of tri-angles ACC1 and BCC2 of exercise 2 is 1

4that of ABC . (Use integrals

to compute the areas between lines.)

10. Use calculus to prove directly that the area of the region bounded byy D sx2 and the line through A D

�a; sa2

�and B D

�b; sb2

�is 4

3

times the area of the triangle with vertices at A; B , and C D�c; sc2

�where c D 1

2.aC b/.

1.2 Newton and Leibniz: deeper reflections

11. According to Newton’s formula,p1 � x2 D

P1kD0

�1=2k

�.�1/k x2k .

(a) ExpandP5kD0

�1=2k

�.�1/k x2k .

(b) Verify that�1 � 1

2x2 � 1

8x4 � 1

16x6�2D 1 � x2 CO

�x8�:

(c) Verify that�1 � 1

2x2 � 1

8x4 � 1

16x6 � 5

128x8�2D 1�x2CO

�x10

�.

Here O .xn/ is used to indicate that the power of x in the remainingterms is n or greater.

12. Use the ratio test for series to verify that

1XkD2

1 � 3 � � � .2k � 3/

2kkŠ .2k C 1/

converges.


1.3 Cauchy, Riemann, and Darboux: deeper reflections13. Suppose that the function

f .x/ D

�0; x 2 Œ0; 1/

3; x 2 Œ1; 5�

represents the current flowing through a line if a switch is closed at timet D 1. If the battery to which the line is connected maintains a voltageof 2 volts, then the power consumed over the time period Œ0; t �, where0 � t � 5, is given by

R t0 2f .x/ dx:

(a) Explain why this integral does not exist in the Newton-Leibnizsense.

(b) Explain why the Riemann integral RR t1f .x/ dx does exist for 1 �

t � 5.

14. Let

F .x/ D

�x2 sin �

x; x ¤ 0

0; x D 0:

(a) Compute the derivative F 0 of F .(b) Explain why Cauchy’s version of the fundamental theorem of calcu-

lus cannot be used to conclude that CR 10 F

0 D 0.(c) Explain why Riemann’s version of the fundamental theorem of cal-

culus can be used to conclude that RR 10 F

0 D 0.

1.4 Lebesgue: deeper reflections15. Many of the more interesting questions related to Fourier series involve

discontinuous functions. Explain why the Fourier series of a discontinu-ous function cannot converge uniformly.

16. One of the hurdles in defining the Lebesgue integral is the problem ofmaking sense of the “length” of sets that might arise as inverse im-ages of an interval. For the purpose of thinking about how to con-struct a general definition, what is your intuitive sense for an appropriatevalue for the “length” of the following sets? How do you arrive at yourvalue?

(a) Œ0; 1�(b) Œ0:2; 0:3� [ .0:4; 0:5/ [ Œ0:7; 0:9/(c) Œ0:2; 0:4� [ .0:3; 0:5/ [ Œ0:7; 0:9/(d)

˚12; 13; 14

�(e)

˚1nW n 2 N

�

1.7. Exercises 21

(f) the rational numbers in Œ0; 1�(g) the irrational numbers in Œ0; 1�

What properties do you expect “length” to have?

1.5 Henstock and Kurzweil: deeper reflections17. Define

f .x/ D

(1px; x > 0

0; x � 0:

(a) Given 0 D x0 < x1 < x2 < � � � < xn D 1, explain how theevaluation points can be chosen so that the corresponding Riemannsum is larger than any predetermined value v. Explain why you canconclude that f is not Riemann integrable over Œ0; 1�.

(b) Set

ı .t/ D

�t=2; t > 0

1; t � 0

and suppose that we have values satisfying 0 D x0 < x1 < x2< � � � < xn D 1; ti 2 Œxi�1; xi �, and xi � xi�1 < ı .ti / for 1 � i �n:

Prove thatPniD1 f .ti / .xi � xi�1/ < 3. (Combine the following

hints into a well presented argument.)i. Explain why t1 D 0. (What is the minimal value of x0 if 0 < t1?)

ii. Prove that g .x/ D 3�px �pa�� 1p

a.x � a/ is increasing for

x 2 Œa; 2a�.iii. Use (ii) to prove that

PniD1 f .ti / .xi � xi�1/ <

3PniD2

�pxi �pxi�1

�.

1.6 Extensions: deeper reflections18. Suppose that the random variableX has a 2

3probability of being 0 and 1

3

probability of being 1. Define a function g W R! Œ0; 1� so that g .xi / �g .xi�1/ is the probability that the value ofX is in the interval .xi�1; xi �.

19. Suppose that the random variable X has equal likelihood of taking onany value in the interval Œ0; 1� and never takes on a value outside ofŒ0; 1�. Define a function g W R! Œ0; 1� so that g .xi / � g .xi�1/ is theprobability that the value of X is in the interval .xi�1; xi �.

20. Let

g .x/ D

8<:0; x < 0

3x; 0 � x � 2

8; 2 < x:

Compute

(a) R-SR 10 1 dg:

(b) R-SR 5�2 1 dg:

(c) R-SR 20 x dg:

(For (c), write out the Riemann-Stieltjes sums and then reinterpret themas Riemann sums.)

21. Suppose that

� .A/ D

8<:0; 1; 2 62 A

1; 1 2 A; 2 62 A

2; 1 62 A; 2 2 A

3; 1; 2 2 A:

Compute

(a) MR 20 1 d�:

(b) MR 10 x d�:

(c) MR 20x d�:

(Partition the interval and consider which terms contribute to the sum.)

22. Let

P2 D1

2

�1 1

1 1

and

P4 D1

2

�1 �1

�1 1

:

Then P2 and P4 are projections onto the spaces spanned by

��1

1

and��

1

�1

respectively. Suppose that

� .A/ D

8<:

0; 2; 4 62 A

P2; 2 2 A; 4 62 A

P4; 2 62 A; 4 2 A

P2 C P4; 2; 4 2 A:

(a) Use the ideas from the previous exercise and the definitions of matrixoperations to determine the matrix values of

i. MR 30 1 d�:

1.8. References 23

ii. MR 50 1 d�:

iii. A D MR 50x d�:

iv. B D MR 50 x

2 d�:(b) Compute A2.(c) Write an integral expression that you think is likely to represent

pA.

(d) Evaluate your integral from part (c) and compute its square.

1.8 References

Baker, Alan (1975). Transcendental Number Theory, Cambridge UniversityPress.

Berkeley, G. (1734). The Analyst. Edited by David Wilkins (2002)http://www.maths.tcd.ie/pub/HistMath/People/Berkeley/Analyst/

Bradley, R.E. and C.E. Sandifer, C.E. (2009). Cauchy’s Cours d’analyse, AnAnnotated Translation. Sources and Studies in the History of Mathe-matics and Physical Sciences. Springer.

Bressoud, D.M. (2006). A Radical Approach to Real Analysis (2nd ed.).Mathematical Association of America.

Burk, F.E. (2007). A Garden of Integrals. Mathematical Association ofAmerica.

Edwards Jr., C.H. (1994). The Historical Development of the Calculus (3rded.). Springer.

Fourier, J. (1822). The Analytical Theory of Heat. Translated byAlexander Freeman (1878, re-released 2003). Dover Publications.books.google.com/books?id=No8IAAAAMAAJ.

Hawkins, T. (1975). Lebesgue’s Theory of Integration (2nd ed.). Chelsea.Heath, T.L. (2011). The Works of Archimedes (2nd ed.). CreateSpace.Henstock, R. (1968). A Riemann type integral of Lebesgue power, Canadian

Journal of Math., 20: 79–87.Lebesgue, H. (1903). Sur une condition de convergence des series de Fourier,

C.R. Acad. Sci, Paris 140: 229–242.Lebesgue, H. (1903). Recherches sur la convergence des series de Fourier,

Math. Ann. 61: 251–280.Lindemann, F. (1882). Uber die Zahl � , Mathematische Annalen 20:

pp. 213–225.Nauenberg, M. (Forthcoming) Barrow and Leibniz on the Fundamen-

tal Theorem of the Calculus, Annals of Science arXiv:1111.6145[math.HO].


Newton, I. (1687). The Mathematical Principles of Natural Philosophy,Translated by B. Motte (1724). Middle-Temple-Gate, Fleetstreet.books.google.com/books?id=Tm0FAAAAQAAJ.

Riemann, B. (1868). Uber die Darstellbarkeit einer Function durch einetrigonometrische Reihe, H. Weber (ed.), B. Riemann’s GesammelteMathematische Werke, Dover, reprint (1953) pp. 227–271 (Original:Gottinger Akad. Abh. 13).

Stein, S.K. (1999). Archimedes: What Did He Do Besides Cry Eureka?Mathematical Association of America.

Stillwell, J. (2004). Mathematics and its History (2nd ed.). Springer.Struik, D.J. (1969). A Source Book in Mathematics, 1200–1800. Harvard

University Press.Swain, G. and D. Thomas (1998). Archimedes’ quadrature of the parabola

revisited, Mathematics Magazine 71 (2): 123–30. JSTOR 2691014.Truesdell, C.A. (1980). The Tragicomical History of Thermodynamics,

1822–1854. Springer.

CHAPTER 2

The Riemann Integral

Modern students of mathematics are typically introduced to integrationthrough the Riemann integral, so you are no doubt familiar with its defi-nition. But, since the definition requires a significant amount of supportingnotation that may vary from text to text, we will provide a review. Moreover,for ease of transition and to make later comparisons more straightforward,we will use a unified terminology and notation throughout this text. WhileRiemann did not use this notation and it may be unfamiliar initially, the ideasand approaches behind the terminology remain Riemann’s.

Figure 2.1. A Riemann Term

Given an interval Œa; b�, a partition of Œa; b� is a finite set of contiguousintervals Œxk�1; xk� with a D x0 < x1 < x3 < � � � < xn D b. The endpointsof the intervals interior to Œa; b�, fxig

n�1iD1 , are called the division points of

the partition. A tagged partition is a partition together with a set of tags

ftkgnkD1 where tk 2 Œxk�1; xk�. For example,

nhk�1n; kn

ionkD1

is a partition

25

26 CHAPTER 2. The Riemann Integral

of Œ0; 1� for any value of n 2 N and

P1 D˚�0;�0; 1n

��;�1n;�1n; 2n

��;�2n;�2n; 3n

��; : : : ;

�n�1n;�n�1n; 1��

P2 D˚�1n;�0; 1n

��;�2n;�1n; 2n

��;�3n;�2n; 3n

��; : : : ;

�1;�n�1n; 1��

; and

P3 D˚�

12n;�0; 1n

��;�32n;�1n; 2n

��;�52n;�2n; 3n

��; : : : ;

�1 � 1

2n;�n�1n; 1��

are tagged partitions where the tags are respectively the left endpoints, rightendpoints, and midpoints of the subintervals. These three tagged partitionsuse equal-length subintervals, but that is not necessary.

P4 D˚�0;�0; 12

��;�12;�12; 23

��;�23;�23; 34

��;�34;�34; 1��

is also a tagged partition of Œ0; 1�. On the other hand, the set˚�0;�0; 12

��;�12;�12; 23

��;�23;�23; 34

��; : : :

�is not a tagged partition of Œ0; 1� because the set is not finite, nor does itterminate with a right endpoint of 1.

Given a partition P DfŒxk�1; xk�gnkD1 of Œa; b�, we denote the width ofthe kth subinterval by �xk and the mesh of P by

kPk D max1�k�n

�xk D max1�k�n

.xk � xk�1/ :

The tagged partitions P1, P2, and P3 all have a mesh of 1n

. The mesh of P4is kP4k D 1

2.

Given a real-valued function f defined on Œa; b�, we compute the Rie-mann sum for a tagged partition P D

˚�tk; Œxk�1; xk �

��nkD1D f.tk ; Ik/g of

Œa; b� as

SR.f;P/ DnXkD1

f .tk/ .xk � xk�1/ D

nXkD1

f .tk/�xk DXPf �x:

Notice the varying levels of detail in the notation introduced above. Goingforward, we will use the most compact notation that is unambiguous in thecontext. For example, we will use f.tk ; Ik/g rather than


��nkD1

when we do not need to reference the endpoints of the intervals andPP f �x rather than

PnkD1 f .tk/�xk when the particular tags are not

important. This will help us reason at higher levels of abstraction.For future reference, note that

PP �x D

PnkD1 .xk � xk�1/which tele-

scopes to xn � x0 D b � a: This fact will be used frequently and withoutreference.

2.1. Riemann integrability 27

Definition 2 (Riemann Integral). Let f be a real-valued function definedon Œa; b�. Then f is Riemann integrable over Œa; b� if there is a real numberA such that for any " > 0 we can find a ı > 0 so that

jSR.f;P/ � Aj < " (2.1)

whenever P is a tagged partition of Œa; b� with kPk < ı. In this case, A isthe integral of f over Œa; b� and we write R

R ba f D

RR ba f .x/ dx D A.

Note that (2.1) must hold for all partitions with mesh kPk < ı (evenlyspaced or not) and for all choices of the tags (left, right, center, or otherwise).

Though stated here for the Riemann integral, we will address the followingthree questions for all the integrals we will study:

1. What functions are Riemann integrable?

2. When is it true that ddx

RR xa f D f and R

R ba f

0 D f .b/ � f .a/?

3. Under what circumstances is RR ba

Pn fn D

Pn

RR ba fn or,

equivalently, under what circumstances is RR ba limn!1 fn D

limn!1RR ba fn?

We address each question in turn, obtaining only partial answers initially.

2.1 Riemann integrabilityThe first thing to note about the Riemann integral is that any Riemann inte-grable function must be bounded. You are asked to provide a proof of thisfact in exercise 2.

Before addressing the general questions about Riemann integrable func-tions, we present several examples that will be considered for each new typeof integral. Pay attention to both the results and the means by which theresults are obtained.

Example 1 (Constant functions). Let f .x/ D c on Œa; b�. Let P D

f.tk ; Ik/g be any tagged partition of Œa; b�. Since f .tk/ D c, we find that

SR.f;P/ DXPc�x D c

XP�x D c .b � a/ :

Hence we see that

jSR.f;P/ � c .b � a/j D 0 < "

for any " > 0 and any tagged partition P of Œa; b�. Thus the constant functionf .x/ D c is integrable over Œa; b� with

RZ b

a

f D c .b � a/:

Example 2 (Dirichlet). The Dirichlet function

d .x/ D

�1; x 2 Q

0; x 62 Q;

where Q is the set of rational numbers, is not Riemann integrable over anynon-degenerate interval Œa; b�. To see why, suppose that P is a partition ofŒa; b�. Create two tagged partitions from P . Choose rational numbers for thetags of P1 and irrational tags for P2. Then, irrespective of the mesh of P , wesee that

SR.d;P1/ DXP1

1 �x D b � a

andSR.d;P2/ D

XP2

0 �x D 0:

If d .x/ were integrable with a value of A, we would be able to find aı > 0 so that any tagged partition Q with a mesh less than ı would sat-isfy jSR.d;Q/ � Aj < b�a

4. In that case we could begin with a partition P

with mesh kPk < ı and use the triangle inequality to show that

b � a D jSR.d;P1/ � SR.d;P2/j

< jSR.d;P1/ � Aj C jA � SR.d;P2/j <b � a

2.

This clearly false conclusion implies that d .x/ is not Riemann integrableover Œa; b�.

Example 3 (Identity function). Let f be the identity function f .x/ D x

on Œ0; 2� and take Pn Dn

2kn;h2k�2n; 2kn

i�onkD1

. Then kPnk D 2n

and

SR.f;Pn/ DnXkD1

2k

n

2

nD

4

n2

nXkD1

k D4

n2n .nC 1/

2D 2C

2

n:

As n gets larger and the mesh decreases, SR.f;Pn/ decreases to 2. We con-clude that if f is Riemann integrable, then R

R 20f D 2.


How can we show that f is Riemann integrable? This task is rather moredifficult than identifying 2 as the potential value of the integral. We willshow f to be Riemann integrable by connecting the Riemann sums as-sociated with arbitrary tagged partitions to the Riemann sums associatedwith Pn of example 3. To that end, begin with an arbitrary tagged parti-tion P Df.tk ; Œxk�1; xk�/g and create a new (untagged) partition Qn thathas

˚x0; x1; x2; : : : ; xn;

2n; 4n; 6n; : : : ; 2n

n

�as its division points. Place these

division points in order, relabel them as fy0; y1; y2; : : : ; ymg.This type of construction occurs frequently enough to merit its own no-

tation. Recognizing that the set of division points of Qn is the union of thedivision points of Pn and P , we use the notation Qn D Pn [ P . To avoidpossible confusion, even when the two partitions are tagged, the resultingpartition is not. The union of two partitions (tagged or untagged) is alwaysuntagged. One natural way to attach tags to an untagged partition is to useone of the endpoints of each subinterval. We will use subscripts of L and Rto designate the tagged partition created by using the left and right endpoints.

We claim that

1. SR.f;PL/ � SR.f;P/ � SR.f;PR/;

2. SR.f;PL/ � SR.f;Qn;L/ � SR.f;Qn;R/ � SR.f;PR/;

3. SR.f;Pn;L/ � SR.f;Qn;L/ � SR.f;Qn;R/ � SR.f;Pn;R/,

4. 0 � SR.f;PR/ � SR.f;PL/ � 2 kPk ; and

5. 0 � SR.f;Pn;R/ � SR.f;Pn;L/ � 2 kPnk D 4n

.

Assuming these results for the moment, note that Pn D Pn;R so that (3)and (5) imply that

jSR.f;Qn;R/ � SR.f;Pn/j � 2 kPnk D4

n:

With a bit more effort, we observe that (1), (2), and (4) imply

jSR.f;P/ � SR.f;Qn;R/j � 2 kPk :

This is most easily seen by considering the values of SR.f;PL/ andSR.f;PR/ to be endpoints of an interval. Points known from (1) and (2)to be in that interval cannot be further from each other than the length of theinterval.

Now suppose that we are given an " > 0. Choose n so that 6=n < "=2 andsuppose that our tagged partition P has a mesh kPk < ı D "=4: Then

jSR.f;P/ � 2j� jSR.f;P/ � SR.f;Qn;R/jC jSR.f;Qn;R/ � SR.f;Pn/j C jSR.f;Pn/ � 2j

< "=2C 4=nC 2=n D "=2C 6=n < "=2C "=2 D ":

This means that the identity function is integrable over Œ0; 2� withRR 20 f D 2.

Now what about the five claims? There are some general principles operat-ing behind these statements that apply more broadly. So instead of verifyingthe claims for the identity function, we will prove the more general results.

Lemma 1. Suppose that f is an increasing1 function on the interval Œa; b�.Then SR.f;PL/ � SR.f;P/ � SR.f;PR/ for any tagged partition P ofŒa; b�.

Proof. The proof of this lemma is left as an exercise (exercise 3).

Lemma 1 establishes claim (1) and the inner inequalities of claims (2) and(3). Before turning to the next lemma, note that the critical feature of Qn isthat its set of division points includes the division points of both P and Pn.The following definition places this idea in a more general setting.

Definition 3 (Refinement). Let P and Q be partitions of an interval Œa; b� :We say that Q is a refinement of P if the division points of P are includedin the division points of Q. Alternatively, Q is a refinement of P if eachsubinterval of Q is contained in a subinterval of P .

Lemma 2. Suppose that f is an increasing function on the interval Œa; b�,that P is a partition of Œa; b�, and that Q is a refinement of P . ThenSR.f;PL/ � SR.f;QL/ � SR.f;QR/ � SR.f;PR/:

Proof. The inner inequality is a consequence of lemma 1.To prove the first inequality, assume that Q is created from P by the ad-

dition of a single division point, y; that falls in the interval�xj�1; xj

�of

1 We say that a function with domain X is increasing if whenever x; y 2 X with x � y, wehave f .x/ � f .y/. We will use strictly increasing when x < y implies f .x/ < f .y/.Decreasing is defined analogously.


P . Since all the other subintervals of PL and QL are equal, their corre-sponding terms will cancel when the Riemann sums are subtracted. Thenon-cancelling terms are generated from the intervals

�xj�1; y

�and

�y; xj

�from Q and

�xj�1; xj

�from P . Since f is increasing,

SR.f;QL/ � SR.f;PL/D�f�xj�1

� �y � xj�1

�C f .y/

�xj � y

�� f

�xj�1

� �xj � xj�1

��f�xj�1

� �y � xj�1

�C f

�xj�1

� �xj � y

�� f

�xj�1

� �xj � xj�1

�D 0.

The general result now follows by induction. The right inequality is left asan exercise.

Claims (4) and (5) are consequences of the next lemma.

Lemma 3. Suppose that f is an increasing function on the interval Œa; b� :Then 0 � SR.f;PR/ � SR.f;PL/ � .f .b/ � f .a// kPk for any taggedpartition P of Œa; b�.

Proof. The left inequality follows from lemma 1: For the right-hand side,note that 0 � f .xk/ � f .xk�1/ so that

SR.f;PR/ � SR.f;PL/ DnXkD1

f .xk/�xk �

nXkD1

f .xk�1/�xk

D

nXkD1

.f .xk/ � f .xk�1//�xk

�

nXkD1

.f .xk/ � f .xk�1// kPk

D .f .b/ � f .a// kPk :

The five claims on page 29 are thus established.What we have just done properly strikes most students as a tremendous

amount of work to verify that f .x/ D x is integrable over Œ0; 2�. Newtonand Leibniz did not worry about such concerns. For them, the integral obvi-ously existed and the only question was how to evaluate it. As a cautionarynote, consider the Dirichlet function. Had we simply evaluated the Riemannsums for the partitions Pn D

˚�k�1n; kn

��using left-, right-, or midpoints, we

would have easily, but incorrectly, deduced that the value of the integral ofthe Dirichlet function over Œ0; 1� is 1:


The need for the careful analysis we have just completed arose from con-siderations of trigonometric series. It is not at all clear that such functionswill be better behaved than the Dirichlet function. In fact, in Chapter 4 wewill show how trigonometric series can produce rather ill-behaved functions.One might be tempted to ignore the issues raised by the Dirichlet functionas issuing from an annoying, but irrelevant, pathological example. However,trigonometric series functions arise as solutions to differential equations andcannot be so ignored. We do, indeed, need to expend the effort to developcriteria for integrability for poorly-behaved functions rather than assumingthat any naturally arising function is integrable.

Happily, the effort we devoted to verifying that RR 20 x D 2 is rewarded by

providing us with insights into ways of modifying the definition of the inte-gral that allow us to apply similar techniques to a wider range of functions.We will do this in the next chapter, but we do not need to wait until then tosee some return on our effort. We can get additional mileage from our workright now by considering general increasing functions. If you reflect on thestructure of example 3, you will see that we have most of the tools needed toprove that increasing functions are integrable. What we lack is a method toidentify a value for the integral.

Theorem 4. If f is an increasing function defined on Œa; b� then f is inte-grable over Œa; b�.

Proof. To identify the value of the integral, we will use the fact that anybounded increasing sequence of real numbers converges. Consider the parti-tion Pn generated by the division points fa C k

2n.b � a/g2

n

kD1. Since PnC1

is a refinement of Pn, we can use lemma 2 (page 30) to conclude that

SR.f;P1;L/ � SR.f;P2;L/ � SR.f;P3;L/ � � � � � SR.f;P1;R/:

Let A be the limit of the bounded, increasing sequence fSR.f;Pn;L/g. Theproof can now be completed in a manner similar to example 3 (page 28).(See exercise 6b.)

There is another general class of functions whose members are integrable:continuous functions.

Theorem 5. Suppose that f is continuous on the interval Œa; b�. Then f isintegrable over Œa; b�.

Proof. Since Œa; b� is a compact set, f is bounded and uniformly continu-ous2 on Œa; b�. In particular, there is a real number B so that jf .x/j < B forall x 2 Œa; b�. Take �x D .b � a/ =n and let Pn be the partition defined bythe division points faC k�xgn�1kD1. Since

�B .b � a/ � SR.f;Pn;L/ � B .b � a/

by exercise 7, the sequence fSR.f;Pn;L/g1nD1 must have a cluster point A.

We claim that RR baf D A.

Let " > 0 and use the uniform continuity of f to choose ı > 0

so that jf .t/ � f .s/j < " whenever s; t 2 Œa; b� with js � t j < ı.Choose n so that jSR.f;Pn;L/ � Aj < " and kPnk D b�a

n< ı. Now

let P Dftk ; Œxk�1; xk�gnkD1 be a tagged partition of Œa; b� with kPk < ı.Finally, define Q D Pn [ P and write Q D

˚�yj�1; yj

��mjD1

.Focus for the moment on a single subinterval Œxi�1; xi � of P . Since Q is a

refinement of P , there are values j and k so that xi�1 D yj�1 < � � � < yk Dxi . In other words, Œxi�1; xi � is made up of one or more subintervals of Q.Since ti ; yj ; yjC1; : : : ; yk are all in the interval Œxi�1; xi � and jxi � xi�1j <ı, we can conclude that

ˇf .ti / � f

�yp�ˇ< " for j � p � k.

Now consider the difference of the terms in the Riemann sums corre-sponding to these intervals.ˇf .ti / .xi � xi�1/ �

�f�yj� �yj � yj�1

�C � � � C f .yk/ .yk � yk�1/

�ˇDˇ�f .ti /

�yj � yj�1

�C � � � C f .ti / .yk � yk�1/

��f�yj� �yj � yj�1

�C � � � C f .yk/ .yk � yk�1/

�ˇ�ˇf .ti / � f

�yj�ˇ �yj � yj�1

�C � � � C

ˇf .ti / � f .yk/

ˇ.yk � yk�1/

< "�yj � yj�1

�C � � � C " .yk � yk�1/

D "�yk � yj�1

�D " .xi � xi�1/ .

Since the same analysis applies to all the subintervals of P and everysubinterval of Q is accounted for this way, we see that

jSR.f;P/ � SR.f;QR/j< " .x1 � x0/C " .x2 � x1/C � � � C " .xn � xn�1/ D " .b � a/ .

2 Statements of definitions and important results from real analysis can be found in the appen-dices.

Because Q is also a refinement of Pn the same analysis proves that

jSR.f;Pn;L/ � SR.f;QR/j < " .b � a/

as well. Pulling the various pieces together, we see that

jSR.f;P/ � Aj� jSR.f;P/ � SR.f;QR/jC jSR.f;QR/ � SR.f;Pn;L/j C jSR.f;Pn;L/ � Aj

< 2" .b � a/C ".

This is sufficient to conclude that f is integrable with RR ba f D A. (See

exercise 8.)

A function need not be continuous everywhere to guarantee Riemann inte-grability. A function f is also Riemann integrable when it has only isolateddiscontinuities.

Theorem 6. Suppose that f is bounded and has a finite number of discon-tinuities in Œa; b�. Then f is Riemann integrable over Œa; b�.

Proof. Suppose that f is continuous on Œa; b� except at x D a. Since f isbounded, there is a B so that jf .x/j 1b�a

:

Moreover, when m > n,

jsm � snj D

ˇˇ RZ aC 1n

aC 1m

f

ˇˇ � R

Z aC 1n

aC 1m

jf j � RZ aC 1n

aC 1m

B D B

�1

n�1

m

� 0 and choose n0 so thatˇsn0 � A

ˇ< "=6, 1

n0< "=3B , and

n0 >1b�a

. Set a0 D aC 1n0

. Since f is integrable over Œa0; b�, we can find

a ı so thatˇSR.f;P/ � sn0

ˇ< "=6 for any tagged partition P of Œa0; b� with

kPk < ı.Set ı0 D min

˚ı; "6B

�and suppose that P is a tagged partition of Œa; b�

with mesh kPk < ı0. If a0 is not a division point of P , create a new taggedpartition P� by inserting a0 as a division point and using the left endpointsas the tags of the newly created subintervals. By construction, P and P�will agree except for the subintervals from P and P� containing a0. Since

the values of f at any pair of tags for a given interval can differ by at most2B , we can use the techniques of lemma 3 (page 31) to conclude that

jSR.f;P/ � SR.f;P�/j � 2B � ı < "=3.

(See exercise 9b.) If a0 is a division point of P , then take P� D P so thedifference of the Riemann sums is zero.

Split P� into two tagged partitions P�1 and P�2 of Œa; aC 1n0� and ŒaC 1

n0; b�

respectively. Then

jSR.f;P�1 /j D

ˇˇXP�1

f �x

ˇˇ �X

P�1

B �x D B �1

n0< "=3

and, since the mesh of P�2 is less than ı;

jSR.f;P�2 / � Aj �ˇSR.f;P�2 / � sn0

ˇCˇsn0 � A

ˇ< "=3.

Thus

jSR.f;P/ � Aj� jSR.f;P/ � SR.f;P�/j C jSR.f;P�/ � Aj� jSR.f;P/ � SR.f;P�/j C jSR.f;P�1 /j C jSR.f;P�2 / � Aj< ".

We conclude that f is integrable over Œa; b�.The case where the right endpoint is a point of discontinuity is treated

similarly. (See exercise 9d.) The proof extends to a finite number of dis-continuities located anywhere in Œa; b� by using the results of the next sec-tion to break the original interval into a finite number of subintervals. (exer-cise 9e.)

One might be tempted to conclude, on the basis of the Dirichlet func-tion, that any function with an infinite number of discontinuities fails to beRiemann integrable. This conclusion would be false as illustrated by the fol-lowing two examples.

Example 4. Define

f .x/ D

�1; x D 1

n, n 2 N

0; otherwise:

Suppose that P is a tagged partition of Œ0; 1� with kPk < 1N2

. The subin-tervals of P contained in

�0; 1N

�can contribute at most 1

Nto SR.f;P/. The

intervals intersecting�1N; 1�

contribute at most N nonzero terms. Since thewidth of these subintervals is less than 1

N2, 0 � SR.f;P/ < 1

NC N 1

N2D

2N

. Thus f is Riemann integrable over Œ0; 1� with RR 10f D 0.

Not only can a Riemann-integrable function have an infinite set of discon-tinuities, the set of discontinuities can be dense.

Example 5. Define

sc .x/ D

�0; x < c

1; c � x:

(See Figure 2.2.) Let frig1iD1 be an enumeration of the rational numbers in

the interval Œ0; 1� and define f by

f .x/ D

1XnD1

1

2nsrn .x/ :

The function f is discontinuous on the dense set of rational points of Œ0; 1�(exercise 10) but is Riemann integrable since f is increasing.

1

c

Figure 2.2. sc .x/

These examples make evident the fact that we do not yet have a clearlydemarcated boundary between functions that are Riemann integrable andthose that are not. We will set that question aside temporarily and take it upagain in the next chapter.

Reflections. Before proceeding, it is worthwhile spending some time re-flecting on the proofs. There are several common features that will reappearin our subsequent work. Moreover, some of the techniques will illuminatethe advantages of alternative definitions of the integral.

Probably the most obvious observation is that the proofs thus far have gen-erally consisted of two parts: (1) identifying a potential value for the integraland (2) verifying that the Riemann sums corresponding to partitions of smallmesh are close to this value. The first phase is generally accomplished by

2.2. Subintervals 37

examining a specific sequence of Riemann sums. One verifies that the corre-sponding sequence of values is bounded or Cauchy and uses the propertiesof sequences to identify a potential value for the integral.

The second phase of the proof typically consists of three parts. The firstcritical component is to find a way to relate the values of Riemann sums inthe special sequence to the values of Riemann sums for arbitrary partitionswith small mesh. This is accomplished using refinements. At this point, thetechniques seem rather ad hoc; the approaches used for increasing functions,continuous functions, and functions with a single discontinuity are ratherdifferent. The need to connect the values of the Riemann sums of a partitionand of one of its refinements is nonetheless present.

The second common feature can be described as bound-and-telescope.This technique is used to show that two Riemann sums are close in value byconsidering the difference of the two sums. In the case of increasing func-tions (lemma 3, page 31), the length of the subintervals was bounded and thevalue of the function was telescoped. For continuous functions (theorem 5,page 32), the difference in the values of the function at the tags is boundedand the subintervals are telescoped. Later, we will see proofs that combineboth types of bound-and-telescope.

The third ingredient is the triangle inequality. A member of the specialsequence of Riemann sums is selected to be close to the potential value ofthe integral. Then a generic partition with sufficiently small mesh is chosen.A refinement of the two partitions is constructed and its sum is shown to beclose to the Riemann sums of both of the other two partitions. An applicationof the triangle inequality then completes the proof.

You are strongly encouraged to review the proofs in this section. Clas-sify the steps in each proof according to the taxonomy just outlined. Goingforward, both in this and subsequent chapters, watch for these moves. Thisdiscipline will help you make sense of the proofs.

2.2 Subintervals

Suppose f is integrable over Œa; b� and Œb; c�. Is f integrable over Œa; c�?If so, is R

R caf D R

R baf C R

R cbf ? While this may seem obvious, the

conclusion fails for the Riemann-Stieltjes integral introduced in Chapter 8.The result is true for the Riemann integral.

Theorem 7. Suppose f is Riemann integrable over Œa; b� and Œb; c�. Thenf is integrable over Œa; c� and R

R caf D R

R baf C R

R cbf .

Proof. For notational efficiency set A1 D RR ba f , A2 D R

R cb f , and A D

A1 C A2.Now suppose that " > 0. Since f is integrable over Œa; b� and Œb; c�,

we can conclude that there is a value B such that jf .x/j 0 so that whenever P1is a tagged partition of Œa; b� with kPk < ı1 then jSR .f;P1/ � A1j < "=4.We also can find a corresponding ı2 > 0 for the interval Œb; c�. Set ı Dmin fı1; ı2; "=4Bg and suppose that P Df.ti ; Œxi�1; xi �/g is a tagged parti-tion of Œa; c� with kPk < ı. In exercise 20 you are asked to prove that if b isa division point of P then jSR .f;P/ � Aj < "=2:

If b is not a division point of P , create a new tagged partition P� byadding b as a division point and taking b as the tag for both newly createdsubintervals. Now xi�1 < b < xi for some i between 1 and n. Since all theother tagged intervals are the same in P and P�, their corresponding termsin the Riemann sums cancel leavingˇ

SR .f;P/ � SR�f;P�

�ˇD jf .ti / .xi � xi�1/ � .f .b/ .b � xi�1/

Cf .b/ .xi � b//j

D j.f .ti / � f .b// .xi � xi�1/j

� 2B .xi � xi�1/ < 2Bı < "=2:

To complete the proof, apply the result of exercise 20 (the case when b is adivision point) to P� and conclude that

jSR .f;P/ � Aj �ˇSR .f;P/ � SR

�f;P�

�ˇCˇSR

�f;P�

�� A

ˇ< ".

What about the other direction? If f is integrable over Œa; b� and Œc; d � �Œa; b� must f be integrable over Œc; d �? Yes!

Theorem 8. If f is integrable over Œa; b� and Œc; d � � Œa; b�, then f isintegrable over Œc; d � :

Proof. As in the proof of theorem 5 (page 32), for each n 2 N create apartition Pn of Œc; d � using the division points fc C k

n.c � d/gn�1

kD1and

let A be a cluster point of fSR .f;Pn/g1nD1. Let " > 0 be given andchoose n0 so that jSR.f;Pn0/ � Aj < "=2. Also choose a ı > 0 so thatˇSR .f;P�/ � R

R ba f

ˇ< "=4 whenever P� is a tagged partition of Œa; b�

with kP�k < ı. Then if P� and Q� are tagged partitions of Œa; b� withmeshes less than ı, jSR .f;P�/ � SR .f;Q�/j < "=2.

Let Ra, P , and Rb be tagged partitions of Œa; c�, Œc; d �, and Œd; b� respec-tively with meshes less than ı. (Ra or Rb will be empty if Œa; c� or Œd; b� is

2.3. The fundamental theorems 39

degenerate.) The partitions Ra, P , and Rb combine to create a partition P�of Œa; b� with kP�k < ı. Similarly, Ra, Pn0 , and R create a partition P�n0 .Since P� and P�n0 share Ra and Rb in common,ˇ

SR .f;P/ � SR�f;Pn0

�ˇDˇSR

�f;P�

�� SR

�f;P�n0

�ˇ< "=2.

Hence

jSR .f;P/ � Aj �ˇSR .f;P/ � SR

�f;Pn0

�ˇCˇSR

�f;Pn0

�� A

ˇ< ".

Thus f is Riemann integrable over Œc; d � with RR dc f D A.

2.3 The fundamental theoremsOne of the important ideas of elementary calculus is the fundamental theo-rem of calculus that connects the concepts of the Newton-Leibniz and Rie-mann integrals. Intuitively, the fundamental theorem of calculus says thatintegration and differentiation are inverse operations. As we shall see, therelationship is not quite that simple. We begin with statements and proofs ofthe two forms of the fundamental theorem.

Theorem 9 (FTC-1). If F is a differentiable function on the interval Œa; b�and F 0 is continuous on .a; b/ then

1. F 0 is Riemann integrable on Œa; b� and

2. RR xa F

0 D F .x/ � F .a/ for all x 2 Œa; b�.

Proof. The first statement follows from theorem 5 (page 32).To verify the value of the integral, let P DfŒxi�1; xi �gniD1 be any partition

of Œa; x� : Applying the mean value theorem to F on the interval Œxi�1; xi �,we can select tags ftig

niD1 so that F 0 .ti / .xi � xi�1/ D F .xi / � F .xi�1/.

Then, by telescoping the sum,

SR�F 0;P

�D

nXkD1

F 0 .tk/ .xk � xk�1/

D

nXkD1

.F .xk/ � F .xk�1// D F .x/ � F .a/ :

Since we know that F 0 is integrable over Œa; x�, we can conclude thatRR xa F

0 D F .x/ � F .a/ for all x 2 Œa; b�. (See exercise 23b.)


The requirement that F 0 be continuous is somewhat annoying. If, as Pois-son believed, differentiation and integration truly are inverse operations, thenwe should have R

R xa F

0 D F .x/ � F .a/ whenever F is differentiable onŒa; b� and x 2 Œa; b�. But the theorem fails when the assumption of continuityis removed.

Example 6. Let

f .x/ D

(x2 sin 1

x2; x ¤ 0

0; x D 0:

Then f is differentiable everywhere, but f 0 is not bounded. Thus f 0 is notRiemann integrable. (See exercise 24.)

Unbounded derivatives are not the only barrier to a generalized version ofFTC-1. In Chapter 4 we will construct a function with a bounded derivativefor which FTC-1 fails.

If you review the proof of FTC-1, you will see that the role played by theassumption that F 0 is continuous is to ensure that F 0 is Riemann integrable.In light of this fact, the push to determine exactly which functions are Rie-mann integrable (and, if possible, to extend this set) takes on a heightenedimportance.

On the other hand, if there is a single point at which F fails to be differ-entiable, the conclusion of the fundamental theorem may be false even whenthe point where F is not differentiable is a removable discontinuity of F 0.

Example 7. Define

F .x/ D

�0; x 2 Œ0; 1=2/

1; x 2 Œ1=2; 1� :

Then F is differentiable on Œ0; 1� except at x D 1=2. Since F 0 is zeroexcept at x D 1=2, the discontinuity of F 0 at x D 1=2 is removable. ThenRR 10F 0 D 0, but F .1/ � F .0/ D 1.

There is, of course, a second form of the fundamental theorem of calculus.

Theorem 10 (FTC-2). Suppose that f is Riemann integrable on Œa; b�.Define F on Œa; b� by F .x/ D R

R xa f:

1. Then F is continuous on Œa; b�.

2. If f is continuous at x0 2 .a; b/ ; then F is differentiable at x0 andF 0 .x0/ D f .x0/.

2.4. Convergence theorems 41

Proof. First note that, in order to be Riemann integrable, f must be boundedby some value B . Thus if x < y, we can use theorems 7 and 8 and exercise1 to show that

jF .y/ � F .x/j D

ˇRZ y

x

f

ˇ� R

Z y

x

B D B .y � x/ .

The continuity of F follows easily. (See exercises 25b and 25c.)Now suppose that f is continuous at x0. For any " > 0 we can find a

ı > 0 so that f .x0/ � " < f .x/ < f .x0/C " for any x 2 Œx0; x0 C ı� \Œa; b�. Since constants are Riemann integrable, monotonicity of the Riemannintegral (exercise 1c) implies that

RZ x

x0

.f .x0/ � "/ �RZ x

x0

f � RZ x

x0

.f .x0/C "/

so that

.f .x0/ � "/ .x � x0/ � F .x/ � F .x0/ � .f .x0/C "/ .x � x0/ :

Rearranging,

�" �F .x/ � F .x0/

x � x0� f .x0/ � ":

A similar argument applies to values to the left of x0 so that F is differen-tiable at x0 and F 0 .x0/ D f .x0/.

2.4 Convergence theoremsWe now take up the driving question behind the intense historical investiga-tion of integration. How does the Riemann integral interact with limits? Webegin with a set of examples that illustrate some of the issues.

Example 8. Let frng1nD1 be an enumeration of the rational numbers in Œ0; 1�.

Define

fn .x/ D

�1; x 2 fr1; r2; : : : ; rng

0; otherwise:

Then fn has a finite number of discontinuities and thus is Riemann inte-grable. With a bit more work, one can verify that R

R 10fn D 0 (exercise 26).

On the other hand, if x 2 Œ0; 1� then limn!1 fn .x/ D d .x/, the Dirichletfunction of example 2 (page 28). Since the Dirichlet function is not Riemannintegrable, R

R 10 limn!1 fn ¤ limn!1

RR 10 fn. The left-hand side does not

exist.

Example 9. Define fn on Œ0; 1� by

fn .x/ D

�n; x 2 .0; 1=n/

0; otherwise:

Then fn is Riemann integrable with RR 10 fn D 1. In this case, the

limit function, f .x/ D limn!1 fn .x/ D 0, is Riemann integrable, butRR 10 limn!1 fn ¤ limn!1

RR 10 fn. Both sides exist, but they do not agree.

Example 9 is the more troubling of the two examples. Since bothRR 10 limn!1 fn and limn!1

RR 10 fn exist, there is less of an indication

that something may be amiss. Of course, often things work out “right” inthe sense that the equation R

R 10 limn!1 fn D limn!1

RR 10 fn is true.

When is this the case? The problem in the two examples is that, whilefn .x0/ ! f .x0/ for any x0 2 Œ0; 1�, no matter how large the value ofn, there remain other values of x 2 Œ0; 1� for which fn .x/ is far from f .x/.The key to getting the Riemann integral to interact “nicely” with limits isuniform convergence.

Definition 4 (Uniform Convergence). A sequence of functions ffng con-verges uniformly to a function f on a set S if for any " > 0, it is possible tofind an N 2 N such that jfn .x/ � f .x/j < " for all n > N and all x 2 S .

The modifier “uniformly” refers to the fact that the value of N can bechosen without regard for the value of x. One value of N will serve for allx 2 S . The sequence of functions in example 9 converges (pointwise) butdoes not converge uniformly.

Theorem 11. Suppose that ffng is a sequence of functions that are Rie-mann integrable over Œa; b� and converge uniformly on Œa; b� to f . Then f isRiemann integrable over Œa; b� and

RZ b

a

limn!1

fn DRZ b

a

f D limn!1

RZ b

a

fn:

Proof. Since fn is Riemann integrable on Œa; b�, fn is bounded on Œa; b�.Because ffng converges uniformly, we can find a value B so that jfn .x/j <B for all x 2 Œa; b� and all n 2 N. (See exercise 27.) Hence

�B .b � a/ � RZ b

a

fn � B .b � a/ :

Consequently, f RR ba fng, being a bounded sequence, must have a cluster

point A.We will show that f is integrable with R

R ba f D A: To that end, suppose

that " > 0 and select n so that j RR ba fn � Aj < "=3 and jfn .x/ � f .x/j <

"=3 .b � a/ for all x 2 Œa; b�. (Why does such an n exist?) Then for anytagged partition P Df.tk ; Œxk�1; xk�/gmkD1 ,

jSR .f;P/ � SR .fn;P/j DˇˇmXkD1

.f .tk/ � fn .tk//�k

ˇˇ

�

mXkD1

"

3 .b � a/�k D "=3.

As fn is integrable over Œa; b� ; we can find a ı > 0 so thatjSR.fn;P/ � R

R ba fnj < "=3 for any tagged partition P of Œa; b� with mesh

kPk < ı. But then for any such tagged partition P of Œa; b�,

jSR .f;P/ �Aj

�

ˇˇSR .f;P/�SR .fn;P/

ˇˇCˇˇSR .fn;P/ � R

Z b

a

fn

ˇˇC

ˇˇ RZ b

a

fn � A

ˇˇ<".

We conclude that f is integrable over Œa; b� with RR ba f D A D

limn!1RR ba fn.

To complete the proof, we need to verify thatn

RR ba fn

oconverges to A.

By the uniqueness of the integral (exercise 1) the sequence can have only

one cluster point. Thusn

RR ba fn

omust converge to A (exercise 29).

We close out this chapter with a final example showing that, while uniformconvergence is sufficient to ensure that R

R ba limn!1 fn D limn!1

RR ba fn,

uniform convergence is not necessary.

Example 10. Let fn .x/ D xn for x 2 Œ0; 1�. Then ffng converges point-wise, but not uniformly, to

f .x/ D

�1; x D 1

0; x 2 Œ0; 1/ :

Nevertheless, RR 10 fn D

1nC1

converges to 0 D RR 10 f .


2.5 Exercises

2.1 Riemann integrability: filling the gaps1. Verify the standard integral properties for the Riemann integral. These

properties will be used without comment in the text.

(a) Uniqueness. The value of the Riemann integral is unique (if itexists).

(b) Linearity. Let c 2 R. If f and g are Riemann integrableover the interval Œa; b�, then so are f C g and cf . Moreover,RR ba.f C g/ D R

R baf C R

R bag and R

R bacf D c R

R baf:

(c) Monotonicity. If f and g are Riemann integrable over the intervalŒa; b� with f .x/ � g .x/ for x 2 Œa; b�, then R

R ba f �

RR ba g:

(d) Triangle inequality. If f and jf j are Riemann integrable over

Œa; b�, thenˇ

RR ba f

ˇ� R

R ba jf j.

2. Show that if f is unbounded over the interval Œa; b�, then f is not in-tegrable over Œa; b�. (Show that if f is not bounded above, given anypartition P and any potential value A for the integral, tags can be cho-sen so that jSR .f;P/ j > AC 1.)

3. Let f be an increasing function defined on Œa; b� and let P be a taggedpartition P of Œa; b�. Prove that SR.f;PL/ � SR.f;P/ � SR.f;PR/.

4. Prove that the two definitions of refinement given in Definition 3(page 30) are equivalent.

5. In the proof of lemma 2 (page 30),

(a) Complete the induction proof of the left inequality.(b) Prove the right inequality.(c) Where and how would the proof fail if f were not increasing.

6. In the proof of theorem 4 (page 32)

(a) Why is PnC1 a refinement of Pn?(b) Use example 3 (page 28) as a guide to complete the proof of theo-

rem 4 (page 32).

7. Suppose that f is bounded by B on Œa; b�; in other words, jf .x/j � Bfor all x 2 Œa; b� : Show that �B .b � a/ � SR .f;P/ � B .b � a/ forany tagged partition P of Œa; b�.

2.5. Exercises 45

8. Suppose that for any " > 0 you can select a ı > 0 so thatjSR.f;P/ � Aj < 15" for any partition P with mesh kPk < ı.Explain how this fact can be used to prove that for any " > 0 it ispossible to select a ı > 0 so that for any partition P with kPk < ı thecondition jSR.f;P/ � Aj < " is satisfied.


(a) The proof that f is Riemann integrable over Œa; b� if f is boundedand continuous except at the left endpoint, a, implicitly used theo-rem 7. Identify where and how this theorem was used.

(b) Explain why jSR.f;P/ � SR.f;P�/j � 2B �ı. (Most terms cancel.)(c) Why is j

PP�1f �xj � B � 1

n0?

(d) Provide a proof that f is Riemann integrable over Œa; b� if f isbounded and continuous except at the right endpoint, b.

(e) Use theorem 7 (page 37) to prove that if f is bounded on the intervalŒa; b� and continuous except for a finite number of points, then f isRiemann integrable over Œa; b�.

10. Define

sc .x/ D

�0; x < c;

1; c � x:

Let frig1iD1 be an enumeration of the rational numbers in the interval

Œ0; 1� and define f by f .x/ DP1nD1

12nsrn .x/.

(a) Why is f defined? In other words, how do you know thatP1nD1

12nsrn .x/ converges for x 2 Œ0; 1�?

(b) Prove that f is strictly increasing.Let c 2 Œ0; 1�.

(c) Prove that limx!cC f .x/ DPrk�c

1

2k.

(d) Prove that limx!c� f .x/ DPrk<c

1

2k.

(e) Conclude that f is continuous at c if and only if c is an irrationalnumber.

(f) Prove that RR 10

P1nD1

12nsrn D

P1nD1

12n

RR 10 srn .

(g) Express the value of RR 10

P1nD1

12nsrn without using an integral.

11. Use theorems 7 and 8 (page 37) to explain why RR c� 1ma

f � RR c� 1na

f D

RR c� 1mc� 1n

f whenever m > n > 1c�a

: (See theorem 6 on page 37.)

2.1 Riemann integrability: deeper reflections12. Give an example of a function f that is not Riemann integrable but for

which jf j is Riemann integrable.

13. Give an alternate proof that f .x/ D x is Riemann integrable overŒ0; 2� by verifying and using the following facts about a partitionP D


��nkD1

of Œ0; 2�.

(a)PnkD1

�x2k� x2

k�1

�D 4:

(b) tk �xkCxk�1

2when kPk is small.

14. Prove that if f is a decreasing function on Œa; b� then f isRiemann integrable. (Work smart, not hard.)

15. Prove that if f is Riemann integrable over Œa; b� and g agrees with f ex-cept at c 2 Œa; b�, then g is also Riemann integrable and R

R ba g D

RR ba f:

16. Suppose that for any " > 0 you can select a ı > 0 so thatjSR.f;P/ � Aj � " for any partition P with kPk < ı. Explain howthis fact can be used to prove that for any " > 0 it is possible to selecta ı > 0 so that for any partition P with mesh kPk < ı we are assuredthat jSR.f;P/ � Aj < ".

17. Explain how the proof of theorem 5 (page 32) can be modified to arriveat the conclusion that jSR.f;P/ � Aj < ". Why do you think this wasnot done in the provided proof?

18. Prove that

f .x/ D

�sin 1

x; x ¤ 0

0; x D 0

is Riemann integrable over Œ0; 1�.

19. Prove that if f is Riemann integrable over Œa; b� then so is jf j.(Hint: Use the fact that for any x and y in the domain of f ,jjf .x/j � jf .y/jj � jf .x/ � f .y/j.)

2.2 Subintervals: filling the gaps20. Fill in the gap in the proof of theorem 7 (page 37) by showing that if b

is a division point of P , then jSR .f;P/ � Aj < "=2. (Hint: Split P intotagged partitions of Œa; b� and Œb; c�.)


(a) Why must fSR .f;Pn/g1nD1 have a cluster point?

2.5. Exercises 47

(b) Why must we have jSR .f;P�/ � SR .f;Q�/j < "=2?(c) Explain how Ra, P , and Rb can be combined to create a partition

P� of Œa; b� :

2.2 Subintervals: deeper reflections22. Suppose that f is bounded on Œa; b�, that f is Riemann integrable over

Œt; b� for all t 2 .a; b/, and that limt!aCRR bt f exists. Prove that f is

Riemann integrable over Œa; b� with RR ba f D limt!aC

RR bt f .

2.3 Fundamental theorems: filling the gaps23. In the proof of FTC-1

(a) Explain how the mean value theorem is used to select the tags.(b) Complete the final piece of the proof of FTC-1 by using the defini-

tion of integrability to show that if A D RR xa F

0 then A D F .x/�

F .a/.

24. Let

f .x/ D

(x2 sin 1

x2; x ¤ 0

0; x D 0:

(a) Show that

f 0 .x/ D

(2x sin 1

x2� 2x

cos 1x2; x ¤ 0

0; x D 0:

(Be sure to address the x D 0 case.)(b) Explain why FTC-1 fails for this function.


(a) Why is F .x/ defined for x 2 Œa; b�?(b) Explain how theorems 7 and 8 (page 37) and exercise 1 can be used

to show that jF .x/ � F .y/j Dˇ

RR yx f

ˇ� B .y � x/ when x < y.

(c) Explain how jF .x/ � F .y/j � B jy � xj implies that F is contin-uous.

(d) Explain how

.f .x0/ � "/ .x � x0/ < F .x/ � F .x0/ < .f .x0/C "/ .x � x0/

follows from

RZ x

x0

.f .x0/ � "/ <RZ x

x0

f < RZ x

x0

.f .x0/C "/ :

(e) Supply the details for the proof of the differentiability of F in FTC-2when x is to the left of x0.

2.4 Convergence theorems: filling the gaps

26. Prove that g .x/ D

�1; x 2 fr1; r2; : : : ; rng

0; otherwiseis Riemann integrable

with RR bag D 0.

27. Prove that if ffng is a sequence of bounded functions that convergesto f uniformly on a set S , then ffng is uniformly bounded on S . Inother words, there is a value B so that jfn .x/j N ?

(b) Why can we select n so thatˇRR ba fn � A

ˇ< "=3 and

jfn .x/ � f .x/j < "=2 .b � a/ for all x 2 Œa; b�?(c) How does the uniqueness of the integral imply that

˚RR ba fn

�can

have only one cluster point?

29. Prove that any bounded sequence with a unique cluster point convergesto that cluster point.

30. Prove that the sequence ffng of example 10 (page 43) converges point-wise but not uniformly to f .

31. Define

fn .x/ D

(1; x 2

�0; 1n

�0; x 2

�1n; 1�:

Show that ffng does not converge uniformly. Nevertheless,RR ba limn!1 fn D limn!1

RR ba fn.

2.4 Convergence theorems: deeper reflections32. Prove that

r .x/ D

(1n; x D m

nwith m

nin lowest terms

0; otherwise

is Riemann integrable with RR 10r D 0.

2.5. Exercises 49

33. Give a counterexample to the conclusion of exercise 27 if ffng onlyconverges to f pointwise rather than uniformly. In other words, finda convergent sequence of bounded functions that is not uniformlybounded.

34. Provide an alternate proof that limn!1RR ba fn exists in theorem 11

(page 42) by showing that˚RR b

a fn�

is a Cauchy sequence.

35. Modify exercise 31 to give an example of a sequence of Riemann-integrable functions that is not uniformly bounded but for whichRR ba

limn!1 fn D limn!1RR bafn:

2.5 Related ideas: deeper reflections36. Let P be a partition of Œa; b� and let " > 0: Explain how one could select

tags frkgnkD1 and fskg

nkD1 so that if Pr and Ps are the tagged partitions

using those tags, then SR .f;Pr /�"=4 < SR .f;Pt / < SR .f;Ps/C"=4for any other tagged partition Pt also based on P .

37. Prove the following generalization of exercise 15. If f is Riemann in-tegrable over Œa; b� and g .x/ D f .x/ except for finitely many points inŒa; b�, then g is Riemann integrable over Œa; b� and R

R 10 g D

RR 10 f .

Prior to Riemann’s work, Cauchy introduced a definition of the integral.In Cauchy’s definition, only left endpoints are used.

Definition (Cauchy Integral) Let f be a real-valued function definedon Œa; b�. Then f is Cauchy Integrable over Œa; b� if there is a numberA such that for any " > 0 we can find a ı > 0 so that whenever P is apartition of I with kPk < ı;

jSR.f;PL/ � Aj < ":

In this case, A is the Cauchy integral of f over Œa; b� and we writeCR ba f D

CR ba f .x/ dx D A.

38. Prove that if f is Riemann integrable over Œa; b�, then f is Cauchy in-tegrable over Œa; b� and the two integrals agree.

39. Prove that the Dirichlet function is not Cauchy integrable over Œ0; 1�.

40. Prove that every function that is Cauchy integrable over Œa; b� is boundedon Œa; b�.


2.6 ReferencesBressoud, D.M. (2006). A Radical Approach to Real Analysis (2nd ed.).

Mathematical Association of America.Burk, F.E. (2007). A Garden of Integrals. Mathematical Association of

America.DePree, J. and C. Swartz (1988). Introduction to Real Analysis. John Wiley

& Sons.Gelbaum, B.R. and J.M.H. Olmsted (2003). Counterexamples in Analysis.

Dover.Riemann, B. (1990). Gesammelte Mathematishe Werke, reprinted with com-

ments by Raghavan Narasimhan. Springer.Simmons, G.F. (2007). Calculus Gems. Mathematical Association of

America.

CHAPTER 3

The Darboux integral

While the Riemann integral is relatively straightforward to understand, prov-ing theorems, particularly existence theorems, using its definition is ratherawkward. Determining that there is a limiting value for the Riemann sumstypically requires the use of special sequences of Riemann sums. Once apotential value for the integral is determined, we need a way to connect ele-ments of the special sequence of Riemann sums to any generic Riemann sumwith small mesh. The connection is almost always established by means ofrefining partitions. The process is relatively straightforward for increasingfunctions, but can be difficult for more general functions.

In 1870, sixteen years after Riemann published his definition, GastonDarboux defined an integral that makes many of these issues take care ofthemselves. The cost (and you should expect there to be a cost) is the need towork with supremums and infimums. For mathematicians who have workedwith these concepts for a while, this cost seems almost trivial. For studentsencountering the notion of supremum and infimum for the first time, thecost appears rather more substantial. If you are not relatively comfortableworking with these ideas, you would be well advised to spend some timereviewing Appendix A.2 (page 311 and following) which provides a quickoverview.

An important reason supremums are useful for our purposes is that thesupremum, supS , of a non-empty, bounded set S of real numbers alwaysexists. Among other benefits, the supremum’s existence eliminates the needto use special sequences to establish a value for the integral. Supremums andinfimums also make it possible to prove results like lemmas 1 and 2 (page30) for non-monotonic functions. The key to proving the two lemmas is thefact that, when f is an increasing function, f .xi�1/ � f .t/ � f .xi / forany t 2 Œxi�1; xi �. This suggests that we might try selecting tags mi andMi so that f .mi / and f .Mi / are the minimum and maximum values of f

51

52 CHAPTER 3. The Darboux integral

over the interval. While the idea is appealing, it turns out to be problematic.In addition to introducing an additional layer of notation, the sums corre-sponding to tags fmig and fMig will not telescope like those in lemma 3(page 31). Maybe we could work around these obstacles were it not for aneven more fundamental difficulty. The function f may not have a maximumor minimum value over Œxi�1; xi � (see exercise 1). Supremums and infimumssolve this problem as well.

Definition 5 (Darboux Integral). Let f be a bounded function defined onŒa; b� and let P DfŒxk�1; xk �gnkD1DfIkg be a partition of Œa; b�. The lowerand upper Darboux sums of f over P are, respectively,

SD .f;P/ DnXiD1

infŒxi�1;xi �

f � .xi � xi�1/ DXP

infIkf �xk

and

SD .f;P/ DnXiD1

supŒxi�1;xi �

f � .xi � xi�1/ DXP

supIk

f �xk .1

The lower and upper Darboux integrals of f over Œa; b� are

DZ b

a

f D supP

SD .f;P/ D sup˚SD .f;P/ W P is a partition of Œa; b�

�and

DZ b

a

f D infPSD .f;P/ .

If DR ba f D

DR ba f , then f is Darboux integrable over Œa; b� and the

Darboux integral of f over Œa; b� is

DZ b

a

f D DZ b

a

f .

In the exercises, you are asked to verify that all the relevant supremumsand infimums are defined, that SD .f;P/ � SD .f;P/ ; and that D

R ba f �

DR ba f . (See exercises 2 and 9.)Given any partition P of Œa; b�, Riemann’s approach to the integral must

consider an infinite number of Riemann sums: one for each choice of tags.

1 infIk f D inf ff .x/ j x 2 Ikg and supIk f D sup ff .x/ j x 2 Ikg.

3.1. Darboux integrability 53

In contrast, the Darboux approach only considers two sums, the upper andlower. When a tagged partition is used in a Darboux sum, the tags areignored. The drawback is that the two Darboux sums, because they in-volve infimums and supremums, are conceptually more complicated than areRiemann sums.

On the other hand, the Riemann definition of the integral requires an un-specified value A and, as we have seen, it can take some effort to verify itsexistence. The Darboux integral provides two potential values for the inte-gral, D

R baf and D

R baf , and f is Darboux integrable if these two values

match. Consequently, there is no mention of mesh, ", or ı in the Darbouxdefinition. So, if we are willing to overlook the complication of workingwith supremums and infimums, the Darboux definition is much simpler thanthe Riemann definition.

3.1 Darboux integrability

How does this translate to computations and proofs? To address this ques-tion, we will consider the same examples we used to introduce the Riemannintegral. Keep your finger or a bookmark in Chapter 2 and compare the work.

Example 11 (Constant functions). Let f .x/ D c on Œa; b�. Let P DfIkgbe any partition of Œa; b�. Since f is constant, both the supremum and infi-mum over any subinterval are c. Thus

SD.f;P/ DXP

supIk

f �xk DXPc �xk D c

XP

�xk D c .b � a/ :

Similarly, SD .f;P/ D c .b � a/. Since c .b � a/ is the only possible valueof a Darboux sum,

DZ b

a

f D supP

SD .f;P/ D c .b � a/ D DZ b

a

f .

This means that the constant function f .x/ D c is Darboux integrable overŒa; b� with D

R ba f D c .b � a/ :


d .x/ D

�1; x 2 Q

0; x 62 Q;

where Q is the set of rational numbers, is not Darboux integrable over anynon-degenerate2 interval Œa; b�. To see why, suppose that P DfIkg is a par-tition of Œa; b�. Over any non-degenerate subinterval the supremum of d is 1and the infimum is 0. Thus

SD.d;P/ DXP

supIk

d �xk DXP�xk D b � a

and

SD .d;P/ DXP

infIkd �xk D 0:

Again, the upper and lower Darboux sums have only one possible value each.Thus

DZ b

a

d D infPSD.d;P/ D .b � a/

and

DZ b

a

d D supP

SD .d;P/ D 0:

The Dirichlet function, d .x/, is not Darboux integrable.

Example 13 (Identity function). Let f be the identity function f .x/ D x

on Œ0; 2� and take PnDfIkg Dnh2k�2n; 2kn

ionkD1

. Since f is increasing,

infIkf D f

�2k � 2

n

�D2k � 2

n

and

supIk

f D f

�2k

n

�D2k

n:

Thus

SD .f;Pn/ DnXkD1

2k � 2

n

2

nD 2 �

2

n

and

SD.f;Pn/ DnXkD1

2k

n

2

nD 2C

2

n:

2 In this context, non-degenerate means a < b so that the interval consists of more than a singlepoint.

Now, for all n,

2 �2

nD SD .f;Pn/ � sup

PSD .f;P/ D D

Z 2

0

f

� DZ 2

0

f D infPSD.f;P/ � SD.f;Pn/ D 2C

2

n.

Hence DR 20f D D

R 20f D 2 so that f is Darboux integrable over Œ0; 2� with

DR 20 f D 2.

Compare what we just did for the Darboux integral to the amount of ef-fort required to prove the same results for the Riemann integral. While theconcepts we are working with are a bit more abstract, the proof itself is farmore compact.

In example 13, we used a specific partition to show that the upper andlower integrals are the same. We will be well served by an integrability cri-terion that helps us do this more generally. Cauchy provided a relevant cri-terion for Riemann integrals that we adapt here for Darboux integrals. Noteits connection to a key idea in the previous example.

Theorem 12 (Cauchy criterion for integrability). A bounded function fon the interval Œa; b� is Darboux integrable if and only if for any " > 0 it ispossible to find a partition P DfIkg for which SD.f;P/ � SD .f;P/ < "

or, equivalently, for whichP

P�supIk f � infIk f

��xk < ".

Proof. Suppose that a bounded function satisfies the criterion. Let " > 0 andsuppose that P" is a partition guaranteed by the criterion. Since

DZ b

a

f D supPSD .f;P/ � SD .f;P"/

andDZ b

a

f D infPSD.f;P/ � SD.f;P"/;

we see that

0 � DZ b

a

f � DZ b

a

f � SD.f;P"/ � SD .f;P"/ < ".

Because the inequality holds for any " > 0, we can conclude thatDR baf D D

R baf and that f is Darboux integrable over Œa; b�.

Conversely, suppose that f is Darboux integrable over Œa; b� so thatDR baf D D

R baf . Let " > 0. Since D

R baf is the least upper bound of

˚SD .f;P/ W P a partition of Œa; b�

�, we know that D

R ba f � "=2 is not an

upper bound. Thus there must be a partition P1 so that DR ba f � "=2 <

SD .f;P1/. Similarly, there is a partition P2 satisfying DR ba f C "=2 >

SD.f;P2/: Let Q D P1 [ P2 and use lemma 13 (below) to observe that

SD.f;Q/ � SD .f;Q/ � SD.f;P2/ � SD .f;P1/

<

DZ b

a

f C "=2

!�

DZ b

a

f � "=2

!D "

as required.

Lemma 13 (below) is an analog of lemma 3 (page 31). While the proofthat f .x/ D x is Darboux integrable did not need the five sets of inequali-ties used in the Riemann-integrability proof, analogous results still have animportant role to play. Lemma 13 is more broadly applicable since the as-sumption that f is increasing is dropped.

Lemma 13 (Refinements). Suppose that f is a bounded function on theinterval Œa; b�, that P is a partition of Œa; b�, and that Q is a refinement of P .Then SD .f;P/ � SD .f;Q/ � SD.f;Q/ � SD.f;P/:

Proof. The inner inequality is a consequence of exercise 2.For the first inequality, begin by assuming that Q is created from P by

the addition of a single division point y that falls in the interval�xj�1; xj

�of P . Since all the other subintervals of P and Q are equal, their corre-sponding terms will cancel when the lower Darboux sums are subtracted.The remaining terms come from the intervals

�xj�1; y

�and

�y; xj

�for Q

and�xj�1; xj

�for P . Since infA � infB whenever A B (why?),

SD.f;Q/ � SD.f;P/ D

infŒxj�1;y�

f ��y � xj�1

�C infŒy;xj �

f ��xj � y

�!� infŒxj�1;xj �

f ��xj � xj�1

��

inf

Œxj�1;xj �f � .y � xj�1/C inf

Œxj�1;xj �f � .xj � y/

!� infŒxj�1;xj �

f ��xj � xj�1

�D 0.

Hence SD .f;P/ � SD .f;Q/.

The general result for the first inequality now follows by induction. Theproof of the right inequality is similar and is left as an exercise.

We close out this section by addressing the Darboux integrability of in-creasing and continuous functions.

Theorem 14 (Increasing functions). If f is an increasing function definedon Œa; b� then f is Darboux integrable over Œa; b�.

Proof. Let " > 0 and let P DfŒxk�1; xk�gnkD1 be any partition of Œa; b� withkPk < "

f .b/�f .a/. Since f is increasing, infŒxk�1;xk � f D f .xk�1/ and

supŒxk�1;xk � f D f .xk/. Hence

SD.f;P/ � SD .f;P/ DnXkD1

f .xk/�xk �

nXkD1

f .xk�1/�xk

�

nXkD1

.f .xk/ � f .xk�1// kPk

D .f .b/ � f .a// kPk < ".

Darboux integrability follows from the Cauchy criterion.

Theorem 15 (Continuous functions). Suppose that f is continuous on theinterval Œa; b�. Then f is Darboux integrable over Œa; b�.

Proof. Let " > 0. Since Œa; b� is a compact set, f is uniformly continuouson Œa; b� and we can find a ı > 0 so that jf .x/ � f .y/j < "

b�afor all

x; y 2 Œa; b� satisfying jx � yj < ı. Now choose P DfIkg to be a partitionof Œa; b� with kPk < ı. Then for all k between 1 and n,

supIk

f � infIkf �

"

b � a

(see exercise 11). Hence

SD.f;P/ � SD .f;P/ DXP

supIk

f ��xk �XP

infIkf ��xk

DXP

supIk

f � infIkf

!��xk

�"

b � a

XP�xk D ".

We conclude that f is Darboux integrable on Œa; b� since f satisfies theCauchy criterion.

If you have not already done so, you should take some time to comparethe proofs for the Riemann and Darboux integrals. In every case, you willfind that by working with the slightly more abstract concepts of infimum andsupremum, the proofs are shorter and often easier to understand.

3.2 Comparing Riemann and Darbouxintegration

Thus far, we have essentially followed the same path for the Riemann andDarboux integrals. The proofs are different, but the results are the same. Hereis a tantalizing result for the Darboux integral that has no counterpart for theRiemann integral.

Theorem 16. Let F be differentiable on Œa; b� with a bounded derivative.Then

DZ b

a

F 0 � F .b/ � F .a/ � DZ b

a

F 0:

Proof. Exercise 15.

This theorem generalizes the fundamental theorem of calculus for the Rie-mann integral. When F 0 is Darboux integrable, we conclude that D

R ba F

0 D

F .b/ � F .a/. When F 0 is not Darboux integrable, we know that the upperand lower Darboux integrals straddle F .b/ � F .a/. There is no analogousresult for the Riemann integral since, in the Riemann context, nothing playsa role similar to the upper and lower Darboux integrals.

So, are there any functions that are Darboux integrable but not Riemannintegrable? It turns out that the answer is no. But before we establish thisfact, we prove a result in the spirit of lemma 2 (page 30):

Lemma 17 (Refinement bounds). Suppose that f is defined on Œa; b� withjf j < B . Let P be a partition of Œa; b�. If Q is a refinement of P created byintroducing N additional division points, then

1. 0 � SD.f;P/ � SD.f;Q/ � 2BN kPk and

2. 0 � SD .f;Q/ � SD .f;P/ � 2BN kPk.

Proof. Lemma 13 shows that both differences are at least zero.To verify the right side of the second inequality, consider for the moment

a single subinterval Œxi�1; xi � of P . Since Q D˚�yj�1; yj

��mjD1

is a refine-ment of P , there are values j and k so that xi�1 D yj�1 < � � � < yk D xi .

3.2. Comparing Riemann and Darboux integration 59

In other words, Œxi�1; xi � is made up of one or more subintervals of Q. Ifj D k, then no additional division points were added in Œxi�1; xi � and

infŒxi�1;xi �

f � .xi � xi�1/ � infŒyk�1;yk �

f � .yk � yk�1/ D 0.

Alternatively, if j < k, then"inf

Œyj�1;yj �f �

�yj � yj�1

�C � � � C inf

Œyk�1;yk �f � .yk � yk�1/

#� infŒxi�1;xi �

f � .xi � xi�1/

D

inf

Œyj�1;yj �f � inf

Œxi�1;xi �f

!��yj � yj�1

�C � � �

C

�inf

Œyk�1;yk �f � inf

Œxi�1;xi �f

�� .yk � yk�1/

� 2B � .yj�yj�1/C � � � C2B � .yk�yk�1/D2B.xi�xi�1/� 2B kPk.

Since N division points were added to make Q, there are at most N inter-vals from P for which the difference is nonzero. Thus

SD .f;Q/ � SD .f;P/ � 2BN kPk .

A similar analysis (exercise 16) proves the right-hand side of the firstinequality.

Note the similarities between the preceding proof and the proof of theo-rem 5 (page 32). In both cases we bound the function differences and tele-scope the intervals.

Theorem 18 (Riemann-Darboux equivalence). Let f be defined on Œa; b�.Then f is Riemann integrable if and only if f is Darboux integrable. More-over, R

R baf D D

R baf . In other words, the Riemann and Darboux integrals

are the same.

Proof. First, suppose that f is Riemann integrable. To prove that f is Dar-boux integrable, we need to control the upper and lower Darboux sums usingtagged partitions.

Let " > 0 be given and choose ı > 0 so thatˇ

RR ba f � SR .f;P/

ˇ< "

for any tagged partition P of Œa; b� with mesh kPk < ı. Fix a particularpartition P0 D fIkg with kP0k < ı and select tags fskg and ftkg so that

f .sk/ � " < infIkf � sup

Ik

f < f .tk/C ":

If we denote the corresponding tagged partitions by Ps and Pt , then

SD .f;P0/ � SD .f;P0/ DXP

supIk

f ��xk �XP

infIkf ��xk

<XP.f .tk/C "/ ��xk�

XP.f .sk/ � "/ ��xk

D SR .f;Pt / � SR .f;Ps/C 2" .b � a/ :

Now SR .f;Pt / and SR .f;Ps/ are both within " of RR ba f , so they are

within 2" of each other. Thus

SD .f;P0/ � SD .f;P0/ < 2"C 2" .b � a/ :

Since there is a partition satisfying the Cauchy criterion (see exercise 17b),we conclude that f is Darboux integrable.

To prove the converse, suppose that f is Darboux integrable and let " > 0be given. Since f is Darboux integrable, f is bounded by some value Band we can find a partition P0 so that SD.f;P0/ � SD .f;P0/ < "=2. Forfuture reference, let N be the number of division points in P0. Suppose thatP is any tagged partition of Œa; b� with mesh kPk < ı, where ı will bedetermined later, and let Q D P0 [ P . By lemma 13 (page 56),

SD.f;Q/ � SD .f;Q/ � SD.f;P0/ � SD .f;P0/ < "=2

and, since

SD .f;Q/ � DZ b

a

f � SD .f;Q/ ;

we can conclude thatˇ

DR baf � SD .f;Q/

ˇ< "=2 as well. Then by

lemma 17 (page 58),

0 � DZ b

a

f � SD .f;P/

�

ˇˇ DZ b

a

f � SD .f;Q/ˇˇC ˇSD .f;Q/ � SD .f;P/ˇ

< "=2C 2BN kPk .

Taking kPk < ı D "4BN

, the last expression is less than " and we canconclude that

DZ b

a

f � " < SD .f;P/ :

3.3. Additional integrability results 61

By a similar argument,

SD .f;P/ < DZ b

a

f C ":

Then

DZ b

a

f � " < SD .f;P/ � SR .f;P/ � SD .f;P/ < DZ b

a

f C "

implies that ˇˇSR .f;P/ � D

Z b

a

f

ˇˇ < ":

Hence f is Riemann integrable with RR ba f D

DR ba f .

Since the Riemann and Darboux integrals are the same, what is the advan-tage of having the two integrals? In fact, since the two integrals are not dif-ferent, a better question is: What is the advantage to having two approachesto the Riemann-Darboux integral? This question has a fairly straightforwardanswer. The Riemann approach is conceptually more natural and thereforeeasier to understand. The Darboux integral has more efficient tools for veri-fying the properties of the integral.

Lest you fear that the integrals to be introduced later in the text are alsodifferent approaches to the same integral, be assured that this is not the case.Later integrals will extend the Riemann-Darboux integral. In addition to em-ploying different approaches, these other integrals will produce a larger setof integrable functions and more robust convergence theorems.

3.3 Additional integrability resultsWe will complete our investigation of the Riemann-Darboux integral by de-veloping a pair of theorems that provide a precise set of integrability criteriafor the Riemann integral. It is strongly recommended that you complete exer-cise 14 before proceeding. Doing this exercise will prepare you to understandthe issues in the proof of theorem 19 (below).

By the Cauchy criterion, if f is integrable thenP

P.supIkf�infIkf /�xkmust be small for a suitably chosen partitionP . The idea in theorem 19 is thatfor this to occur, supIk f � infIk f can only be “large” on a set of subinter-vals with very small total length. The proof essentially combines the bound-and-telescope method used for continuous functions with a modification ofthe technique used for increasing functions. The variables h and l below are

meant to suggest respectively the height of the difference supIk f � infIk fand the total length of the subintervals where this difference is “large.”

Theorem 19 (Height-width bounds). Let f be a bounded function onŒa; b�. Then f is Riemann-Darboux integrable if and only if for any choiceof positive values h and l , we can find a partition P of Œa; b� for which thesum of the lengths of the subintervals where supIk f � infIk f exceeds h isless than l .

Proof. Begin by assuming that f is Darboux integrable and suppose thath; l > 0 are given. By the Cauchy criterion, we can find a partition P DfIkgfor which SD.f;P/ � SD .f;P/ < hl . Split P into two disjoint sets P>,the intervals Ik 2 P for which supIk f � infIk f exceeds h, and P�, theintervals for which supIk f � infIk f is bounded by h. Then

hl > SD.f;P/ � SD .f;P/

DXP

supIk

f � infIkf

!�xk

DXP�

supIk

f � infIkf

!�xk C

XP>

supIk

f � infIkf

!�xk

�XP>

supIk

f � infIkf

!�xk

� hXP>

�xk .

We conclude that the sum of the lengths of the subintervals where supIk f �infIk f exceeds h is less than l .

For the converse, suppose that f is bounded by B and that, for any choiceof positive values h and l , we can find a partition P DfIkg of Œa; b� forwhich the sum of the lengths of the subintervals for which supIk f � infIk fexceeds h is less than l . Let " > 0 and take h D "

2.b�a/and l D "

4B. Using

the same notation as above,

SD.f;P/�SD.f;P/ DXP�

supIk

f �infIkf

!�xkC

XP>

supIk

f �infIkf

!�xk

�XP�

h�xk CXP>

2B�xk

� h .b � a/C 2Bl D ".

Hence f is Darboux integrable by the Cauchy criterion.

3.3. Additional integrability results 63

What could cause supIk f � infIk f to exceed h? One possibility is thatthe interval is sufficiently long that f can change gradually from one valueto another. This case is not of particular interest since we can easily elimi-nate this behavior by choosing smaller intervals for the partition. The otherpossibility is that Ik contains a discontinuity of f . Theorem 19 tells us thatin some sense the set of discontinuities cannot be very large. The next defi-nition and theorem make this intuition precise.

Definition 6 (Measure zero). A set S has measure zero if for any " > 0 wecan find a countable set of open intervals fIkg so that

1. S � [kIk and

2.Pk l .Ik/ < "

where l .Ik/ is the length of Ik . In other words, there is a countable coverof S by open intervals of total length less than ".

Example 14. Any finite set has measure zero.Suppose that S D fa1; a2; : : : ; ang. Let " > 0 be given and take Ik D�ak �

"3n; ak C

"3n

�. Then S � [n

kD1Ik and

PnkD1 l .Ik/ D

PnkD1

2"3nD

23".

Example 15. The set S D˚1n

�1nD1

has measure zero.

For any " > 0, take In D�1n� "4�2n

; 1nC "

4�2n

�. Then S � [1nD1In andP1

nD1 l .In/ DP1nD1

"2�2nD 1

2".

Over 30 years after Darboux’s integral was introduced to the mathematicalworld, Henri Lebesgue used the concept of sets of measure zero to provide aprecise description of the functions that are Riemann-Darboux integrable.

Theorem 20 (Lebesgue, 1902). Let f be a bounded function on Œa; b�. Thenf is Darboux integrable over Œa; b� if and only if the set of discontinuitiesof f in Œa; b� has measure zero. (Alternatively, we say that f is continuousalmost everywhere on Œa; b�. This is often abbreviated to f is continuous a.e.on Œa; b� :)

Proof. Let D be the set of discontinuities of f in Œa; b� and suppose thatx 2 D. If I is an interval containing x as an interior point of I , then supI f �infI f > 0. Thus supI f � infI f > 1

nfor sufficiently large values of n:

Alternatively, if x is a boundary point between two subintervals, then byconsidering the interval I formed by taking the union we can find an integern such that supI f � infI f > 1

nfor at least one of the subintervals. (See

exercise 19.)Suppose that f is Darboux integrable. Given any " > 0 and an integer n,

theorem 19 tells us that we can find a partition Pn so thatPI2Pn;> l .I / <

"2�2n

where Pn;> is the set of subintervals I from Pn for which supI f �infI f > 1

n. Since the subintervals of Pn;> are closed rather than open,

modify them to become open intervals by extending them on each end by"

4m�2nwhere m is the number of intervals in Pn;>.

Denote the set of expanded open intervals by P�n;>. Then the union ofthe closed subintervals in Pn;> is contained in the union of the open inter-vals in P�n;>. Moreover,

PI2P�n;> l .I / <

"2n

. Together, all the subintervalsin [1nD1P�n;> form a countable set of open intervals whose union containsD. Computing the sum of the lengths of all the open subintervals, we findthat X

n

XI2P�n;>

l .I / <

1XnD1

"

2nD ":

Thus D, the set of discontinuities of f in Œa; b�, has measure zero.For the converse, suppose that f is bounded by B and that the set D of

discontinuities of f in Œa; b� has measure zero. Let " > 0 and choose acountable set of open intervals fIkg whose union contains D[fa; bg andfor which

Pk l .Ik/ < "=4B . Let C be the set of points in .a; b/ where

f is continuous. Then for any x 2 C we can find an open interval Jx contain-ing x and for which supJx f � infJx f < "

4.b�a/. Since fIkg[ fJx W x 2 Cg

is an open cover of the compact set Œa; b�, we can find a finite sub-cover. Express the subcover as fIk W k 2 FDg [ fJx W x 2 FCg where FDand FC are finite sets of indices. If needed, remove excess intervals un-til no point of Œa; b� falls into more than two intervals of the cover. (Seeexercise 21.)

Ignore any endpoints of intervals in the finite subcover that fall out-side of .a; b/ and use the endpoints that fall inside Œa; b� as divisionpoints of a partition P of Œa; b�. We will show that P fulfills the Cauchycriterion.

Now every subinterval of P will be contained in the closure of at least oneof the intervals in the finite subcover and no point in Œa; b� will be in more

3.4. Exercises 65

than two intervals. (See exercise 22.) Hence

SD.f;P/ � SD .f;P/

DXI2P

�supI

f � infIf

�l .I /

�Xk2FD

supIk

f � infIkf

!l .Ik/C

Xx2FC

supJx

f � infJxf

!l .Jx/

� 2BXk

l .Ik/C"

4 .b � a/

Xx2FC

l .Jx/

< 2B"

4BC

"

4 .b � a/2 .b � a/ D ".

We conclude that f is Darboux integrable by the Cauchy criterion.

The power of this theorem is illustrated in the following examples.

Example 16. Any bounded function f with a finite number of discontinu-ities in Œa; b� is Riemann-Darboux integrable over Œa; b� since any finite sethas measure zero.

Example 17. The function

f .x/ D

�1n; x 2

�1nC1

; 1n

�; n 2 N

0; otherwise

is Riemann-Darboux integrable over Œ0; 1� since the set of discontinuities off is

˚1; 12; 13; 14; : : :

�which is a set of measure zero.

Example 18. The Dirichlet function is not Riemann integrable since it isdiscontinuous everywhere.

3.4 Exercises3.0 Darboux integral: filling the gaps

1. Give an example of a bounded function on Œ0; 1� that has no maximum orminimum value. What are the supremum and infimum of your functionover Œ0; 1�? (Your function cannot be continuous.)

2. Assume that f is defined and bounded on Œa; b�.

(a) Explain why supI f and infI f exist for any interval I � Œa; b�.(Why are the relevant sets nonempty and bounded?)


(b) Explain why SD .f;P/ and SD .f;P/ exist for any partition ofŒa; b�.

(c) Explain why SD .f;P/ � SD .f;P/ for any partition of Œa; b�.(d) Explain why infP SD .f;P/ and supP SD .f;P/ are defined in the

definition of the Darboux integral.

3. Prove that if P is any tagged partition of Œa; b� then SD .f;P/ �SR .f;P/ � SD.f;P/:

4. Use exercise 3 to prove that if f is both Riemann and Darboux inte-grable over Œa; b�, then R

R ba f D

DR ba f .

5. Verify the standard integral properties for the Darboux integral.

(a) Uniqueness. The value of the Darboux integral is unique (if itexists).

(b) Linearity. If f and g are Darboux integrable over the inter-val Œa; b� and c 2 R, then so are f C g and cf . Moreover,DR ba.f C g/ D D

R ba f C

DR ba g and D

R ba cf D c

DR ba f:

(c) Monotonicity. If f and g are Darboux integrable over the intervalŒa; b� with f .x/ � g .x/ for x 2 Œa; b�, then D

R ba f �

DR ba g:

(d) Triangle inequality. If f and jf j are Darboux integrable over Œa; b�,

thenˇ

DR ba f

ˇ� D

R ba jf j.

3.1 Darboux integrability: filling the gaps6. In example 3 (page 28)

(a) Verify thatPnkD1

2k�2n

2nD 2 � 2

nand

PnkD1

2kn2nD 2C 2

n.

(b) Explain why DR 20 f D

DR 20 f D 2.

7. Verify thatSD.f;P/ � SD .f;P/ D

PP�supŒxi�1;xi � f � infŒxi�1;xi � f

��x:

8. In the proof in lemma 13 (page 56)

(a) Why is infA � infB whenever A B?(b) Use induction on the number of inserted division points to complete

the proof that SD .f;P/ � SD .f;Q/.(c) Prove that SD .f;Q/ � SD .f;P/.

9. Use lemma 13 (page 56) to explain why DR ba f �

DR ba f .

10. Suppose that f is an increasing function on Œa; b�. Verify that f .b/satisfies the definition of supŒa;b� f .

3.4. Exercises 67

11. Assume that jf .x/ � f .y/j < B for all x; y 2 S .

(a) Prove that supS f � infS f � B . (Use contraposition.)(b) Give an example of a function for which supS f � infS f D B .

3.1 Darboux integrability: deeper reflections12. Suppose f is Darboux integrable over Œa; b� and Œb; c�. Prove that f is

Darboux integrable over Œa; c� and DR caf D D

R baf C D

R cbf . (First show

that f is Darboux integrable over Œa; c� using the Cauchy criterion. Then

verify the value of DR ca f by showing that

ˇDR ba f C

DR cb f �

DR ca f

ˇ<

" for any " > 0.)

13. Suppose f is Darboux integrable over Œa; b� and a < c < b. Provethat f is Darboux integrable over Œa; c� and Œc; b�. (Use the Cauchycriterion.)

14. Using only the Cauchy criterion and theorem 15 (page 57), prove that iff is bounded with a finite number of discontinuities on Œa; b�, then f isDarboux integrable over Œa; b�. (Use the boundedness of f to constrainthe difference between upper and lower sums near points of discontinu-ity. Use the integrability of f on intervals of continuity to control thedifference between the upper and lower sums elsewhere.)

3.2 Relationship between integrals: filling the gaps15. Let F be differentiable on Œa; b�. Prove that D

R ba F

0 � F.b/ � F.a/ �DR ba F

0: (Use exercise 3 and the ideas in the proof of theorem 9 onpage 39.)

16. Complete the proof of lemma 17 by proving that 0 � SD.f;P/ �SD.f;Q/ � 2BN kPk.


(a) Use the definitions of supremum and infimum to explain whyit is possible to select tags fskg and ftkg so that f .sk/ � "

< infŒxk�1;xk � f � supŒxk�1;xk � f < f .tk/C ".(b) The proof shows that for any " > 0 we can find a partition P0 for

which SD .f;P0/�SD .f;P0/ < 2"C2" .b � a/. Explain why thisis sufficient. Why do you suppose that a proof was not be adjustedto make the expression come out as SD .f;P0/ � SD .f;P0/ < "?

(c) Supply the details to prove that 0 � SD .f;P/ � DR baf < ".

(d) When proving the converse, we selected an arbitrary partitionP withmesh kPk < ı where ı > 0 was to be selected later. What are thepros and cons of using this type of exposition?

(e) Supply the details to prove that SD .f;P/ < DR ba f C ":

(f) We never explicitly proved that DR ba f D

RR ba f when f is Riemann

integrable. Explain why this is not a problem.

3.3 Additional integrability results: filling the gaps18. In the first part of the proof of theorem 19, under what circumstances

couldP

P>�supIk f � infIk f

��xk and

PP> h�xk be equal?

19. Suppose that f is discontinuous at x.

(a) Prove that supI f � infI f > 0 for any interval I containing x inits interior. (Use contraposition.)

(b) Prove that, for any nondegenerate intervals of the form I1 D Œ˛; x�

and I2 D Œx; ˇ�, either supI1 f �infI1 f > 0 or supI2 f �infI2 f >

0 (or both). (Consider I1 [ I2.)

20. Suppose that f W Œa; b� ! R is continuous at x 2 .a; b/. Ex-plain why there is an interval Jx with x 2 Jx .a; b/ such thatsupJx f � infJx f < "

2.b�a/.

21. Let C be a finite cover of Œa; b� by open intervals.

(a) Suppose that .x1; y1/ ; .x2; y2/ ; .x3; y3/ 2 C and that z 2

.x1; y1/ \ .x2; y2/ \ .x3; y3/. Prove that one of the three intervalsis not needed in the cover of Œa; b�.

(b) Explain why there must be a subset of intervals from C that is still acover of Œa; b� but for which no point in Œa; b� lies in more than twointervals.

22. The set C Df.�0:1; 0:3/ ; .0:2; 0:9/ ; .0:5; 0:8/ ; .0:75; 1:1/g is an opencover of Œ0; 1�.

(a) What is the partition P that would be generated from this cover inthe proof of theorem 20 (page 63)?

(b) For each subinterval I of P find an interval J from C so that I iscontained in the closure of J .

23. In the proof of theorem 20 (page 63), why isPx2FC

l .Jx/ � 2 .b � a/?

3.3 Additional integrability results: deeper reflections24. Prove that any countable set has measure zero.

25. Prove that the union of two sets of measure zero has measure zero.

26. Using the two previous exercises, prove that the set of irrational numbersin Œ0; 1� does not have measure zero.

3.4. Exercises 69

27. Prove that if A is a set of measure zero and B � A, then B has measurezero.

28. Give an alternative proof of the fact that any function f that is increasingon Œa; b� is Darboux integrable over Œa; b�. (Use the following ideas.)

(a) Explain why f can only have jump discontinuities. In other words,limx!d� f .x/ < limx!d� f .x/ at any point d of discontinuity.

(b) Explain why f can only have a countable number of discontinuities.

29. Let

f .x/ D

�12n; x D k

2nwith k odd, n 2 N

0; otherwise:

Give four proofs that f is Riemann-Darboux integrable over Œ0; 1�.

(a) Use the definition of the Riemann integral. Suppose that P is atagged partition with mesh kPk � 1

2N. At most, how many times

can a tag produce the value 12N

? At most, what is the value of theterm f .tk/�xk associated with each subinterval in the partition?Explain why changing the tags or using smaller subintervals canonly decrease the Riemann sum. (This is a hard exercise.)

(b) Use the Cauchy criterion. Choose a particular partition Pn andbound SD.f;Pn/�SD .f;Pn/ above by n2

2nC 1n

or by some similarbound that goes to zero as n ! 1. Explain why this guaranteesintegrability.

(c) Use theorem 20 (page 63). Prove that f is continuous at every irra-tional number.

(d) Use theorem 11 (page 42).

30. Prove that

r .x/ D

�1n; x D m

nwith m

nin lowest terms,

0; otherwise;

is continuous except on a set of measure zero. (Use theorems 11 and 20on pages 42 and 63.)

31. We did not revisit the fundamental theorems for the Darboux integralsince they the same as for the Riemann integral. However, FTC-1 can beextended a bit from the statement given on page 39. Prove the followinggeneralization: Suppose that F is differentiable on Œa; b� and that F 0 isbounded and continuous on Œa; b� except on a set of measure zero. ThenDR xaF 0 D F .x/ � F .a/ for all x 2 Œa; b�.

3.4 Related ideas: deeper reflections32. Use theorem 19 (page 62) to prove the converse of exercise 38 from

Chapter 2: If f is Cauchy integrable over Œa; b� then f is Riemann-Darboux integrable over Œa; b�. (Blend the following ideas into a well-written proof.)

(a) Given ı; h > 0, explain how to modify any partition P D

fŒxi�1; xi �gniD1 with xi � xi�1 D �x D b�a

n< ı=2 to make

partitions Pl D fŒsi�1; si �gniD1 and Pu D fŒti�1; ti �g

niD1 with

kPlk ; kPuk < ı and such that, for at least half of the intervals forwhich supŒxk�1;xk � f � infŒxk�1;xk � f exceeds h;

i. f .sk/ � infŒxk�1;xk � f ,ii. f .tk/ � supŒxk�1;xk � f ,

iii. skC1 D tkC1 D xkC1, andiv. both skC1 � sk D �sk and tkC1 � tk D �tk are at least �x.

(b) Assume that f is not Riemann-Darboux integrable. Use h and l toselect " such that kPlk ; kPuk < ı while the Cauchy sums differ bymore than ".

3.5 ReferencesBressoud, David M. (2006). A Radical Approach to Real Analysis (2nd ed.).


America.Darboux, G. (1875). Memoire sur les functions discontinues. Ann. Sci. Ecole

Normale Superieure 4 (2): 57–112.DePree, J. and C. Swartz. (1988). Introduction to Real Analysis. John Wiley

& Sons.Gelbaum, B.R. and J.M.H. Olmsted (2003). Counter examples in Analysis.

Dover.Lay, S. (2004). Analysis with an Introduction to Proof (4th ed.). Prentice

Hall.

CHAPTER 4

A Functional zoo

As has been noted previously, the study of integration since the time ofNewton and Leibniz has been driven by a recognition that there are manyfunctions that are not as well behaved as polynomials. This chapter intro-duces you to a set of such functions. Even in the absence of questions ofintegration, this collection of functions illustrates the wide variety of some-what unexpected behaviors that functions can exhibit.

4.1 Dirichlet and friends

We begin with functions related to the Dirichlet function. As a reminder, theDirichlet function is defined as

d .x/ D

�1; x 2 Q

0; x 62 Q.

The Dirichlet function is bounded and discontinuous everywhere. In point offact, d .x/ probably should not be called the Dirichlet function since Dirich-let actually defined a class of related functions which take on different valuesfor rational and irrational inputs. However, the 0–1 function has become sucha standard that it has garnered the title of “The Dirichlet Function.”

There are two closely related functions. The first is sometimes called thesnowflake function though one can also find instances where it is referred toas the Dirichlet function. The snowflake function is defined by

s .x/ D

(1n; x D m

nwith m

nin lowest terms

0; otherwise.

The snowflake function is continuous at every irrational number and dis-continuous at every rational number. (See exercise 1.) This provides us with

71

72 CHAPTER 4. A Functional zoo

an example of a non-monotone, bounded function with a dense set of dis-continuities. Since its set of discontinuities has measure zero, s is Riemann-Darboux integrable.

If we take the reciprocal of every non-zero value of s, we obtain thefunction

r .x/ D

(n; x D m

nwith m

nin lowest terms

0; otherwise.

The function r , like the Dirichlet function, is discontinuous everywhere.Moreover, r is unbounded on every nondegenerate1 interval.

4.2 Trigonometric seriesSince questions about trigonometric series triggered much of the mathemat-ical research on integration, it is fitting that we include two examples oftrigonometric series.

4.2.1 Fourier’s functionSuppose that you want to model the steady state temperature of the infinitelamina bounded above and below by x D 0 and x D 1 and on the left byw D 0. (Here we follow Fourier’s convention of using z for the temperature,w for the horizontal axis, and x for the vertical axis.) The top and bottomboundary are held at a constant temperature of 0 and the temperature onthe left boundary is described by the function f .x/. Assuming no loss ofheat from the face of the lamina, the lamina’s temperature is modeled by theinitial value problem

@2z

@w2C@2z

@x2D 0;

z .w; 0/ D z .w; 1/ D 0; w > 0;

z .0; x/ D f .x/ :

Joseph Fourier proposed a method for solving this type of differentialequation.2 Begin by noting that functions of the form z .w; x/ D e�cw sin cx

1 By nondegenerate we mean that the interval has non-empty interior. The interval Œ1; 1� is anexample of a degenerate interval.

2 We have modified Fourier’s heat problem slightly to make the computations simpler. Fourier’soriginal problem used upper and lower boundaries of x D �1 and x D 1: To satisfy theoriginal boundary conditions, Fourier used terms of the form e�.2n�1/

�w2 cos .2n� 1/ �x

2

from which the series f .x/ DP1nD1

.�1/n4.2n�1/�

cos�.2n� 1/ �x

2

�was derived.

4.2. Trigonometric series 73

satisfy the differential equation @2z@w2C @2z

@x2D 0. If we restrict our attention

to the case where c has the form c D n� , then these functions also sat-isfy z .w; 0/ D z .w; 1/ D 0; w > 0. Fourier’s idea is that a solution toa particular initial value problem can be generated by taking infinite linearcombinations of such functions so that

z .w; x/ D

1XnD1

ane�n�w sin .n�x/

where the coefficients an are chosen so thatP1nD1 an sin .n�x/ D

z.0; x/ D f .x/. Similar ideas had been suggested previously, but the pro-cess remained theoretical until Fourier provided a method for computing thesequence fang of coefficients. Fourier’s approach involved what we wouldnow call inner product techniques.

Recall that for integers n and m;Z 1

0

sin .n�x/ sin .m�x/ dx D

�1=2; n D m

0; n ¤ m:

Suppose that f .x/ DP1nD1 an sin .�nx/. ThenZ 1

0

f .x/ sin .m�x/ dx DZ 1

0

1XnD1

an sin .n�x/

!sin .m�x/ dx

D

1XnD1

an

Z 1

0

sin .n�x/ sin .m�x/ dx

D a1 � 0C a2 � 0C � � � C am�1 � 0

C am �1

2C amC1 � 0C � � �

D am=2.

Thus the equation

an D 2

Z 1

0

f .x/ sin .n�x/ dx

provides a method to compute the coefficients we need to generate oursolution

z .w; x/ D

1XnD1

ane�n�w sin .n�x/ :


When the left boundary is held at a constant temperature of f .x/ D 1,we find that an D 0 when n is even and that an D 4

n�when n is odd. Thus

f .x/ D

1XnD1

4

.2n � 1/ �sin ..2n � 1/ �x/

and the steady state temperature of the lamina is given by

z .w; x/ D

1XnD1

4

.2n � 1/ �e�.2n�1/�w sin ..2n � 1/ �x/ :

The practiced eye will recognize multiple issues with the work we havejust done.

1. While our physical experience tells us that heat distribution solutionsshould exist, how do we know that our mathematical model has a so-lution? While unproven, the assumption that a mathematical solutionexists at least seems reasonable.

2. If the mathematical problem does have a solution, by what right dowe expect the solution to take the form

P1nD1 ane

�n�w sin .n�x/?More generally, why should we be able to express the solution as asum of functions that factor so that each part depends on only one ofthe variables? Moreover, how do we know that integer values for c ine�cw sin cx are sufficient? How do we know that we do not also need touse functions like e�cw cos cx or some other types of functions? Unlikethe assumption of the existence of a mathematical solution, the expec-tation that f can be expressed as f .x/ D

P1nD1 an sin .�nx/ smacks

of wishful thinking.

3. What justifies interchanging the infinite summation and the integration?We have seen multiple examples that tell us that interchanging limitsand integrals can produce erroneous results.

4. Even if we can legitimately interchange the series and the integral tocompute the coefficients, how do we know that, except for some obvi-ous special cases such as x D 0, the series even converges? It seemsreasonable that when the series converges, it should converge to f .x/,but how do we know?

Such are the questions that drove the study of real analysis in general andintegration in particular. We will not address these questions here, but somerelated ideas are investigated in the exercises and in the final chapter. (Seefor example exercise 9.)

4.2. Trigonometric series 75

If f .x/ DP1nD1

4.2n�1/�

sin ..2n � 1/ �x/ evaluates to 1 on the interval.0; 1/ and evaluates to 0 at the endpoints, then f must also satisfy

f .x/ D

8<:1; x 2 .�2;�1/ [ .0; 1/

0; x D �2;�1; 0; 1; 2

�1; x 2 .�1; 0/ [ .1; 2/ :

This type of behavior in a function was quite unexpected at the time thatFourier did his work. Functions were expected to have smooth graphs likethose of polynomials or rational functions. Lagrange thought that perhapsthis “un-function-like” behavior was a consequence of the fact that the seriesdoes not converge for all values of x and therefore does not determine atrue function. This line of thinking is given credence by noticing that thecoefficients are multiples of the odd terms of the diverging harmonic series.In fact, Lagrange’s conjecture was wrong: the series converges for all valuesof x. But that is a story for another time and place.

The issues that arose when studying the heat equations are not confinedto that problem. They also appear in the study of mathematical models ofother physical phenomena such as the shape of a vibrating string. (See exer-cise 10.)

4.2.2 Weierstrass’s functionAs a second illustration of a trigonometric series function with unexpectedbehavior we consider a function related to an 1872 example of Karl Weier-strass. Let

g .x/ D

1XnD1

1

2ncos .cn�x/

where c is an integer. This trigonometric series converges uniformly to acontinuous function since the individual terms are bounded by 1

2n. When

c > 2, the series

�

1XnD1

c2

�n� sin .cn�x/

corresponding to term-by-term differentiation converges for all values of xof the form x D k�

cmand diverges for all x of the form x D .kC1=2/�

cm.

It seems reasonable to conjecture that g is differentiable where x D k�cm

and not differentiable where x D .kC1=2/�cm

. Other values of x seem moreproblematic.

In fact, when c is an odd integer greater than 7, g is nowhere differentiable.To show that g is not differentiable at x0 we will prove that given any ı > 0,

we can find a value x1 satisfying jx1 � x0j < ı for whichˇg.x1/�g.x0/x1�x0

ˇis

larger than any specified bound.To that end, select m 2 N such that 3

2cm< ı. Then we can find an integer

N so that1 � jcmx0 �N j �

32

.

In this case, cos .� .cmx0 �N// � 0. Moreover, if we choose x1 so thatcmx1 D N , then jx1 � x0j � 3

2cmand

cos .cm�x1/ � cos .cm�x0/ D cos�N � cos .�N C � .cmx0 �N//

D .�1/N Œ1 � cos .� .cmx0 �N//�

with 1 � cos .� .cmx0 �N// � 1:Now suppose that n is an integer greater thanm. Because we are assuming

c is an odd integer,

cos .cn�x1/ � cos .cn�x0/

D cos cn�m�N � cos .cn�m�N C cn�m� .cmx0 �N//

D .�1/N Œ1 � cos .cn�m� .cmx0 �N//�

with 1 � cos .cn�m� .cmx0 �N// � 0. Thus all the terms in

1XnDm

12n

cos .cn�x1/ � 12n

cos .cn�x0/

x1 � x0

have the same sign. Since jcos .cm�x1/ � cos .cm�x0/j � 1 and jx1�x0j �32cm

, we see thatˇˇ1XnDm

1

2ncos .cn�x1/ � cos .cn�x0/

x1 � x0

ˇˇ � 2

2m1

jx1 � x0j�4

3

c2

�m:

For n less than m, the mean value theorem guarantees an x2 between x1and x0 for whichˇ

cos .cn�x1/ � � cos .cn�x0/

x1 � x0

ˇD j��cn sin�cnx2j � �c

n:

Thus ˇˇm�1XnD0

1

2ncos .cn�x1/ � cos .cn�x0/

x1 � x0

ˇˇ

�

m�1XnD0

� c2

�nD �

�c2

�m� 1

c2� 1

< �

�c2

�mc2� 1

:

4.3. Friends of Cantor 77

Putting the two halves together,ˇˇ1XnD0

12n

cos .cn�x1/ � 12n

cos .cn�x0/

x1 � x0

ˇˇ � 4

3

c2

�m� �

�c2

�mc2� 1

D c2

�m �43�

2�

c � 2

:

As long as c > 7, we note that 43� 2�c�2

> 0 and�c2

�mcan be made arbi-

trarily large. Thusˇg.x1/�g.x0/x1�x0

ˇis unbounded on any neighborhood of x0.

Consequently, g cannot be differentiable at x0.

4.3 Friends of CantorMost real analysis texts include Georg Cantor’s proof that the real numbersare uncountable. In keeping with one of the recurring themes of this text,Cantor’s work had its origins in the study of convergence sets of trigonomet-ric functions. Our purpose here is to introduce some additional results of thisinvestigation: the Cantor set, its relatives, and some associated functions.

4.3.1 The basicsThe Cantor set is generated by starting with the unit interval C0 D I0 D

Œ0; 1� and successively removing the middle thirds of all of the intervals inthe previous stage.

C0 D I0;1 D Œ0; 1�

C1 D I1;1 [ I1;2 D�03; 13

�[�23; 33

�C2 D I2;1 [ I2;2 [ I2;3 [ I2;4 D

�09; 19

�[�29; 39

�[�69; 79

�[�89; 99

�: : :

Cn D In;1 [ In;2 [ � � � [ In;2n D�03n; 13n

�[�23n; 33n

�[ � � � [

h3n�13n

; 3n

3n

i: : :

The intermediate sets Cn each consist of 2n closed intervals of length 13n

.(See Figure 4.1.) The Cantor set is the intersection of all these sets: C D\nCn.

The Cantor set has many surprising properties.

1. The Cantor set is closed as the intersection of closed sets. (OK—notso surprising.)

n = 0

n = 1

n = 2

Figure 4.1. The first three steps in constructing the Cantor set

2. The Cantor set is nowhere dense. In other words the closure of C(in this case, C itself) contains no intervals. To see why, suppose thatx; y 2 C . Then there is an n large enough that 1

3n< jx � yj. Since x

and y cannot both belong to the same interval in Cn, there must be anopen interval I in the complement of Cn located between x and y. Asthe complement of C contains the complement of Cn, I is also in thecomplement of C . Thus C cannot contain the interval .x; y/.

3. Every point x 2 C is a limit of a sequence of points in the comple-ment of C and of a sequence of points from Cn fxg. If x 2 C , thenx 2 Cn for every natural number n. The subinterval of Cn to whichx belongs will have its middle third removed at the next stage. Themidpoint of the removed interval is in the complement of C and bothendpoints of the removed interval belong to C . Thus for every n, wecan chose an 62 C and bn 2 Cn fxg so that both jan � xj and jbn � xjare less than 1

3n.

4. The Cantor set is uncountable. (See exercise 16.)

5. The Cantor set has measure zero. Since Cn consists of 2n (closed)intervals each of length 1

3n, we can cover Cn with open intervals of

total length at most 2 � 2n

3n. As C is contained in Cn for every n, we

can cover C with a finite set of open intervals of arbitrarily small totallength.

The Cantor set can also be defined as the set of numbers in Œ0; 1� that canbe expressed in ternary notation using only 0 and 2. (See exercise 15.)

4.3.2 VariationsSuppose that instead of removing intervals of length 1

3nat the nth stage,

intervals of length ˛n are removed. Then in the limit a total length ofP1nD0 2

n˛nC1 D ˛1�2˛

will be removed from the unit interval. The re-maining Cantor-like set will have all of the same properties as the basic

4.3. Friends of Cantor 79

Cantor set but, rather than being a set of measure zero, will have measure3

1 � ˛1�2˛

D 1�3˛1�2˛

. By suitably choosing 0 < ˛ � 13

, we can arrange for theCantor-like set to have any measure between zero and 1.

4.3.3 The Cantor function

As with the Cantor set, the Cantor function is built up iteratively. Begin bydefining f0 by setting f0 .0/ D 0, f0 .1/ D 1 and using linear interpolationin between. In this base case, we have taken a somewhat circuitous route todefining f0 .x/ D x on Œ0; 1� D C0:

At each stage in the construction of the Cantor set, Cn consists of 2n in-tervals from which the middle third will be removed to make the next setCnC1. To move from fn to fnC1, we set fnC1 D fn on the complement ofCn and modify the function on each of the subintervals of Cn. Consider one

of the 2n subintervals, say In;k Dhk3n; kC13n

i, from Cn. The function fnC1

is defined to agree with fn at the endpoints of In;k . On the interval to beremoved, fnC1 is assigned the average of the values at the two endpoints.Linear interpolation is used on the remaining subintervals. We illustrate theprocess by constructing f1 and f2.

Following the construction process, we begin with the endpoints: f1 .0/ Df0 .0/ D 0 and f1 .1/ D f0 .1/ D 1. The average of the values 0 and 1 is 1

2

so f1 .x/ D 12

for x 2�13; 23

�. On

�03; 13

�, f1 interpolates between 0 and 1

2

and on�23; 33

�, f1 interpolates between 1

2and 1.

To create f2, we set f2 D f1 on�13; 23

�, the complement of C1, and mod-

ify f1 on the subintervals�03; 13

�and

�23; 33

�:Make f2 constant (1

4and 3

4) on

the removed subintervals�19; 29

�and

�79; 89

�and interpolate on the remaining

intervals of C1:�09; 19

�,�29; 39

�;�69; 79

�; and

�89; 99

�.

Another way to think about the process is that fnC1 agrees with fn on theintervals where fn is constant and fnC1 is made constant on the middle thirdof each interval where fn is strictly increasing. The first three iterations ofthe process are displayed in Figure 4.2.

When x is in the complement of C , the sequence ffn .x/g is constantfrom some point on so ffn .x/g converges. The vertical distance betweensuccessive plateaus of fn is 1

2n. These facts can be used to show that ffng

converges uniformly on all of Œ0; 1�. (See exercise 17.) The function to whichffng converges is called the Cantor function which we denote by c.

3 The concept of measure will be more fully developed in the next chapter.

0.0 0.5 1.00.0

0.5

1.0

0.0 0.5 1.00.0

0.5

1.0

0.0 0.5f2

f0

f3

f1

1.00.0

0.5

1.0

0.0 0.5 1.00.0

0.5

1.0

Figure 4.2. Building Cantor’s function

Back in Chapter 2 we saw that

F .x/ D

(0; 0 � x < 1=2

1; 1=2 � x � 1

is differentiable on Œ0; 1� except at x D 1=2. Since (re)defining a functionat a single point will not change its Riemann integral, we can make senseof R

R x0 F

0 but RR x0 F

0 ¤ F .x/ � F .0/ for x 2�12; 1�. This failure is not a

complete surprise since, by FTC-2, RR xa f must be a continuous function of

x when f is Riemann integrable.Now consider

G .x/ D

(0; 0 � x < 1=2

x � 12; 1=2 � x � 1:

G is also differentiable on Œ0; 1� except at x D 1=2. In this case, G is con-tinuous and R

R x0 G

0 D G .x/ � G .0/ for all x 2 Œ0; 1�. (See exercise 18.)The function G illustrates a more general phenomenon. Suppose that F iscontinuous on the interval Œa; b� and is differentiable except on a finite set.As long as F 0 is Riemann integrable when we assign values at the missingpoints, R

R xaF 0 D F .x/ � F .a/ for all x 2 Œa; b�. (See exercise 19.)

Can we extend this process to allow an infinite number of points where thefunction is not differentiable? More specifically, suppose that we can assign

4.4. Volterra’s example 81

values to F 0 at those points where F is not differentiable in a manner that(1) is consistent with the values of F 0 at points of differentiability and (2)guarantees that F 0 is Riemann integrable. Is the continuity of F sufficient toguarantee that R

R xaF 0 D F .x/ � F .a/ for all x 2 Œa; b�‹

No. The Cantor function provides a counterexample. Since the Cantorfunction is constant on the complement of C , the Cantor function, c, is dif-ferentiable with c0 .x/ D 0 on the complement of the Cantor set. Since Chas measure zero, it is natural to extend c0 to be zero there as well. Whenthis is done, R

R 10c0 D 0 while c .1/� c .0/ D 1. In fact, as long as the values

chosen for c0 on the Cantor set are bounded, c0 will be Riemann integrablewith R

R x0 c0 D 0 for x 2 Œ0; 1� : (See exercise 20.)

4.4 Volterra’s exampleIn 1881, Vito Volterra provided an example of a function f that is differen-tiable on all of Œ0; 1� and for which f 0 is bounded but not Riemann integrableon Œa; b�. According to Lebesgue’s criterion (theorem 20, page 63), such afunction will have a derivative that is defined and bounded everywhere whilebeing discontinuous on a set of positive measure.

Volterra’s function is built on a Cantor-like setK constructed by removingintervals of length ˛n at stage n with 0 < ˛ < 1

3. Critically, the set K will

have a measure of 1�3˛1�2˛

> 0. Most of our effort will be spent in defining thefunction V on the complement of K.

The basic building block of the construction is the function

g .x/ D

(x2 sin 1

x; x > 0

0; x D 0:

Figure 4.3. The function g .x/ bounded by �x2 and x2

The key properties of g are

1. The function g is differentiable for all x � 0:

2. The derivative is bounded by jg0 .x/j � 3.

3. Given any ˛ > 0, there are x; z 2 .0; ˛/ for which g0 .x/ D 1 andg0 .z/ D 0.

Now suppose that we are given an interval .a; b/. Then there is a largestvalue 0 < c < b�a

2so that g0 .c/ D 0. Define ha;b .x/ on .a; b/ by

ha;b .x/ D

8<:g .x � a/ ; x 2 .a; aC c/

g .c/ ; x 2 ŒaC c; b � c�

g .b � x/ ; x 2 .b � c; b/ :

The left portion of ha;b consists of a copy of g that has been shifted to be-gin at x D a. The right portion is a mirror image ending at b. We havespliced a constant function in the middle in such a way that the left andright derivatives at a C c and b � c are all zero. (See Figure 4.4.) Thusˇha;b .x/

ˇ� min

n.x � a/2 ; .x � b/2

oand ha;b is differentiable on .a; b/

withˇh0a;b.x/ˇ� 3.

ba+ c b – ca

Figure 4.4. ha;b

The Volterra function V is defined by taking V .x/ D 0 for all x 2 K andV .x/ D ha;b .x/ for x in any interval .a; b/ removed to form K.

4.5. Exercises 83

By its construction, V is continuously differentiable on any interval in thecomplement of K. Suppose then that x0 2 K. If x 2 K, then

V .x/ � V .x0/

x � x0D 0.

If x 62 K, then x belongs to some interval .a; b/ in the complement of K.Since x 2 .a; b/ and x0 62 .a; b/, one of .x � a/2 or .x � b/2 is smaller than.x � x0/

2. Hence,

jV .x/ � V .x0/j Dˇha;b .x/

ˇ� min

n.x � a/2 ; .x � b/2

o< .x � x0/

2 .

Since ˇV .x/ � V .x0/

x � x0

ˇ� jx � x0j ,

we conclude that V 0 .x0/ D 0 for all x0 2 K.To see that V 0 is not continuous at any x0 2 K, note that x0 is the limit

of endpoints of removed subintervals. Moreover, arbitrarily close to eachendpoint of a removed subinterval there is a point where V 0 is 1. Thus, whileV 0 .x0/ D 0; there is a sequence of points fxng approaching x0 for whichV 0 .xn/ D 1. Since V 0 is discontinuous on K and K has positive measure,V 0 is not Riemann integrable.

We will soon see definitions of the integral for which V 0 is integrable withR x0V 0 D V .x/.

4.5 Exercises

4.1 Dirichlet and friends: filling the gaps1. Prove that Dirichlet’s snowflake function is discontinuous at every

rational number and continuous at every irrational number. (In any finiteinterval, there are only finitely many rational numbers with denominatorless than 1=".)

2. Prove that the function

r .x/ D

(n; x D m

nwith m

nin lowest terms

0; otherwise

is unbounded on every nondegenerate interval. (Explain why, given anyB , a nondegenerate interval must contain an x for which f .x/ > B .)

4.2 Trigonometric series: filling the gaps3. Show that if f .x/ D 1 for x 2 .0; 1/ then, according to the computa-

tional technique of Fourier,

(a) the even coefficients in f .x/ DP1nD1 an sin .n�x/ are zero

(b) an D 4n�

when n is odd.

4. Suppose that f .x/ DP1nD1

4.2n�1/�

sin ..2n � 1/ �x/ evaluates to 1on the interval .0; 1/. Prove that f must also satisfy

f .x/ D

8<:1; x 2 .�2;�1/ [ .0; 1/

0; x D �2;�1; 0; 1; 2

�1; x 2 .�1; 0/ [ .1; 2/ :

5. Prove that g .x/ DP1nD1

12n

sin .cnx/ is well defined and continuousby showing that the series converges uniformly.

6. Prove thatP1nD1

�c2

�n� sin .cn�x/ converges for x D k�

cmand diverges

for x D .kC1=2/�cm

.

7. Suppose that c;m;N 2 N, 1 � jcmx0 �N j � 32; and cmx1 D N .

(a) Fill in the details to prove that

cos .cm�x1/ � cos .cm�x0/ D .�1/N Œ1 � cos .� .cmx0 �N//� :

(b) If n is an integer greater than m, provide the details to show that

cos .cm�x1/�cos .cm�x0/ D .�1/N Œ1�cos.cn�m�.cmx0�N//�:

8. Prove that if f is differentiable at x0, then f .x/�f .x0/x�x0

is bounded onsome neighborhood of x0.

4.2 Trigonometric series: deeper reflections9. Functions of the form e�cw cos cx also satisfy @2z

@w2C @2z

@x2D 0. Show

that if we use the techniques of Fourier to express f .x/ D 1; x 2

.0; 1/ ; as f .x/ DP1nD1 an cos .n�x/, then all of the coefficients are

zero. Comment.

10. A vibrating string can be modeled by the initial value problem

@2y

@x2D

1

c2@2y

@t2;

y .x; 0/ D f .x/

where y .x; t/ is the y-coordinate of the string x units from the originat time t . The parameter c depends on the weight and tension of the

4.5. Exercises 85

string and f .x/ represents the starting position from which the string isreleased at time t D 0.

(a) Show that functions of the form cosac�t sin a�x satisfy this differ-ential equation.4

(b) Ignoring potential issues related to convergence, use the ideas ofFourier to find a trigonometric series that describes the shape of astring of length 1 at time t if the string begins in the shape of atriangle

f .x/ D

(2x; x 2 Œ0; 1=2�

2 � 2x; x 2 Œ1=2; 1� :

(c) Use a computerized graphics system to animate an approximation toyour solution.

(d) Show that functions of the form sinac�t cos a�x also satisfy thedifferential equation. What happens when we try to use functions ofthis form to construct a solution?

11. Prove that if f is bounded on Œa; b� and differentiable at x0, thenf .x/�f .x0/

x�x0is bounded on Œa; b�.

12. Trigonometric series are not the only series that exhibit bad behavior.Let

gn .x/ Dx3

.1C nx2/ .1C .n � 1/ x2/

and define

G .x/ D

1XnD1

gn .x/ .

(a) Use partial fractions and telescoping series to verify thatG .x/ D x.(b) Compute g0n .x/.(c) Show that

P1nD1 g

0n .0/ converges but is not equal to G0 .0/.

This behavior is more problematic than a series for whichP1nD1 f

0n .c/

does not converge as convergence can give one a false sense that all iswell.

4.3 Friends of cantor: filling the gaps13. In property (5) of the Cantor set, why is the total length of the covering

open intervals 2 � 2n

3nrather than 2n

3n?

4 Note that c is a parameter in this initial value problem so that a plays the role analogous tothat of c in Fourier’s original problem.

14. Prove that the Cantor function is monotone increasing.

15. While we commonly express numbers in the decimal system, it is pos-sible to use other bases. The Cantor set can be understood using theternary (base-three) system. With ai 2 f0; 1; 2g, we define

.0:a1a2a3 : : :/3 D

1XnD1

an

3n.

(a) Show that .0:1111 : : :/3 D12

.(b) Show that both .0:1000 : : :/3 and .0:02222 : : :/3 represent 1

3.

(c) Find two ternary representations for 23

.(d) Use the ternary representation to describe the sets that are removed

to form C1 and C2; the first and second stages in constructing theCantor set.

(e) Give a ternary description (with proof of equivalence) of the Cantorset.

(f) Use this description to find an element of C that is not the endpointof a removed interval.

16. Use the description of C in exercise 15 to prove that C is uncountable.(Use a diagonalization argument similar to the one that proves Œ0; 1� isuncountable.)

17. Prove that the sequence of functions that define the Cantor function con-

verges uniformly. (How far apart are fn k3n

�and fn

kC13n

�? Use this

fact to boundˇfm .x/ � fp .x/

ˇwhen x 2

hk3n; kC13n

i.)

18. Let

G .x/ D

�0; 0 � x < 1=2

x � 12; 1=2 � x � 1:

(a) Compute G0 .x/ for x 2 Œ0; 1�. Explain any special considerations.(b) Compute R

R x0 G

0 for x 2�0; 12

�:

(c) Compute RR x12G0 for x 2

�12; 1�:

(d) Compute RR x0 G

0 for x 2�12; 1�:

19. Suppose that F is continuous on Œa; b� and differentiable except at afinite number of points. Suppose further that, by suitably defining F 0

at the points where F is not differentiable, F 0 is Riemann integrableon Œa; b�. Prove that R

R xaF 0 D F .x/ � F .a/ for all x 2 Œa; b�. (Use

induction.)

4.5. Exercises 87

20. Let c .x/ be the Cantor function. Define c0 on the Cantor set in sucha way that c0 remains bounded. Prove that c0 is Darboux integrablewith D

R x0 c0 D 0 for all x 2 Œ0; 1�. (Given " > 0; use Cn to show

that it is possible to select a partition P of Œ0; x� in such a way thatSD .f;P/ < ".)

4.3 Friends of Cantor: deeper reflections21. Prove that the length of the curve y D c .x/ is 2. (Use the fact thatp

�x2 C�y2 � �x C �y to conclude that the length of the Cantorfunction is bounded above by 2. Explain why the length of the Cantorfunction is greater than the length of any fn used in its construction. Finda lower bound for the length of fn by replacing the diagonal segmentswith vertical segments.)

As with the Cantor set, the Cantor function can be defined using ternaryexpansions of the numbers in Œ0; 1�. (See exercise 15.)Given x 2 Œ0; 1�

(a) Express x in ternary notation.(b) If the ternary expansion of x contains a 1, replace every digit after

the first 1 with 0.(c) Replace every 2 with a 1.(d) Interpret the result as a binary number.

This process defines a function c�. The following two exercises investi-gate c� and its relationship to c.

22. Use the preceding definition of c� for the following computations.

(a) Verify that .0:111111 : : :/3 D12

and use this information to computec��12

�.

(b) Verify that .0:020202 : : :/3 D14

and use this information to computec��14

�.

(c) The value 13

can be expressed both as .0:100000 : : :/3 and as.0:022222 : : :/3. Verify that both expressions produce the samevalue for c�

�13

�.

(d) Verify that c� is well defined. In other words, show that in thosecases where x has two possible ternary expansions, both expansionsproduce the same value for c� .x/.

23. Show that c� D c on Œ0; 1�.

(a) Verify that c� is increasing (but not strictly increasing).

(b) Verify that c� is constant on every interval in the complement of theCantor set.

(c) Verify that c� .x/ D c .x/ for x in the complement of the Cantor set.(If x is in the complement of C , then x is in the complement of Cnfor some n. Verify that fn .x/ D c� .x/ where ffng is the sequenceof functions used to define the Cantor function in Section 3.)

(d) Use parts (a) and (c) to verify that c� .x/ D c .x/ on Œ0; 1�.

24. There is a third way of defining the Cantor function on Œ0; 1�.

(a) Let g0 .x/ D x.(b) For any integer n � 1, define

gnC1 .x/ D

8<:

12gn .3x/ ; x 2

�0; 13

�12; x 2

�13; 23

�12gn .3x/C

12; x 2

�23; 1�:

Prove that gn .x/ D fn .x/ where ffng is the sequence of func-tions used to define the Cantor function in Section 3. Conclude thatlimn!1 gn .x/ D c .x/, the Cantor function. (Use induction.)

4.4 Volterra’s example: filling the gaps25. Define

g .x/ D

(x2 sin 1

x; x > 0

0; x D 0:

(a) Use the definition of the derivative to prove that g is differentiableat x D 0:

(b) Compute g0 .x/ for x > 0 and verify that jg0 .x/j � 3 for x 2 Œ0; 1�.(c) Explain why any interval of the form .0; ˛/ contains points x and z

for which g0 .x/ D 1 and g0 .z/ D 0.

4.4 Volterra’s example: deeper reflections26. Suppose that x belongs to a subinterval .a; b/ of length ˛n or more that

is removed in the construction of K. Define Vn .x/ D ha;b .x/ for suchx and define Vn to be 0 elsewhere.

(a) Explain why Vn is differentiable on all of Œ0; 1�.(b) Prove that R

R x0 V

0n D Vn .x/ for all x 2 Œ0; 1�.

(c) Prove that Vn converges uniformly to V on Œ0; 1�. (Hint: How largecan jV .x/j be if x comes from a removed interval of length less than˛n?)

4.6. References 89

(d) Prove that V 0n converges to V 0 on Œ0; 1�.(e) Discuss limn!1

RR x0 V

0n and R

R x0 limn!1 V

0n.

4.6 ReferencesAllen, E.S. (1941). The scientific work of Vito Volterra. Amer. Math.

Monthly 48: 516–519. JSTOR 2303385.Boyce, W.E. & DiPrima, R.C. (2005). Elementary Differential Equations

and Boundary Value Problems (8th ed.). John Wiley & Sons.Burk, F.E. (2007). A Garden of Integrals. Mathematical Association of

America.Dauben, J.W. (1979). Georg Cantor: His Mathematics and Philosophy of the

Infinite. Harvard University Press.Dirichlet, G.L. (1969). Werke, reprint. Chelsea.Fourier, J. (1822). The Analytical Theory of Heat. translated by Alexander

Freeman (1878, re-released 2003). Dover Publications.Gelbaum, B.R. and J.M.H. Olmsted (2003). Counter examples in Analysis.

Dover.Gonzalez-Velasco, E.A. (1992). Connections in mathematical analysis: the

case of Fourier series. American Mathematical Monthly 99 (5): 427–441. JSTOR 2325087.

Goodstein, J. R. (2007). The Volterra Chronicles: The Life and Times of anExtraordinary Mathematician 1860–1940. History of Mathematics 31,American Mathematical Society.

Zygmund, A. (2002). Trigonometric Series (3rd ed.). Cambridge UniversityPress.

CHAPTER 5

Another Approach: MeasureTheory

I have to pay a certain sum, which I have collected in mypocket. I take the bills and coins out of my pocket and givethem to the creditor in the order I find them until I havereached the total sum. This is the Riemann integral. But Ican proceed differently. After I have taken all the money outof my pocket I order the bills and coins according to iden-tical values and then I pay the several heaps one after theother to the creditor. This is my integral.

— Henri Lebesgue in a letter to Paul Montel

In his 1902 Ph.D. thesis, Henri Lebesgue introduced an approach to integra-tion that resolves many of the convergence issues that we have noted withthe Riemann-Darboux integral. Lebesgue partitioned the y-axis rather thanthe x-axis.1 Figure 5.1 shows an interval from a partition of the range of f .The set of points that f sends into this subinterval is marked on the x-axis.The effect of this approach is to gather together those points for which f hasapproximately the same value.

Preimages play a critical role in this development. Given any set A, thepreimage of A under the function f is the set f �1 .A/ D fx 2 R W

f .x/ 2 Ag. For the time being, we will only consider preimages ofintervals.

Now suppose that we are given a function f that is defined on Œa; b� andfor which ˛ < f < ˇ. Take any partition P DfŒyk�1; yk�gnkD1 of Œ˛; ˇ� andset Ek D f �1 ..yk�1; yk �/, k D 1; 2; : : : ; n. Note that fEkg

nkD1 consists

of disjoint sets whose union is all of Œa; b�. The lower and upper Lebesgue

1 What follows is in the spirit of Lebesgue’s work but, for the sake of comparison, notationfrom previous chapters is used.

91

92 CHAPTER 5. Another Approach: Measure Theory

Figure 5.1. The preimage of an interval

sums are

SL .f;P/ DnXkD1

yk�1 �m.Ek/ and SL .f;P/ DnXkD1

yk �m.Ek/

where m.A/ is the measure of the set A.2

Example 19 (Constant function). Suppose that f .x/ D c is a constantfunction on Œa; b�. Noting that c � 1 < f < c C 1, let P DfŒyk�1; yk �gnkD1be a partition of Œc � 1; c C 1�. Taking kc to be the unique integer for whichykc�1 < c � ykc , the preimages associated with this partition are Ekc DŒa; b� and Ek D ¿ for k ¤ kc . Thus SL .f;P/ D ykc�1 � .b � a/ andSL .f;P/ D ykc �.b � a/. As the mesh kPk of the partition becomes smaller,both the upper and lower sum will approach c �.b � a/. Even though we havenot yet defined the Lebesgue integral, we conclude that L

R ba f D c � .b � a/.

Example 20 (Dirichlet function). Let d .x/ be the Dirichlet function

d .x/ D

�1; x 2 Q

0; x 62 Q

on Œ0; 1�. Note that �1 < d < 2 and take a partition P DfŒyk�1; yk �gnkD1of Œ�1; 2�. Now there are unique integers k0 and k1 for which yk0�1 <0 � yk0 and yk1�1 < 1 � yk1 . As long as there is a division point in.0; 1/, the associated preimages are Ek0 D Œ0; 1� nQ and Ek1 D Œ0; 1�\

Q. For all other values of k, Ek D ¿. As we have seen previously,

2 For now, think of measure as a generalization of length.

5.1. Measurable sets I 93

m�Ek1

�D m.Œ0; 1� \Q/ D 0 so that m

�Ek0

�D m.Œ0; 1� nQ/ D 1.3 Thus

SL .f;P/ D yk0�1 �1Cyk1�1 �0 D yk0�1 and SL .f;P/ D yk0 �1Cyk1 �0 Dyk0 . Both of these values are within kPk of 0. Hence d is Lebesgue inte-

grable with LR ba d D 0.

Given a function f W Œa; b�! .˛; ˇ/, it is fairly straightforward to verifythat for any pair of partitions P1 and P2 of Œ˛; ˇ�, we have SL .f;P1/ �SL .f;P2/. (See exercises 2 and 3.) Also, for any partition P of Œ˛; ˇ�,

SL .f;P/ � SL .f;P/ DnXkD1

yk �m.Ek/ �

nXkD1

yk�1 �m.Ek/

D

nXkD1

.yk � yk�1/ �m.Ek/

� kPknXkD1

m.Ek/ D kPk .b � a/ .

Since we can arrange for kPk to be arbitrarily small, we conclude thatthe upper and lower Lebesgue sums will converge and that every boundedfunction has an integral in the sense of Lebesgue. This seems too good to betrue.

It is.

5.1 Measurable sets IWhile the process of computing the measure of an arbitrary set may be com-plicated, the idea that every set should have a measure seems rather straight-forward. Moreover, the measure should satisfy some basic properties.

1. Intervals. For any interval Œa; b�, m.Œa; b�/ D b � a. In particular, themeasure of a single point is zero.

2. Monotonicity. For any sets A � B , m.A/ � m.B/.

3. Translation invariance. If B D A C t D faC t W a 2 Ag for somet 2 R (in other words,B is a translation by t of the setA), thenm.B/ Dm.A/.

4. Countable additivity. If a set E is the disjoint union of a countablecollection of sets fEkg then m.E/ D

Pk m.Ek/.

3 We have not yet justified the statement thatm.Œ0; 1� nQ/ D 1, but we will.


The restriction to countable disjoint unions is necessary. In addition tothe problem of making sense of an uncountable sum, the combination ofproperties (1) and (4) for unrestricted sums would imply that all sets havezero measure.

While all appears to be in good order, there are significant issues lurkingbelow the surface.

Example 21 (Vitali, 1905). Define two real numbers x and y to be ratio-nally equivalent if x � y 2 Q. For example, � and � C 1

2are rationally

equivalent while � andp2 are not. This relation is an equivalence rela-

tion that divides R into disjoint subsets of rationally equivalent real num-bers called equivalence classes. (See exercise 6.) Given an element x of anequivalence class, adding (or subtracting) any rational number will result inanother member of the same equivalence class. Thus every equivalence classwill have infinitely many members in Œ0; 1�. For each equivalence class, se-lect a single representative element from the interval Œ0; 1�. For example, wemight represent the class containing � by � � 3 and the class containing

p2

byp2 � 5

4. Denote the set of representatives by V .4

We make the following observations.

1. If s and t are distinct rational numbers, then V C s and V C t

are disjoint. Suppose that s and t are rational numbers with x 2

.V C s/\ .V C t /. Then x � s and x � t are elements of V and, sinces and t are rational numbers, both x � s and x � t represent the equiv-alence class that contains x. As the equivalence class of x has only onerepresentative in V , we conclude that x � s D x � t so that s D t .

2. For any x 2 Œ0; 1� there is a rational number s 2 Œ�1; 1� such thatx 2 V C s. Given x 2 Œ0; 1�, denote the representative of x by rx .Since both x and rx belong to Œ0; 1�, we know that jx � rxj � 1 or,equivalently, s D x � rx 2 Œ�1; 1�. Now x and rx differ by a rationalnumber so that s is rational with x D rx C s 2 V C s.

3. If x 2 Œ0; 1� and s 2 Œ�1; 1�, then x C s 2 Œ�1; 2�.

Now let fsng be an enumeration of all of the rational numbers in Œ�1; 1�.Since V C sn and V C sm are disjoint when n ¤ m, countable additivity and

4 If you are familiar with the concept, you may have noticed that we have just used the Axiomof Choice.

5.2. Outer measure 95

the translation invariance of measure imply that

m.[n .V C sn// DXn

m.V C sn/ DXn

m.V / .

At this point there are two options. Either m.V / D 0 (in which casem.[n .V C sn// D 0) or m.V / > 0 (which implies that m.[n .V C sn//is infinite).

From our previous discussion we know

Œ0; 1� � [n .V C sn/ � Œ�1; 2�

so that monotonicity and the interval property imply that

1 � m.[n .V C sn// � 3.

We have arrived at a contradiction.

What are we to make of this situation? We see that it is impossible forall four of the properties to hold for a measure defined on all sets. Sincewe want our measure to satisfy the four properties, we have to accept thefact that some sets of real numbers, like V , will not be measurable. We willinvestigate which sets are measurable later in this chapter. And as we shallsee, the set of measurable sets has an impact on the set of functions that areLebesgue integrable.

5.2 Outer measureSince the seemingly obvious fact that all sets are measurable is demonstrablyfalse, we turn our attention to the question of computing the measure of a set.We will come back to the problem of identifying which sets are measurablelater. We base the measure of an arbitrary set on the measure of intervals.

Definition 7 (Outer measure). Let A � R. The Lebesgue outer measureof A is

�� .A/

D inf

(Xk

l .Ik/ W fIkg is a countable cover of A by finite open intervals

)

where l .I / is the length of the interval I .

Example 22 (Outer measure of Œa; b�). Consider the interval Œa; b�. Weclaim that �� .Œa; b�/ D b � a. For any " > 0 we can cover Œa; b� by asingle open interval .a � "; b C "/ whose length is b � aC 2". Thus, we seethat �� .Œa; b�/ � b � a.

To show that �� .Œa; b�/ � b � a, suppose that fIkg is a cover of Œa; b�by open intervals. Since Œa; b� is compact,5 we can select a finite subcover˚�aj ; bj

��njD1

: For definiteness, we will also assume that the intervals havebeen ordered by their left endpoints and that (by removing unneeded inter-vals) no left endpoint is repeated and no interval is contained in another.In order to cover Œa; b�, the intervals

˚�aj ; bj

��njD1

must overlap. In otherwords,

a1 < a, a2 < b1; a3 < b2; : : : ; an < bn�1; and b < bn.

These inequalities imply that

Xk

l .Ik/ �

nXjD1

l��aj ; bj

��D .b1 � a1/C .b2 � a2/C � � � C .bn � an/

> .b1 � a/C .b2 � a2/C � � � C .b � an/

> .b1 � a/C .b2 � b1/C � � � C .b � bn�1/

D b � a:

Since b�a is a lower bound on the sums of the lengths of covering intervals,�� .Œa; b�/ � b � a. Combining the two inequalities, �� .Œa; b�/ D b � a.

In fact, the Lebesgue outer measure of any interval I is its length. Thisfact can be verified by modifying the preceding argument, but it is far easierto prove using the monotonicity of ��.

We have verified one of the four properties that a measure should satisfy.The verifications of monotonicity and translation invariance are relativelystraightforward and are left as exercises. This leaves the property of count-able additivity. From Vitali’s example, we know that this property fails forcertain combinations of sets. Can we rescue anything in this regard? Yes. Wecan prove subadditivity.

5 See Appendix A.1 for a review of the definition and basic properties of a compact set.

5.3. Measurable sets II 97

Theorem 21 (Subadditivity). Suppose that fAkg1kD1 is a countable collec-

tion of sets. Then�� .[kAk/ �

Xk

�� .Ak/ .

Proof. If �� .Ak/ is infinite for any k, the inequality automatically holds.So suppose that �� .Ak/ is finite for all k and let " > 0. Since �� .Ak/is defined by an infimum, we can choose a countable cover

˚Ik;j

�j

of Akconsisting of open intervals such that

�� .Ak/ �Xj

l�Ik;j

�< �� .Ak/C

"

2k.

Then˚Ik;j

�k;j

is a countable cover of [kAk by open intervals. As�� .[kAk/ also is defined by an infimum,

�� .[kAk/ �Xk;j

l�Ik;j

�<Xk

�� .Ak/C

"

2k

�DXk

�� .Ak/C ".

But " > 0 is arbitrary, so that

�� .[kAk/ �Xk

�� .Ak/ �

Note that we have strict inequality in the case of Vitali’s example. For theset V constructed in that example, �� .V / > 0 else

1 D �� .Œ0; 1�/ � �� .[n .V C sn// DXn

�� .V / D 0.

Since �� .V / > 0, we conclude that

�� .[n .V C sn// � �� .Œ�1; 2�/ D 3

<1DXn

�� .V / DXn

�� .V C sn/ .

5.3 Measurable sets IIWe now see that the crux of the problem lies in identifying the sets for whichwe have countable additivity. Constantin Caratheodory (1873–1950) pro-vided the key criterion.6 If the measure of the set E is always the sum of

6 Lebesgue took an alternative approach which is outlined in Exercises 71 and following.

the measures of its countable disjoint parts, then all the more will �� .E/ D�� .A/C�� .B/ when A and B are disjoint sets with E D A[B .7 Equiva-lently,�� .E/ D �� .E \ C/C�� .E \ C c/ for any setC:8 Caratheodory’sinsight was to apply the Golden Rule. Instead of asking if every set would“play nicely” with E, we ask if E will “play nicely” with every set.

Definition 8 (Caratheodory’s measurability condition). A set E R is(Lebesgue) measurable if

�� .A/ D �� .A \E/C �� .A \Ec/

for every subset A R. When E is measurable, we write � .E/ D �� .E/.

Note that, by subadditivity, we only need to check that ��.A/ �� .A \E/C �� .A \Ec/.

Example 23 (Every finite set is measurable). Suppose that E D

fx1; x2; : : : ; xng. Let A be a subset of R. Now˚�xi �

"4n; xi C

"4n

��niD1

is acover of E and thus of E \ A with total length "

2. Hence �� .A \E/ < "

2.

Choose a countable cover fIkg of A by open intervals such thatXk

l .Ik/ � �� .A/C

"

2:9

Then fIkg, being a cover of A, is also a cover of A \ Ec so that�� .A \Ec/ � �� .A/C "

2. Hence

�� .A \E/C �� .A \Ec/ � �� .A/C ".

Since the inequality holds for all " > 0, we conclude that �� .A/ �� .A \E/C�� .A \Ec/. The reverse inequality is a consequence of sub-additivity so that the finite set E is measurable.10

Before spending too much energy identifying various measurable sets, weshould check to see that Caratheodory’s definition gets us what we are look-ing for: countable additivity. We don’t even know that additivity works for apair of disjoint measurable sets. Let’s start there.

7 We will see this move repeatedly in what follows. Wanting to know a result for countableunions, we start with the union of a pair of sets.

8 Ac denotes the complement of the set A. In other words, Ac D fx 2 R W x 62 Ag.9 If �� .A/ D1, we cannot have strict inequality.

10 There are easier ways to verify that finite sets are measurable. (See Exercise 18.) This proofwas chosen to illustrate the definition of measurability.

5.3. Measurable sets II 99

If E and F are disjoint, measurable sets, we can use the measurability ofF to conclude that

�� .E [ F / D �� ..E [ F / \ F /C �� ..E [ F / \ F c/

D �� .F /C �� .E/ .

Additivity for any finite set of disjoint, measurable sets follows by induction.Hence, if fEkg

1kD1 is a sequence of disjoint, measurable sets,

nXkD1

�� .Ek/ D ��[nkD1Ek

��

�[1kD1Ek

��

1XkD1

�� .Ek/ .

Since this inequality is true for all values of n,

1XkD1

�� .Ek/ � ��[1kD1Ek

��

1XkD1

�� .Ek/ .

Countable additivity holds for disjoint measurable sets!Unfortunately, we don’t know very many measurable sets nor do we even

know that the countable union of disjoint measurable sets is itself a mea-surable set. In fact, we don’t even know that the union of a pair of disjointmeasurable sets is again measurable. We will deal with that question in thenext section.

Meanwhile, let’s increase our collection of known measurable sets byshowing that .�1; a/ is measurable.

Example 24 (.�1; a/ is measurable). Let A be a subset of R. Given " >0, choose a countable cover fIkg of A by open intervals such thatX

k

l .Ik/ � �� .A/C ":

Since A [kIk , monotonicity and subadditivity imply that

�� .A \ .�1; a//C �� .A \ Œa;C1//

� �� ..[kIk/ \ .�1; a//C �� ..[kIk/ \ Œa;C1//

�Xk

�� .Ik \ .�1; a//CXk

�� .Ik \ Œa;C1// :

Now the intersection of two intervals is again an interval and the measureof any interval is its length. Moreover, all the terms in the summations are


nonnegative so the terms can be reordered. Therefore we can continue thechain of inequalities with

DXk

l .Ik \ .�1; a//CXk

l .Ik \ Œa;C1//

DXk

Œl .Ik \ .�1; a//C l .Ik \ Œa;C1//�

DXk

l .Ik/ � �� .A/C ".

Since " > 0 was arbitrary, we conclude that

�� .A \ .�1; a//C �� .A \ Œa;C1// � �� .A/ .

The reverse inequality is a consequence of subadditivity. Therefore, the in-terval .�1; a/ is measurable.

It is apparent that using Caratheodory’s definition to verify that sets aremeasurable tends to be quite tedious. Think about the increased complicationof using the definition to prove that .a; b/ is a measurable set. We need someadditional tools.

5.4 Sigma algebrasThe concept of a sigma algebra is a powerful tool for verifying measurabil-ity. While the definition of a sigma algebra is stated generally, our primaryconcern is with sigma algebras for which the X in the definition is either R

or a subset of R.

Definition 9 (Sigma algebra). Let X be a set. A collection A of subsets ofX is a sigma algebra if

1. The empty set belongs to A.

2. If A 2 A, then Ac 2 A.

3. Given a countable collection of sets fAkg from A, we have [kAk 2 A.

By using complements, the third property of a sigma algebra also impliesthat \kAk 2 A for any countable collection of sets from A. (See exer-cise 28.)

Before proving that the set of measurable sets is a sigma algebra, we pauseto demonstrate the power of the concept. For purposes of illustration, assumefor now that the set of Lebesgue measurable sets is a sigma algebra.

5.4. Sigma algebras 101

Example 25. The following sets are measurable.

1. Œa;C1/ D .�1; a/c .

2. Œa; b/ D .�1; b/ \ Œa;C1/.

3. .a; b/ D [n�aC 1

n; b�.

4. fag D \n�a � 1

n; aC 1

n

�.

5. Any countable set.

With a bit more effort, one can prove that all open and all closed sets aremeasurable. (See exercise 37.)

Compare the work just expended to the toil that would have been requiredto prove the same facts directly from the definition. The difference in effortjustifies the non-trivial mental energy we will expend to prove that the set ofLebesgue measurable sets is a sigma algebra.

The first two conditions of a sigma algebra are fairly straightforward exer-cises (25 and 26). As one might expect, significantly more effort is requiredto verify that the set of Lebesgue measurable sets is closed under countableunions. We begin by verifying that the union of two measurable sets is againmeasurable.

To that end, assume that E and F are measurable sets. Applying the mea-surability of E to an arbitrary set A,

�� .A/ D �� .A \E/C �� .A \Ec/ :

Then using the measurability of F with the set A \Ec ,

�� .A \Ec/ D �� ..A \Ec/ \ F /C �� ..A \Ec/ \ F c/ :

Combining these results and using subadditivity twice, we find that

�� .A/ D �� .A \E/C �� ..A \Ec/ \ F /C �� ..A \Ec/ \ F c/

� �� ..A \E/ [ .A \Ec \ F //C �� .A \Ec \ F c/

D �� .A \ .E [ F //C ��A \ .E [ F /c

��

�.A \ .E [ F // [

�A \ .E [ F /c

��D �� .A/ .

Since equality must hold throughout the last chain, we conclude that theunion of two measurable sets is measurable. The measurability of finiteunions of measurable sets follows by induction.


Now suppose that we have a sequence fFkg of disjoint, measurable sets.We will show that [kFk is measurable. The proof relies on an intermediateresult.

Let A be an arbitrary subset of the real numbers. We claim that

��A \

�[nkD1Fk

��D

nXkD1

�� .A \ Fk/ :

The claim is trivially true when n D 1. Assuming the result for n � 1, themeasurability of Fn and the fact that the elements of fFkg are disjoint implythat

��A \

�[nkD1Fk

��D ��

�A \

�[nkD1Fk

�\ Fn

�C ��

�A \

�[nkD1Fk

�\ F cn

�D �� .A \ Fn/C �

��A \

�[n�1kD1Fk

��D �� .A \ Fn/C

n�1XkD1

�� .A \ Fk/

D

nXkD1

�� .A \ Fk/ :

Having previously verified that finite unions of measurable sets are measur-able, we see that

�� .A/ D ��A \

�[nkD1Fk

��C ��

A \

�[nkD1Fk

�c��

�A \

�[nkD1Fk

��C ��

A \

�[1kD1Fk

�c�D

nXkD1

�� .A \ Fk/C �� A \

�[1kD1Fk

�c�.

Because this inequality holds for all n, we can use subadditivity to concludethat

�� .A/ �

1XkD1

�� .A \ Fk/C �� A \

�[1kD1Fk

�c��

�[1kD1 .A \ Fk/

�C ��

A \

�[1kD1Fk

�c�D ��

�A \

�[1kD1Fk

��C ��

A \

�[1kD1Fk

�c�.

5.5. Measurable sets III 103

The inequality in the other direction is a consequence of subadditivity.We conclude that the union of countably many disjoint, measurable sets ismeasurable.

What happens when the sets are not disjoint? Suppose that we have a se-quence fEkg of measurable sets. Set Fn D En \

�[n�1kD1

Ek�c

. Then fFkgis a sequence of disjoint, measurable sets with [kEk D [kFk . Since[kFk is measurable, so is [kEk : Any countable union of measurable sets ismeasurable.

The collection of measurable sets is a sigma algebra!

5.5 Measurable sets III

At this point, we know that .�1; a/ is a measurable set and that the collec-tion of measurable sets is a sigma algebra. Given any collection S of subsetsof R, there is a smallest11 sigma algebra containing S . This sigma alge-bra is called the sigma algebra generated by S . (See exercise 45.) Sincewe know that all intervals of the form .�1; a/ are measurable, we knowthat the sigma algebra M of measurable sets must contain the sigma algebragenerated by f.�1; a/ W a 2 Rg. This sigma algebra is important enough tomerit its own name.

Definition 10 (Borel sets). The Borel sigma algebra, B, is the sigma alge-bra generated by f.�1; a/ W a 2 Rg. The elements of B are called Borelsets.

There are many other options for a collection of sets that will generate B:finite open intervals, half-open intervals, open sets, compact sets, and more.(See exercise 46.)

We know that B M. Is B DM? No, but as we shall see, the two sigmaalgebras are almost equal. The key to this investigation is an alternative de-scription of measurable sets. Loosely speaking, theorem 22 (below) tells usthat measurable sets are those sets that can be “closely approximated” byopen or closed sets.12

11 Smallest in this context means contained in any other sigma algebra satisfying the conditionin question.

12 This theorem also connects Caratheodory’s and Lebesgue’s senses of measurability. SeeExercises 71 and following.

Theorem 22 (Measurability). Let E R. The following are equivalent.

1. E is Lebesgue measurable in the sense of Caratheodory.

2. Given " > 0; there is an open setG such that E G and �� .GnE/ <".13

3. Given " > 0; there is an closed set F such that F E and�� .EnF / < ".

Proof. Let E be a measurable set. First suppose that � .E/ < 1 and let" > 0. Then there is a countable cover of E by open intervals fIkg such that

� .E/ � � .[kIk/ �Xk

l .Ik/ < � .E/C ".

If we setG D [kIk , thenE G and � .G/ < � .E/C". The measurabilityof E now implies that

� .E/C " > � .G/ D � .G \E/C � .GnE/ D � .E/C � .GnE/ .

Statement (2) follows by subtraction.If � .E/ D1, then apply the same argument toEk D E\ Œ�k; k� to find

open sets Gk containing Ek such that �� .GknEk/ <"

2k. Then G D [kGk

is an open set containing E with �� .GnE/ < ".To prove that (2) implies (1), suppose that E is a subset of R and that for

any " > 0 there is an open set G containing E with �� .GnE/ < ". Notethat Gc Ec and that G, being an open set, is measurable (exercise 37).We will use the measurability of G to establish the measurability of E.

Let A be an arbitrary subset of R. We need to show that �� .A/ �� .A \E/C �� .A \Ec/. Since G is measurable,

�� .A \Ec/ D �� ..A \Ec/ \G/C �� ..A \Ec/ \Gc/

D �� .A \ .GnE//C �� .A \Gc/

� �� .GnE/C �� .A \Gc/

< "C �� .A \Gc/ .

Again appealing to the measurability of G,

�� .A/ D �� .A \G/C �� .A \Gc/

> �� .A \E/C �� .A \Ec/ � ":

13 Recall that for sets A and B , AnB D A\Bc .

5.6. Measurable functions 105

Since " > 0 is arbitrary, we conclude that ��.A/ � ��.A \ E/C

��.A \Ec/ so that E is measurable.Having established that statements (1) and (2) are equivalent, the equiva-

lence of (1) and (3) follows by taking complements.

By definition, the outer measure of any set E is approximated byPk l .Ik/ where fIkg is a countable cover of E by open intervals. In this

case, the difference between the outer measures of [kIk and E is arbitrarilysmall. If you look at the proof of theorem 22, you will see that E is measur-able exactly when the outer measure of the difference between the sets[kIkand E is also small. (Specifically, �� .[kIk/ � �� .E/ being small shouldimply that �� .[kIknE/ is also small.) Thinking about this another way,the critical sets on which Caratheodory’s condition must hold are unions ofopen intervals (open sets) that barely contain E. A similar comment appliesto closed sets barely contained in E. If we take this idea to the limit, we ar-rive at the following theorem that tells us that Lebesgue measurable sets arealmost Borel sets.

Theorem 23. A subset E of R is Lebesgue measurable if and only if E canbe written as E D B [ Z where B is a Borel set and Z is a set of measurezero.

Proof. Exercise 44:

5.6 Measurable functions

Think back to what triggered this line of investigation. Given a functionf W Œa; b� ! .˛; ˇ/ and a partition P DfŒyk�1; yk �gnkD1 of Œ˛; ˇ�, weset Ek D f �1 ..yk�1; yk �/, k D 1; 2; : : : ; n, and compute the upper andlower Lebesgue sums SL .f;P/ D

PnkD1 yk�1 � m.Ek/ and SL .f;P/ DPn

kD1 yk �m.Ek/wherem.A/ is the measure of the setA. At first, this pro-cess always seems to produce a well-defined value for the integral. A closerlook reveals problems with the notion of the measure of an arbitrary set E.

We have essentially solved the measurement problem by using Lebesgue(outer) measure on measurable sets. This leaves us with the task of applyingthe ideas of measurement in the context of a function. The critical conditionrequired to make our computations work is that Ek D f �1 ..yk�1; yk�/ bemeasurable. This leads us to the definition of a measurable function.

Definition 11 (Measurable function). Let X be a measurable subset14 ofreal numbers and let f W X ! R. Then f is measurable if the preimageunder f of every interval is measurable.

While this definition is straightforward enough, the variety of intervalforms makes it slightly awkward to work with. Even if we were to restrictthe definition to consider only preimages of the form f �1 ..a; b�/, the com-putations would be needlessly complicated. Sigma algebras prove useful insimplifying this situation. Since the collection of measurable sets is a sigmaalgebra and preimages interact well with complements and unions, it suf-fices to verify that preimages of the form f �1 ..�1; a// are measurablesets. When it is more convenient, we can verify instead that all preimages ofthe form f �1 ..�1; a�/ or f �1 ..a;C1// are measurable sets. (See exer-cises 49 and 59.)

Example 26. Let f .x/ D jxj. Then

f �1 ..�1; a// D

�;; a � 0

.�a; a/ ; 0 < a:

Since these preimages are always measurable sets, f is a measurablefunction.

Notice how much more awkward our verification would be if we wererequired to explicitly verify that the preimage of every possible type of in-terval is measurable. Also note the usefulness of knowing that the collectionof measurable sets is a sigma algebra in the next two examples.

Example 27 (Increasing functions are measurable). SupposeX is a mea-surable set and that f W X ! R is an increasing function. Let ˛ Dsup fx 2 X W f .x/ < ag. (It may be that ˛ D �1 in which case we under-stand .�1;�1/ to be the empty set.) Then f �1 ..�1; a// D .�1; ˛/\Xor f �1 ..�1; a// D .�1; ˛�\X: In either case, the preimage of .�1; a/is the intersection of two measurable sets and so is measurable.

Example 28 (Continuous functions are measurable). Suppose X is ameasurable set and that f W X ! R is continuous. Since .�1; ˛/ is open,its preimage under the continuous function f is relatively open in X . Inother words, f �1 ..�1; a// is the intersection of X with an open set. Since

14 It is not strictly necessary to specify thatX is measurable sinceX D f �1 ..�1;C1//.We include the restriction to remind ourselves that a measurable function will no longer bemeasurable if we restrict its domain to a non-measurable subset.

5.6. Measurable functions 107

open sets are measurable, f �1 ..�1; a// is a measurable set. Thus f is ameasurable function.

In a manner similar to the way that measurable sets are closed under com-plements and unions (and so form a sigma algebra), measurable functionsare closed under addition and scalar multiplication (and so form an vectorspace).

Theorem 24 (Measurable functions form vector space). Suppose that fand g are measurable functions defined on Œa; b� and that c is a real number.Then cf , f C g; max ff; gg, and min ff; gg are also measurable.

Proof. The proofs that cf , max ff; gg, and min ff; gg are measurable areleft as exercises.

To show that f C g is measurable, suppose that .f C g/ .x/ < c. Letk D c � .f C g/ .x/ so that g .x/ D c � k � f .x/. Since k > 0, wecan choose a rational number r so that f .x/ < r < f .x/ C k. Theng .x/ D c � k � f .x/ < c � r .

We have just shown that if .f C g/ .x/ < c, there is a rational number rsuch that f .x/ < r and g .x/ < c � r . Hence

.f C g/�1 ..�1; c// [r2Q�f �1 ..�1; r// \ g�1 ..�1; c � r//

�:

The reverse containment is straightforward so, in fact, the two sets are equal.Since f and g are measurable and the collection of measurable sets is asigma algebra, .f C g/�1 .�1; c/, a countable union of measurable sets, ismeasurable. Thus f C g is a measurable function.

The class of measurable functions has an important property that sets itapart from other classes of functions such as continuous and Riemann in-tegrable functions: the set of measurable functions is closed under takinglimits.

Theorem 25 (Limits of measurable functions are measurable). Let ffkgbe a sequence of measurable functions that converges pointwise to the func-tion f . Then f is measurable.

Proof. Let gn .x/ D supk�n fk .x/. Then

g�1n ..�1; a�/ D \k�nf�1k ..�1; a�/

which is a measurable set. Thus all of the fgng are measurable. Now seth .x/ D infn gn .x/. Then

h�1 ..�1; a// D [ng�1n ..�1; a//


so h is measurable. Since ffkg converges to f ,

h .x/ D infn

supk�n

fk .x/ D limkfk .x/ D lim

kfk .x/ D f .x/ ,

establishing the measurability of f .

Note that in the process of proving the previous theorem, we have also ver-ified that for any sequence ffkg of measurable functions, supk fk , infk fk ,and limkfk are measurable when they are finite. Similarly, when finite,limkfk is measurable.

In theorem 20 (page 63), we proved that a bounded function is Riemannintegrable if and only if it is continuous except on a set of measure zero.Because this type of condition is so ubiquitous in the context of the Lebesgueintegral, we introduce a definition to capture the idea.

Definition 12 (Almost everywhere). A property is said to hold almosteverywhere (a.e.) if the property holds except on a set of measure zero.

Theorem 26. If f is a measurable function and g D f a.e., then g ismeasurable.

Proof. Let c 2 R, D D fx W f .x/ ¤ g .x/g ; and E D fx W f .x/ Dg .x/g. By hypothesis, �� .D/ D 0 so both D and E are measurable: Now

f �1 ..�1; c// \E D g�1 ..�1; c// \E

g�1 ..�1; c//

�g�1 ..�1; c// \E

�[D

D�f �1 ..�1; c// \E

�[D:

Thus g�1 ..�1; c// D�f �1 ..�1; c// \E

�[ Y where Y D. As a set

with zero outer measure, Y is measurable and since f �1 ..�1; c//, E, andY are all measurable, so is g�1 ..�1; c//. We conclude that g is a measur-able function.

Corollary 27 (of theorem 25). Let ffkg be a sequence of measurable func-tions that converges almost everywhere to the function f . Then f ismeasurable.

Proof. Exercise 58.

This type of situation is typical. Behavior on a set of measure zero usuallycan be ignored.

5.7. Exercises 109

5.7 Exercises

5.0 Another approach: filling the gapsFor the following three exercises assume that f is a function defined onŒa; b� and that ˛ < f < ˇ. Also assume that the measures of all relevant setsare well defined.

1. Let P DfŒyk�1; yk �gnkD1 be a partition of Œ˛; ˇ� and set Ek Df �1 ..yk�1; yk�/.

(a) Prove that the Ek are disjoint sets.(b) Prove that [n

kD1Ek D Œa; b�.

2. Let P1 and P2 be partitions of Œ˛; ˇ� with P2 a refinement of P1. Provethat SL .f;P1/ � SL .f;P2/ � SL .f;P2/ � SL .f;P1/. (First assumethat P2 adds a single new division point to P1 and compute SL .f;P2/�SL .f;P1/.)

3. Let P1 and P2 be partitions of Œ˛; ˇ�. Use exercise 2 to prove thatSL .f;P1/ � SL .f;P2/.

5.0 Another approach: deeper reflections4. Half-open subintervals are used in the definition of the upper and lower

Lebesgue sums instead of the closed intervals that are used for theRiemann integral.

(a) Why?(b) Could we have bounded f on Œa; b� by ˛ � f � ˇ? Explain.

5. If f is a function defined on Œa; b� with ˛ < f < ˇ, then there areinfinitely many different choices for ˛ and ˇ. The set of partitions ofŒ˛; ˇ� used to define the Lebesgue integral will be different if we changethe values of ˛ or ˇ. Explain why the choice of ˛ and ˇ does notmatter.

5.1 Measurable sets I: filling the gaps6. Prove that the rationally-equivalent relation (two numbers are rationally

equivalent if their difference is rational) is an equivalence relation. Inother words, verify that the relation is

(a) Reflexive. Any real number x is rationally equivalent to itself.(b) Symmetric. If x is rationally equivalent to y, then y is rationally

equivalent to x.


(c) Transitive. If x is rationally equivalent to y and y is rationallyequivalent to z, then x is rationally equivalent to z.

7. Given an equivalence relation on R, an equivalence class is the set of allelements that are related to some fixed element x. Prove that the set ofequivalence classes forms a partition of R. In other words, verify that

(a) If A and B are equivalence classes with A ¤ B , then A \ B D ¿.(Use contraposition.)

(b) Every element of R is in some equivalence class.

5.2 Outer measure: filling the gaps8. Prove that �� is monotone. In other words, verify that if A and B are

sets with A � B then �� .A/ � �� .B/.

9. Prove that the outer measure of any interval is its length. (Use mono-tonicity. Don’t forget infinite intervals.)

10. Prove that �� is translation invariant. In other words, show that if B DAC t D faC t W a 2 Ag then �� .B/ D �� .A/.

11. If �� .Ak/ is infinite for some k, why is �� .[kAk/ �Pk �� .Ak/?

12. Explain why we can conclude that �� .[kAk/ �Pk �� .Ak/ from the

knowledge that �� .[kAk/ �Pk �� .Ak/C " for any " > 0.

5.2 Outer measure: deeper reflections13. Compute the outer measures of the following sets.

(a)˚1nW n 2 N

�:

(b) The rational numbers in Œ0; 1� :(c) The irrational numbers in Œ0; 1� : (Use your work in (b) and the mea-

sure of Œ0; 1� to find a lower bound on the measure of the irrationalnumbers in Œ0; 1�.)

5.3 Measurable sets II: filling the gaps14. Prove that the following statements are equivalent for any subset E

of R.

(a) If A and B are disjoint sets with E D A [ B , then �� .E/ D�� .A/C �� .B/.

(b) For any subset C of R, �� .E/ D �� .E \ C/C �� .E \ C c/.

15. Suppose that I is a finite interval. Explain whyl.I / D l.I \ .�1; a//C l.I \ Œa;C1//.

5.7. Exercises 111

16. Given a sequence fEkg of sets, why is ��[nkD1

Ek��

�[1kD1

Ek�?

5.3 Measurable sets II: deeper reflections17. Prove directly from the definition of measurability that every countable

set is Lebesgue measurable.

18. Prove that any set with outer measure zero is Lebesgue measurable.

19. Prove that Œ0; 1� nQ is measurable and find its measure.

20. Explain why the translation of any measurable set is again measurable.(Use exercise 10.)

21. Prove that �� .Œa; b� [ Œc; d �/ D �� .Œa; b�/ C �� .Œc; d �/ when Œa; b�and Œc; d � are disjoint.

22. Suppose that A and B are sets satisfying A .�1; a/ and B .a;1/for some real number a. Prove that �� .A [ B/ D �� .A/C �� .B/.

23. The set V constructed in Vitali’s example is not measurable accordingto Caratheodory’s definition. Explain how we know this.

24. An alternative approach to the problem identified by Vitali’s examplewould be to give up countable additivity and require only finite additiv-ity. The hope would be that all subsets of R might be measurable in thisless demanding sense. Explain why this hope is doomed.

5.4 Sigma algebras: filling the gaps.

25. Prove that R and ¿ are measurable.

26. Prove that if E is a measurable set, so is Ec .

27. Assuming that�� .E [ F / D �� .E/C�� .F / for disjoint, measurablesets E and F , prove that ��

�[nkD1

Ek�DPnkD1 �

� .Ek/ for any finiteset fEkg

nkD1 of disjoint, measurable sets.

28. Use complements to prove that sigma algebras are closed under count-able intersections. (Use De Morgan’s Law.)

29. Verify that .A \E/ [ ..A \Ec/ \ F / D A \ .E [ F /.

30. Assuming that the union of two measurable sets is measurable, use in-duction to prove that the finite union of measurable sets is measurable.

31. Explain why A\�[nkD1

Fk�\Fn D A\Fn and A\

�[nkD1

Fk�\F cn D

A \�[n�1kD1

Fk�

when the sets fFkg are disjoint.

32. Suppose that fEkg is a countable collection of measurable sets anddefine Fn D En \

�[n�1kD1

Ek�c

.

(a) Verify that fFkg is a collection of disjoint sets.(b) Why is Fk measurable?(c) Prove that [kEk D [kFk .

33. Why is �� A \

�[nkD1

Fk�c��

A \

�[1kD1

Fk�c�‹

34. Show that [1kD1

.A \ Fk/ D A \�[1kD1

Fk�.

5.4 Sigma algebras: deeper reflections35. Give two proofs that .�1; a� is measurable.

(a) Use Caratheodory’s definition of measurability.(b) Use sigma algebras.

36. Give two proofs that Œa; b� is measurable.

(a) Use Caratheodory’s definition of measurability.(b) Use sigma algebras.

37. Prove that every open set G and every closed set F are measurable.(Show that G is the countable union of open intervals.)

38. Let X D f1; 2; 3g.

(a) What is the smallest possible sigma algebra based on X?(b) What is the largest possible sigma algebra based on X?

39. Let X D f1; 2; 3; 4g and let S D ff1g ; f1; 2gg. What other sets must beadded to S to form a sigma algebra on X?

40. Suppose that fA˛g is a collection of sigma algebras on a set X . (We donot assume that the index set is countable.) Prove that \˛A˛ is a sigmaalgebra.

41. Let f be an arbitrary real-valued function defined on a measurable setX . Prove that the collection of subsets of R whose inverse images underf are measurable forms a sigma algebra.

5.5 Measurable sets III: filling the gaps42. Fill in the details of the first part of the proof of theorem 22 (page 104)

when � .E/ D 1. Specifically, prove that if two sequences of setsfAkg

1kD1 and fBkg

1kD1 satisfy Ak Bk and �� .BknAk/ <

"

2k, then

[kAk [kBk and �� ..[kBk/ n .[kAk// < ".

5.7. Exercises 113

43. Using the equivalence of statements (1) and (2), complete the proof theproof of theorem 22 (page 104).

(a) Prove that ifE is Lebesgue measurable in the sense of Caratheodorythen given " > 0 there is a closed set F such that F E and�� .EnF / < ". (If E is measurable, so is Ec .)

(b) Prove that if for any " > 0 there is a closed set F such that F Eand �� .EnF / < "; then E is Lebesgue measurable in the sense ofCaratheodory.

44. Let E � R.

(a) Prove that ifE D B[Z whereB 2 B andZ has measure zero, thenE 2M, the set of Lebesgue measurable sets. (Use exercise 18.)

(b) Prove that if E is Lebesgue measurable, then there are sets B and Zwhere B 2 B and Z has measure zero such that E D B [Z. (Startwith closed sets Fn satisfying Fn E and �� .EnFn/ < 1

n.)

5.5 Measurable sets III: deeper reflections45. In the definition of the Borel sigma algebra, we took for granted that

there is a smallest sigma algebra containing f.�1; a/ W a 2 Rg. Verifythat this is so. Let S be a collection of subsets of R. Prove that there is asmallest sigma algebra containing S .

(a) Why is there a sigma algebra containing S?Let S be the intersection of all the sigma algebras containing S .

(b) Explain why S is a sigma algebra containing S .(c) Explain why S is contained in any other sigma algebra con-

taining S .

46. Prove that the following collections of subsets of R generate the samesigma algebra. In other words, show that any sigma algebra containingone type of sets contains the other types of sets.

(a) The set of all intervals of the form .�1; a/.(b) The set of all intervals of the form .a; b/ :

(c) The set of all open sets.(d) The set of all compact sets.

47. Prove that if fAkg is a sequence of measurable sets for which��[1kD1

Ak�D m > 0, then for any " > 0 there is an N so that

��[NkD1

Ak�> m � ". (Rework the sets to be disjoint and use count-

able additivity.)

48. Prove that if fAkg is a sequence of measurable sets for which��\1kD1

Ak�D 0, then for any " > 0 there is an N so that

��\NkD1

Ak�< ". (One approach is to use the previous exercise and

relative complements. Alternatively, find a way to use countable addi-tivity.)

5.6 Measurable functions: filling the gaps49. Suppose that X is a measurable set, f W X ! R, and f �1 ..�1; a//

is measurable for all a 2 R. Use exercise 59 to prove that the followingsets are measurable for all values a; b 2 R.

(a) f �1 .Œa;C1//(b) f �1 ..�1; a�/(c) f �1 .Œa; b//

50. Suppose that f W X ! R is an increasing function and that ˛ Dsup fx 2 X W f .x/ < ag. Prove that .�1; ˛/\X f �1 ..�1; a// .�1; ˛� \X .

51. Use the "–ı definition of continuity to prove that if f W A ! R iscontinuous andU is an open set of real numbers, then f �1 .U / D V \Afor some open subset V of R.

52. Suppose that f is a measurable function and that c is a real number.Show that cf is measurable.

53. Show that

[r2Q�f �1 ..�1; r// \ g�1 ..�1; c � r//

� .f C g/�1 .�1; c/ :

54. Suppose that f; g W Œa; b� ! R are measurable functions. Prove thatmax ff; gg and min ff; gg are also measurable.

55. Let gn .x/ D supk�n fk .x/. Explain why

g�1n ..�1; a�/ D \k�nf�1k ..�1; a�/ :

56. Let gn .x/ D infk�n fk .x/. Explain why

g�1n ..�1; a// D [k�nf�1k ..�1; a// :

57. In the proof of theorem 26 (page 108), why is E measurable?

58. Prove Corollary 27 (page 108). (Define new functions gk and g so thatgk D fk a.e., g D f a.e., and fgkg converges pointwise to g every-where.)

5.7. Exercises 115

5.6 Measurable functions: deeper reflections59. Let X R and suppose that f W X ! R.

(a) Prove that f �1 .Ac/ D Xnf �1 .A/ for any A R.(b) Prove that f �1 .[˛A˛/ D [˛f �1 .A˛/ for any (not necessarily

countable) collection of subsets fA˛g of R.(c) Prove that f �1 .\˛A˛/ D \˛f �1 .A˛/ for any collection of sub-

sets fA˛g of R.

60. Use the definition of measurable function to prove the following func-tions are measurable.

(a) f W Œ0; 1�! R where f .x/ D 2x C 1.(b) f W R! R where f .x/ D x2.(c) The Dirichlet function d .(d) f W Œ0; 2��! R where f .x/ D sin x.(e)

f .x/ D

8<:�2 x 2 .�1;�1/

x2 � 1 x 2 Œ�1; 2�

3 x 2 .2;C1/ :

61. Give an example of a nonmeasurable function.

62. Give two examples of increasing functions f; g W R ! R such thatf �1 ..�1; 1// D .�1; ˛/ and g�1 ..�1; 1// D .�1; ˇ�, where ˛ Dsup fx 2 X W f .x/ < 1g and ˇ D sup fx 2 X W g .x/ < 1g.

63. Suppose that f is a measurable function.

(a) Prove that jf j is measurable.(b) Prove that f C D 1

2.jf j C f / and f � D 1

2.jf j � f / are both

measurable.(c) Give alternative definitions of f C and f � as piecewise defined

functions.(d) Write f in terms of f C and f �.

64. Suppose that f and g are positive measurable functions with a commondomain. Prove that the product fg is measurable. (Mimic the proof forf C g.)

65. Suppose that f and g are (not necessarily positive) measurable functionswith a common domain. Prove that the product fg is measurable. (Usethe previous two exercises.)

66. Prove that any function f W Œa; b�! R that is continuous a.e. (the set ofdiscontinuities has measure zero) is a measurable function.

67. Give an example of a function that is continuous a.e. but is not equal a.e.to any continuous function.

68. Give an example of a function that is equal a.e. to a continuous functionbut is not itself continuous a.e.

69. Suppose the sequence fakg converges. Explain why supk�n ak andinfk�n ak both exist and are finite.

70. Suppose that ffkg is a sequence of measurable functions which is point-wise bounded. In other words, ffk .x/g is a bounded sequence for anyfixed x. Prove that infk fk and limkfk are measurable.

5.7 Related ideas: an alternative approach to measurabilityLebesgue used an alternate definition of measurability based on bothouter and inner measure. The following exercises outline the basic con-tours of this approach.

The inner measure of a subset A of R is

�� .A/ D sup f�� .F / W F A with F compactg :

71. Prove that �� .A/ � �� .A/ for every subset A of R.

72. Prove that �� .I / D l .I / for any interval I .

A subset A of R with �� .A/ < 1 is measurable (in the sense ofLebesgue) if �� .A/ D �� .A/. In this case we write � .A/ D �� .A/ D�� .A/. If �� .A/ D 1, we say that A is measurable (in the sense ofLebesgue) if A \ Œ�n; n� is measurable for each n 2 N.

73. Prove that every set with zero outer measure is measurable in the senseof Lebesgue.

74. Prove that every compact set is measurable in the sense of Lebesgue.

75. Prove that if˚Kj�

is a finite set of disjoint, compact sets, then��[jKj

�DPj ��Kj�. (Use the fact that disjoint, compact sets

must have a positive distance between them.)

76. Use exercise 75 to prove that if fAig is a countable set of disjoint mea-surable sets then [iAi is measurable and � .[iAi / D

Pi � .Ai /. (First

5.8. References 117

assume that A Œ�n; n�. Show that �� .[iAi / �Pj ��Kj�

for anappropriately chosen finite collection of disjoint, compact sets.)

77. Prove that the collection of sets that are measurable in the sense ofLebesgue is a sigma algebra.

78. Use exercises 73 and 77 to prove that the collection of sets that are mea-surable in the sense of Lebesgue includes all the sets that are measurablein the sense of Caratheodory:

79. Use theorem 22 to prove that the collection of sets that are measurablein the sense of Caratheodory includes all of the sets that are measurablein the sense of Lebesgue.

5.8 ReferencesBartle, R.G. (1966). The Elements of Integration. John Wiley & Sons.Bressoud, D.M. (2006). A Radical Approach to Real Analysis (2nd ed.).


America.Hewitt, E. and K. Stromberg (1975). Real and Abstract Analysis. Springer.Kharazishvili, A.B. (2004). Nonmeasurable Sets and Functions. North-

Holland Mathematics Studies, 195. Elsevier Science B.V.Taylor, A.E. (2010). General Theory of Functions and Integration. Dover.Vitali, G. (1908). Sui gruppi di punti e sulle funzioni di variabili reali,

Atti dell’Accademia delle Scienze di Torino (in Italian) 43: 229–246.archive.org/stream/attidellarealeac43real#page/229.

CHAPTER 6

The Lebesgue Integral

The theory of measurable sets and functions provides all the supportingstructure required to make a formal definition of the Lebesgue integral. Withsome modest modification, the approach suggested at the beginning of thelast chapter (page 92) is ready for development.

Definition 13 (Lebesgue integral). Let f be a (Lebesgue) measurablefunction on Œa; b� satisfying ˛ < f � ˇ < 1. Let P DfŒyk�1; yk �gnkD1be a partition of Œ˛; ˇ� and set Ek D f �1 ..yk�1; yk �/, k D 1; 2; : : : ; n.The (lower) Lebesgue sum of f with respect to P is

SL .f;P/ DnXkD1

yk�1 � � .Ek/

where � .A/ is the Lebesgue measure of the set A. The Lebesgueintegral of f over Œa; b� is

LZ b

a

f D supPSL .f;P/

where the supremum is taken over all partitions of Œ˛; ˇ�.1

This definition may remind you of the definition of the Darboux integralwith yk�1 playing the role of infIk f . This time, however, there is no needfor upper sums or upper integrals–the Lebesgue integral always exists forbounded, measurable functions.

6.1 VariationsThe resemblance of the definition of the Lebesgue integral to that of theDarboux integral suggests a possible modification of the definition. In fact,

1 While SL .f;P/ DPnkD1 yk � � .Ek/ does not appear in the definition of the Lebesgue

integral, it is occasionally used in proofs.

119

120 CHAPTER 6. The Lebesgue Integral

there are several variations on Lebesgue’s approach to integration. We willconsider two of them before investigating the properties of the Lebesgueintegral.

6.1.1 A Darboux-like variationWilliam H. Young (1863–1942) developed an integral based on the obser-vation that the sets Ek D f �1 ..yk�1; yk�/ partition the domain Œa; b� intodisjoint measurable sets. We capture this idea in the concept of a measurablepartition.

Definition 14 (Measurable partition). A measurable partition of a set Sis a finite collection of disjoint, measurable subsets of S whose union is S .

Since intervals are measurable sets, measurable partitions generalizeRiemann-Darboux partitions. But we need to exercise some care here since,unlike in a Riemann partition, the sets in a measurable partition must be dis-joint. While the intervals in a Riemann-Darboux partition are not disjoint,they are non-overlapping in that a pair of intervals can have at most a sin-gle point in common. Moving from non-overlapping to disjoint sets is not aserious obstacle since nothing of significance would have changed had weused partitions of the form fŒx0; x1/ ; Œx1; x2/ ; Œx2; x3/ ; : : : ; Œxn�1; xn�g forthe Riemann integral. With these observations in mind, we drop the restric-tion that f is measurable and make the following definition.

Definition 15 (Lebesgue-Young integral). Let f be a bounded function de-fined on Œa; b�. Let P DfEkgnkD1 be a measurable partition of Œa; b�. Thelower and upper Lebesgue-Young sums of f over P are

SL-Y .f;P/ DXP

infEkf � � .Ek/

andSL-Y .f;P/ D

XP

supEk

f � � .Ek/ .

The lower and upper Lebesgue-Young integrals of f over Œa; b� are

L-YZ b

a

f D supP

SL-Y .f;P/

andL-Y

Z b

a

f D infPSL-Y .f;P/

where the supremum and infimum are taken over the set of all measurablepartitions of Œa; b�.

6.1. Variations 121

If L-YR ba f D

L-YR ba f , then f is Lebesgue-Young integrable over Œa; b�,

and the Lebesgue-Young integral of f over Œa; b� is

L-YZ b

a

f D L-YZ b

a

f .

In this definition we have replaced the restriction that f be measurablewith the requirement that the partitions be measurable.

It is important to understand the difference between the partitions used bythe Lebesgue and Lebesgue-Young integrals. The Lebesgue integral parti-tions the range of f and uses this partition to induce a measurable partitionof the domain. The Lebesgue-Young integral considers an arbitrary measur-able partition of f ’s domain. Of course, when f is measurable, the inducedpartitions formed by the sets Ek D f �1 ..yk�1; yk �/ are measurable so thatthe Lebesgue partitions of the range generate a subcollection of the possi-ble measurable partitions of the domain. As we saw in the introduction tothe previous chapter, when f is bounded and measurable, the gap betweenthe upper and lower sums can be made arbitrarily small using partitionsof the form f �1 ..yk�1; yk �/. Consequently, Lebesgue integrable functionsare Lebesgue-Young integrable.

6.1.2 A simple variationA second variation proceeds from the observation that Lebesgue lower sumscorrespond to functions that take on the value yk�1 on the set Ek . A moreformal way to express this function is as

Pk yk�1 � 1Ekwhere 1A is the

function that is 1 for x in A and is zero elsewhere. In other words,

1A.x/ D

�1; x 2 A

0; x 62 A:

The function 1A is known as the characteristic function of the set A.2

The Dirichlet function, d , is an example of a characteristic function whered D 1Q. Functions of the form � D

PnkD1 ˛k �1Ek where ˛k 2 R are called

simple functions. Alternatively, one can define a simple function as one thattakes on only a finite set of values f˛kg

nkD1. In this case, � D

PnkD1 ˛k �1Ek

where Ek D ��1 .f˛kg/.

2 Some texts use the notation �A instead of 1A.

0.0 0.5 1.00

1

0.0 0.5 1.00

1

2

13

ϕ (x) = 1 34, 1

4

32 · 1 · 1+ 1

234, 1

278,

ψ (x) =

Figure 6.1. Graphs of simple functions

Example 29. Figure 6.1 displays the graphs of the simple functions � D1Œ 13 ;

34 �

and D 32� 1Œ 14 ;

34 �C 1

2� 1Œ 12 ;

78 �

. By considering its distinct values,

the second function can also be expressed as D 32� 1Œ 14 ;

12 /C 2 � 1Œ 12 ;

34 �C

12� 1. 34 ;

78 �

.

Now any respectable definition of an integral for real-valued functionsought to satisfy the basic properties

1.R ba 1A D � .Œa; b� \ A/ when A is a measurable set,

2. Linearity.R ba

�PnkD1 ˛kfk

�DPnkD1 ˛k

R ba fk provided both sides

are defined, and

3. Monotonicity. If f � g on Œa; b� thenR baf �

R bag.

Combining the first two properties tells us that the integral of any measur-able simple function � D

PnkD1 ˛k � 1Ek over Œa; b� should beZ b

a

� D

nXkD1

˛k � � .Œa; b� \Ek/ :

If P DfŒyk�1; yk�gk is a partition of the range of a bounded function f andEk D f

�1 ..yk�1; yk �/, thenXk

yk�1 � 1Ek � f <Xk

yk � 1Ek .

Monotonicity then implies thatZ b

a

Xk

yk�1 � 1Ek

!�

Z b

a

f <

Z b

a

Xk

yk � 1Ek

!.

6.1. Variations 123

These observations are used to extend the definition to more general func-tions using limits.

Definition 16 (Simple-Lebesgue integral). The simple-Lebesgue integralis a type of Daniell integral.3

1. The simple-Lebesgue integral of a measurable simple function � DPnkD1 ˛k � 1Ek is

s-LZ b

a

� D

nXkD1

˛k � � .Œa; b� \Ek/ :

2. For any nonnegative function f , let f�kg be a sequence of positive,measurable simple functions monotonically increasing to f . Then

s-LZ b

a

f D limk

s-LZ b

a

�k :4

3. For arbitrary functions, if the integral of at least one of f C D 12.jf jC

f / or f � D 12.jf j � f / is finite, then

s-LZ b

a

f D s-LZ b

a

f C � s-LZ b

a

f �:

The function f is said to be simple-Lebesgue integrable if s-LR ba jf j <

C1 or, equivalently, if the integrals of both f C and f � are finite.

Using carefully constructed simple functions of the formPk yk�1 � 1Ek

where Ek D f �1 ..yk�1; yk �/, we can show that the Lebesgue and simple-Lebesgue integrals agree for bounded, measurable functions. In other words,Lebesgue integrable functions are simple-Lebesgue integrable and the inte-grals agree.

Notice that the definition of the simple-Lebesgue integral does not requiref to be bounded or measurable. There are, however, two critical issues thatmust be addressed.

3 The Daniell integral (Percy J. Daniell, 1918) begins with (1) a vector space F of elementaryfunctions which is also closed under taking absolute value and (2) a linear function I on Fthat is monotone (f � g implies I .f / � I .g/). In addition, I must satisfy the conditionthat if the monotone sequence ffng decreases to zero, so does fI .fn/g. Monotone increas-ing sequences of functions are then used to extend I to a larger set of functionsLC and thento the set L of functions f that can be expressed as f D f1 � f2 with f1; f2 2 LC.

4 Note that limk s-LR ba �k may be infinite.


1. Part (2) of the definition takes for granted that one can find an increasingsequence of measurable simple functions that converges to f . How canwe guarantee such a sequence exists?

2. Is the simple-Lebesgue integral well defined? In other words, if f�kgand f kg are two increasing sequences of measurable simple functionsthat converge to f; is limk

s-LR ba �k D limk

s-LR ba k?

We will deal with these issues near the end of the next section.

6.2 Reconciling the approaches

In addition to the Riemann-Darboux integral, we currently have three in-tegral definitions in play. How are these integrals related? Specifically, wewant to address the following questions.

1. We know that the Dirichlet function is Lebesgue integrable but notRiemann integrable, so not all Lebesgue integrable functions are Rie-mann integrable. Are there any Riemann integrable functions that arenot Lebesgue integrable?

2. We have sketched arguments that Lebesgue integrable functions areLebesgue-Young and simple-Lebesgue integrable and that all three in-tegrals agree for these functions. Are there functions that are Lebesgue-Young or simple-Lebesgue integrable but not Lebesgue integrable?

3. How are the Lebesgue-Young and simple-Lebesgue integrals related?

A few moments contemplation should lead you to the realization that thefirst two questions are really questions about the measurability of functions.Measurability is the fundamental requirement for a bounded function to beLebesgue integrable.

6.2.1 RiemannLet’s begin with the Riemann integral. By theorem 20 (page 63), the setof discontinuities of a Riemann integrable function has measure zero. Wewill show that all such functions are measurable and therefore Lebesgueintegrable.

Theorem 28 (Riemann H) Lebesgue). If f is Riemann integrable overŒa; b�, then f is also Lebesgue integrable over Œa; b� with L

R baf D R

R baf .

6.2. Reconciling the approaches 125

Proof. If f is Riemann integrable, then f is bounded and the set D of thediscontinuities of f has measure zero. Let C be the set of points in Œa; b�at which f is continuous and let g be the restriction of f to C . Let c 2R. Since g is continuous, g�1 ..�1; c// D U \ C for some open set U .Thus g�1 ..�1; c//, an intersection of measurable sets, is measurable forall values of c. Now

g�1 ..�1; c// f �1 ..�1; c// g�1 ..�1; c// [D;

so that f �1 ..�1; c// D g�1 ..�1; c//[Z for some set Z D: Since Zhas outer measure zero, Z is measurable. Thus f �1 ..�1; c//, the union ofmeasurable sets, is measurable. As f is both bounded and measurable, it isLebesgue integrable.

We will verify that LR ba f D

RR ba f shortly.

6.2.2 Lebesgue-YoungTo address the Lebesgue-Young integral, we make use of several propertiesof upper and lower Lebesgue-Young sums that are analogous to facts aboutupper and lower Darboux sums. Since measurable partitions are not deter-mined by the endpoints of intervals, we first need to adjust the concept of arefinement.

Definition 17 (Refinement). Let P and Q be measurable partitions ofŒa; b�. Then Q is a refinement of P if for each F 2 Q there is an E 2 Psuch that F E. Equivalently, everyE 2 P can be expressed as the disjointunion of sets in Q.

A standard way to create a common refinement of two measurable parti-tions P and Q is to use P [Q DfE \ F W E 2 P , F 2 Q, E \ F ¤ ;g.5

Example 30. P DfŒ0; 0:5� ; .0:5; 1�g and Q DfŒ0; 1� \Q; Œ0; 1� \Qcg aretwo measurable partitions of Œ0; 1�. In this case, P [Q consists of four sets:the rationals in Œ0; 0:5� ; the irrationals in Œ0; 0:5�, the rationals in .0:5; 1�; andthe irrationals in .0:5; 1�.

The proof of the following lemma is similar to that of the correspondingresult for Darboux integrals (lemma 13, page 56).

5 Note that there are two possible interpretations of P [Q depending on the context. If P andQ are partitions of the same interval, P [Q represents the common refinement introducedhere. If P and Q are partitions of non-overlapping intervals I and J , then P [Q is the usualunion of sets. In this case, P [Q is a partition of I [ J .


Lemma 29. Let f be a bounded function on Œa; b� and let P and Q bemeasurable partitions of Œa; b�. If Q is a refinement of P , then SL-Y .f;P/ �SL-Y .f;Q/ � SL-Y .f;Q/ � SL-Y .f;P/.

Proof. We will prove the first inequality and leave the other two as anexercise.

Suppose that P D fEig and Q D fFkg. Since Q is a refinement of P , wecan find an indexing scheme and a set of values fnig such that each Ei canbe expressed as the disjoint union

Ei D [niC1kDniC1

Fk :

Then for k D ni C 1; ni C 2; : : : ; niC1, Fk Ei so that infFk f � infEi f .Thus

SL-Y .f;P/ DXi

infEif � � .Ei /

DXi

infEif �

niC1XkDniC1

� .Fk/

�Xi

niC1XkDniC1

infFkf � � .Fk/

D SL-Y .f;Q/ :

Theorem 30 (Lebesgue = Lebesgue-Young). If f is Lebesgue-Young inte-grable over Œa; b�, then f is also Lebesgue integrable over Œa; b� and the twointegrals agree. (The converse has been established in previous discussion.)

Proof. Our first task is to verify that f is measurable.To any measurable partition P of Œa; b� there corresponds a pair of lower

and upper simple functions defined by �P DPE2P infE f � 1E and �P DP

E2P supE f � 1E . By construction,

�P � f � �P ;

LZ b

a

�P DXE2P

infEf � � .E/ D SL-Y .f;P/ ,

andLZ b

a

�P DXE2P

supE

f � � .E/ D SL-Y .f;P/ .

Since f is Lebesgue-Young integrable, we can find two sequences of par-titions fPng and fQng so that

SL-Y .f;Qn/ � SL-Y .f;Pn/ <1

n.

From these sequences, construct a sequence of successive refinements fRng

by setting Rn D [k�n .Pk [Qk/ and let f�ng and f ng be the sequencesof upper and lower simple functions defined by �n D �Rn and n D �Rn .Then

LZ b

a

.�n � n/ D SL-Y .f;Rn/ � SL-Y .f;Rn/

� SL-Y .f;Qn/ � SL-Y .f;Pn/ <1

n.

Moreover, because RnC1 is a refinement of Rn, f�ng and f ng are respec-tively decreasing and increasing sequences with n � f � �n. If we define� D limn �n and D limn n, then � and ; being limits of measurablefunctions, are measurable (theorem 25, page 107). We claim that, in addition,� D a.e.

Let D be the set of points in Œa; b� for which � ¤ . Then

D D fx 2 Œa; b� W � .x/ � .x/ > 0g

D[k

�x 2 Œa; b� W � .x/ � .x/ >

1

k

.

Set Dk D˚x 2 Œa; b� W � .x/ � .x/ > 1

k

�and Dk;n D fx 2 Œa; b� W

�n .x/ � n .x/ >1k

�. Then, for all n; k 2 N,

� .Dk/ � ��Dk;n

�D L

Z b

a

1Dk;n �LZ b

a

k .�n � n/ <k

n.

Hence � .Dk/ D 0 so

0 � � .D/ �Xk

� .Dk/ D 0

and � D a.e.But � f � �, so that � D f a.e. as well. Hence, by theorem 26 (page

108), f is measurable and, as a Lebesgue-Young integrable function, f isbounded. Consequently, f is Lebesgue integrable.

To show the Lebesgue and the Lebesgue-Young integrals agree, choose apartition P DfŒyk�1; yk�gnkD1 of an interval Œ˛; ˇ� such that ˛ < f < ˇ.Then P generates a measurable partition P � D fEkg D

˚f �1 ..yk�1; yk �/

�of Œa; b� for which yk�1 � infEk f � supEk f � yk . Hence

SL .f;P/ � SL-Y�f;P�

�� L-Y

Z b

a

f � SL-Y�f;P�

�� SL .f;P/ :

Since the Lebesgue integral also falls between SL .f;P/ and SL .f;P/ and

SL .f;P/ � SL .f;P/ � kPk .b � a/ ;

we conclude that

L-YZ b

a

f D LZ b

a

f:

This theorem allows us to take care of a piece of unfinished business fromtheorem 28 (page 124). By modifying Riemann-Darboux partitions of Œa; b�to use half open intervals, we can produce measurable partitions on whichthe lower/upper Darboux and Lebesgue-Young sums agree. Since the setof partitions by half open intervals is contained in the set of measurablepartitions, the properties of supremums and infimums on subsets guaranteethat

DZ b

a

f � L-YZ b

a

f � L-YZ b

a

f � DZ b

a

f:

Hence when f is Riemann integrable over Œa; b�,

RZ b

a

f D DZ b

a

f D L-YZ b

a

f D LZ b

a

f .

6.2.3 Simple-LebesgueTurning to the simple-Lebesgue integral, we first establish that the integralis well defined.

Theorem 31 (The simple-Lebesgue integral is well-defined). Let f be apositive function on Œa; b�. Suppose that f�kg and f kg are two increasingsequences of positive, measurable simple functions on Œa; b� converging tof: Then

limk

s-LZ b

a

�k D limk

s-LZ b

a

k :

Proof. Fix " > 0.First suppose that f < B . By Egoroff’s theorem (theorem 35, page 135 in

the next section), we can find measurable sets E and F with � .E/ < "=4Band � .F / < "=4B such that f�kg and f kg converge uniformly to f onŒa; b� nE and Œa; b� nF respectively. Let G D E [ F . Then � .G/ < "=2B

and both f�kg and f kg converge uniformly to f on Œa; b� nG.ChooseN so that both j�n .x/ � f .x/j and j n .x/ � f .x/j are less than

"=4 .b � a/ for all x 2 Œa; b� nG and n > N . Given n > N , j�n � nj is asimple function satisfying j�n .x/ � n .x/j < "

2.b�a/for all x 2 Œa; b� nG

and j�n .x/ � n .x/j 0 and n > N are arbitrary,

limn

ˇˇ s-LZ b

a

�n �s-LZ b

a

n

ˇˇ D 0

so

limk

s-LZ b

a

�k D limk

s-LZ b

a

k :

Now, remove the restriction that f is bounded and suppose that

L < limk

s-LZ b

a

�k :

Then we can find an integer n0 such that L < s-LR ba �n0 . Set B D

maxŒa;b� �n0 and define f � D min ff;Bg. Define ��k

and �k

analogously.Then

˚��k

�and

˚ �k

�are increasing sequences of positive, measurable sim-

ple functions that converge to the bounded function f �. So by our previouswork,

limk

s-LZ b

a

��k D limk

s-LZ b

a

�k :

Hence

limk

s-LZ b

a

k � limk

s-LZ b

a

�k

D limk

s-LZ b

a

��k

� s-LZ b

a

��n0

D s-LZ b

a

�n0 > L:

Since L < limks-LR ba �k was arbitrary, we conclude that

limk

s-LZ b

a

k � limk

s-LZ b

a

�k :

By symmetry,

limk

s-LZ b

a

�k D limk

s-LZ b

a

k

as claimed.

There are two strategies in the preceding proof that merit comment as theywill reappear repeatedly in the remainder of the chapter. The first strategy isthe Lebesgue integral’s analog to the bound-and-telescope move found in somany Riemann proofs. In the Lebesgue context, the analogous move is touse a general bound on a function on a set with very small measure coupledwith a very small bound on the function over a larger set. We will refer tothis strategy as divide and conquer. This strategy is employed in the previousproof when the integral over Œa; b� is split into the integrals over Œa; b� nG andG. The connection is best seen in theorem 19 (page 62) where the two typesof bound-and-telescope techniques are combined. The analogy is not perfectsince the bound-and-telescope technique is typically used to prove that afunction is Riemann integrable. We rarely prove integrability results for theLebesgue integral since bounded, measurable functions are automaticallyLebesgue integrable. The technique is more often used to prove the equalityof two integrals.

The second strategy is first to prove something under the condition that thefunction(s) is (are) bounded. The bounded case is then applied to a sequenceof truncated functions to obtain the general result. You will encounter thesetwo moves many times in the next two sections.

To round out the analysis of the simple-Lebesgue integral, we note that ifan increasing sequence of measurable simple functions converges to a func-tion f , then by theorem 25 (page 107) f must be measurable. This observa-tion provides the missing piece for the following theorem.

Theorem 32 (LebesgueD simple-Lebesgue). Let f be a bounded functiondefined on Œa; b�. Then f is Lebesgue integrable over Œa; b� if and only if fis simple-Lebesgue integrable over Œa; b�. In this case, the integrals agree.

Proof. Suppose that f is bounded and simple-Lebesgue integrable. Then fis measurable as the limit of measurable functions. Hence f is Lebesgueintegrable.

Conversely, suppose that f is Lebesgue integrable. Given any partitionP DfŒyk�1; yk �g of the range of f , the naturally associated simple function� D

Pk yk�1 � 1Ek , where Ek D f �1 ..yk�1; yk �/, satisfies

0 � f � � < kPk

and

SL .f;P/ D s-LZ b

a

�.

Since LR baf D supP SL .f;P/, we can use a sequence of partitions fPng

to generate a sequence f�ng of measurable simple functions that convergesuniformly to f and so that

LZ b

a

f D limnSL .f;Pn/ D lim

n

s-LZ b

a

�n.

We cannot immediately apply the definition of the simple-Lebesgue integralto conclude that L

R ba f D

s-LR ba f , since f�ng may not be an increasing

sequence.6 However, if we takeb�n D maxk�n �k , then fb�ng is an increasingsequence of measurable simple functions converging to f for which

LZ b

a

f D limn

s-LZ b

a

b�n D s-LZ b

a

f:

Notice that in the process of proving that the Lebesgue and simple-Lebesgue integrals agree for bounded functions, we have shown that anybounded measurable function is the uniform limit of an increasing sequence

6 Exercises 33 through 36 provide an alternate approach to finding an increasing sequence ofmeasurable simple functions that converges to f .

of measurable simple functions. While the Lebesgue, simple-Lebesgue, andLebesgue-Young integrals agree on bounded functions, the simple-Lebesguedefinition is more flexible than the others as it can be applied to unboundedfunctions.

Example 31. Define r W Œ0; 1�! R by

r .x/ D

�n; x D m

nwith m

nin lowest terms

0; otherwise.

Then r is an unbounded function and so is not Lebesgue integrable. How-ever, if we define �N on Œ0; 1� by

�N .x/ D

8<:n; x D m

nwith m

nin lowest terms, n � N

N; x D mn

with mn

in lowest terms, n > N0; otherwise

D

�r .x/ ; r .x/ � N

N; otherwise,

then f�N g is an increasing sequence of measurable simple functions thatconverges to r . When n > 0, ��1N .fng/ is countable. Thus

s-LZ 1

0

�N D 0 � ��1N .f0g/

�C

NXnD1

n � ��1N .fng/

�D 0.

Taking the limit, we find that r is simple-Lebesgue integrable with

s-LZ 1

0

r D limN!1

0 D 0.

6.2.4 Extended definitionsSince we now know that functions that are Lebesgue-Young or simple-Lebesgue integrable must also be measurable, we can modify the Lebesgue-Young and simple-Lebesgue definitions so that all three integral definitionsagree for nonnegative, measurable functions. We are removing the restric-tion that f is bounded.

Definition 18 (Lebesgue integral). Let f W Œa; b� ! Œ0;C1/ be a(Lebesgue) measurable function. Let P be the partition of Œ0;C1/ de-fined by the division points fykg

nkD1 and set Ek D f �1 ..yk�1; yk�/,

k D 1; 2; : : : ; n, and EnC1 D f �1 ..yn;C1//. The (lower) Lebesgue sum

of f with respect to P is SL .f;P/ DPnC1kD1 yk�1 � � .Ek/. The Lebesgue

integral of f over Œa; b� is

LZ b

a

f D supPSL .f;P/

where the supremum is taken over all partitions P of Œ0;C1/. The functionf is said to be Lebesgue integrable if L

R ba f < C1:

Notice the change made to accommodate unbounded functions. The lastinterval in the partition of Œ0;C1/ is .yn;C1/. If f happens to bebounded, then f �1 ..yn;C1// D ; when yn is sufficiently large. Alsonotice that

SL .f;P/ D LZ b

a

nC1XkD1

yk�1 � 1Ek

!:

Definition 19 (Lebesgue-Young integral). Let f be a nonnegative,(Lebesgue) measurable function defined on Œa; b�. Let P DfEkgnkD1 be ameasurable partition of Œa; b�. The lower Lebesgue-Young sum of f over Pis

SL-Y .f;P/ DXP

infEkf � � .Ek/ .

The Lebesgue-Young integral of f over Œa; b� is

L-YZ b

a

f D supP

SL-Y .f;P/

where the supremum is taken over all measurable partitions P of Œa; b�.Wesay that f is Lebesgue-Young integrable if L-Y

R ba f < C1.

Note that, on the basis of our earlier work, we do not need to concernourselves with the upper sum when f is bounded. If f is not bounded thenthe upper sum is infinite and therefore not helpful. Also note that for anymeasurable partition P DfEkgnkD1 of Œa; b�,

SL-Y .f;P/ D LZ b

a

nXkD1

infEkf � 1Ek

!:

Both of these definitions are extended to functions that are not nonnegativeby taking

LZ b

a

f D LZ b

a

f C � LZ b

a

f �

when the integral of at least one of f C D 12.jf j C f / or f � D 1

2.jf j �

f / is finite. We say that f is Lebesgue integrable if LR ba jf j < C1 or,

equivalently, if the integrals of both f C and f � are finite. When thesechanges are made, all three definitions are in complete agreement.

At this point, you may well be thinking “If the three definitions are thesame, why did we bother introducing all of them?” This is an excellent ques-tion and you probably already have some ideas about how to answer it. Letme provide two reasons.

1. It is always a good mathematical practice to explore alternative waysof approaching a problem. In this case, the alternatives did not produceanything new, but often a modified approach is quite fruitful.

2. Having alternative ways of thinking about the same thing can be veryuseful. We have seen this already. When we wanted to prove thatLR baf D R

R baf , we instead proved that L-Y

R baf D D

R baf . The process

would have been far more difficult had we been restricted to using onlythe original Riemann and Lebesgue definitions. In subsequent sections,we will reap additional benefits from having multiple characterizationsof the Lebesgue integral at our disposal.

In texts treating only the Lebesgue integral, it is typically written asR ba f d� or

R ba f .x/ d� .x/. We will retain the notation L

R ba f since we are

interested in comparing the properties of the Lebesgue integral with those ofvarious other types of integrals.

6.3 Convergence theoremsYou will recall that one of the main historical motivations for a more carefuland generalized investigation into integration was the question of conver-gence. Originally, the focus was on convergence of trigonometric series, butother illustrative examples were soon added to the mix. We now turn ourattention to the task of finding conditions under which

LZ b

a

limnfn D lim

n

LZ b

a

fn.

We will build our theorems bit by bit, relaxing the conditions as we proceed.We start with a set of three tools. The first two will be used repeatedly and thethird (Egoroff’s theorem) was used already in the previous section. Noticethe appearance of a version of the divide-and-conquer technique in the proofof the next result.

Theorem 33. If f D 0 a.e., then f is Lebesgue integrable with LR ba f D 0

for any interval Œa; b� in the domain of f .

Proof. The measurability of f follows from theorem 26 (page 108).Let Z D fx 2 Œa; b� W f .x/ ¤ 0g. By assumption, � .Z/ D 0. Now let P

be a measurable partition of Œa; b�. Divide the partition into those sets thatinclude a point x for which f C .x/ D 0 and those that do not. Label the twocollections P0 and P>. For any E 2 P>, E Z so that � .E/ D 0. Thus

SL-Y�f C;P

�D

XE2P0

0 � � .E/CXE2P>

infEf C � � .E/ D 0.

Since the supremum of SL-Y�f C;P

�over all possible measurable partitions

is zero,

LZ b

a

f C D L-YZ b

a

f C D 0:

The same argument shows that LR baf � D 0. Hence L

R baf D 0.

Theorem 34. Suppose that f is a Lebesgue integrable function on Œa; b� andthat g D f a.e. Then g is also Lebesgue integrable with L

R bag D L

R baf .

Proof. First note that g is measurable by theorem 26. Since gC � f C D 0

a.e.,

LZ b

a

gC D LZ b

a

�f C C

�gC � f C

��D L

Z b

a

f C C LZ b

a

�gC � f C

�D L

Z b

a

f C.

The same argument shows that LR ba g� D L

R ba f

� so that LR ba g D

LR ba f .

Theorem 35 (Egoroff, 1911). Let ffkg be a sequence of real-valued, mea-surable functions defined on Œa; b� and suppose that ffkg converges to falmost everywhere. Then for any " > 0 there is a measurable subset E ofŒa; b� with � .E/ < " such that ffkg converges uniformly to f on the com-plement of E.

Proof. For n; k 2 N, define

An;k D

�x 2 Œa; b� W jfm .x/ � f .x/j �

1

n, for some m � k

:

By construction, An;k is measurable, An;kC1 An;k , and, since \kAn;kis contained in the set of points for which ffkg does not converge to f , itfollows that �

�\kAn;k

�D 0. Consequently, for each n we can select an

integer kn so that ��An;kn

�< "

2n. (See exercise 43.)

Let E D [nAn;kn . By subadditivity,

� .E/ �Xn

��An;kn

�<Xn

"

2nD ":

Now for any x 2 Ec , we have x 62 An;kn so that jfk .x/ � f .x/j <1n

fork � kn. Hence ffkg converges uniformly to f on Ec .

With our tools in hand, we are ready to turn to the convergence theorems.The important thing to notice in what follows is that, in the context of theLebesgue integral, uniform convergence is replaced with the weaker condi-tion of pointwise convergence.

The condition in the next theorem that ffkg be uniformly bounded seemslike a new condition that was not required for Riemann integrals. Thisis not the case as the uniform boundedness condition is implied by Rie-mann integrability. Riemann-integrable functions are bounded and whena sequence ffkg of bounded functions converges uniformly, ffkg is uni-formly bounded. The uniform boundedness condition is not redundant forLebesgue-integrable functions as they need not even be bounded.

Note the divide-and-conquer move in the following proof.

Theorem 36 (Bounded convergence). Let ffkg be a uniformly bounded se-quence of Lebesgue measurable functions that converges pointwise to f a.e.on Œa; b�. Then

limk

LZ b

a

fk DLZ b

a

limkfk D

LZ b

a

f .

Proof. By assumption, there is a value B such that jfk .x/j < B for allk 2 N and x 2 Œa; b�. When limk fk .x/ does not exist or is not equal tof .x/, redefine f .x/ D 0. Then jf .x/j � B for all x 2 Œa; b� and, sincethe set of points on which the value of f has been modified has measurezero, L

R baf is unchanged.

Now let " > 0 and use Egoroff’s theorem to select a set E with � .E/ <"4B

such that ffkg converges uniformly to f on Œa; b� nE. Choose anN 2 N

such that jfk .x/ � f .x/j <"

2.b�a/for all k � N and all x 2 Œa; b� nE.

Then for k � N ,ˇˇ LZ b

a

fk �LZ b

a

f

ˇˇ � L

Z b

a

jfk � f j

D LZE

jfk � f j CLZŒa;b�nE

jfk � f j

< 2B � � .E/C"

2 .b � a/� � .Œa; b�/ < ".

Thus limkLR bafk D

LR baf .

This is a very satisfying theorem as it handles situations like the follow-ing two examples which are not explained by convergence theorems for theRiemann integral.

Example 32. Let fk .x/ D xk ; x 2 Œ0; 1�. Since ffkg is bounded by 1 andconverges pointwise to f .x/ D 0 except at x D 1, limk

LR 10 fk D 0.

Example 33. Let frng be an enumeration of the rational numbers in Œ0; 1�.Define dn on Œ0; 1� by

dn .x/ D

�1; x D rk with n � k � 2n0; otherwise.

Then fdng is uniformly bounded by 1 and converges pointwise to f .x/ D 0:Hence limn

LR 10 dn D 0.

Unfortunately, we have no reason to believe that convergent trigonometricseries have uniformly bounded partial sums and, as the following exampleillustrates, we cannot simply drop the condition of a uniform upper bound. Ifwe want to generalize theorem 36, we need to find an alternate, less restric-tive condition to take its place.

Example 34. Define the sequence of functions ffkg on Œ0; 1� by

fk .x/ D

�k; x 2

�0; 1k

�0; x 2

�1k; 1�:

Then fk converges to f .x/ D 0 almost everywhere. However, LR 10 fk D 1

for all k 2 N so that limkLR 10fk ¤

LR 10f:

We’ll see shortly (theorem 39, page 140) that we can replace the uniformupper bound on the sequence of functions with an integrable bounding func-tion (a measurable function with a finite Lebesgue integral). Before we getto the formal statement of the theorem and its proof, we need two additionalstepping stones that are useful in their own right.

Theorem 37 (Monotone convergence, Levi, 1906). Let ffkg be a mono-tone increasing sequence of nonnegative, measurable functions that con-verges a.e. to f on Œa; b�. Then

LZ b

a

f D limk

LZ b

a

fk

where a value ofC1 is permissible.

Proof. By redefining f and each fk on a set of measure zero, we may as-sume that ffkg converges to f for all x 2 Œa; b�.

Since each fk is measurable, there is an increasing sequence of measur-able simple functions

˚�k;j

�1jD1

that converges to fk . (See exercise 36 orthe proof of theorem 32.) Now define a new sequence of measurable simple

functionsnb�no byb�n D max1�k�n �k;n. Then

b�nC1 D max1�k�nC1

�k;nC1 � max1�k�n

�k;nC1 � max1�k�n

�k;n D b�nso that

˚b�n� is also an increasing sequence of measurable simple functions.

Moreover, for all k � n, we have �k;n � fk � fn � f so thatb�n � fn � f .In fact,

˚b�n� converges to f for all x 2 Œa; b�. To see why, fix x 2 Œa; b�and " > 0. Because ffk .x/g converges to f .x/, we can find an N sothat f

�x�� "=2 < fN .x/ � f .x/. Similarly, we can find an M so that

fN .x/ � "=2 < �N;M .x/ � fN .x/ : Then for n � max fN;M g, we havethe inequality

f .x/ � " < �N;M .x/ � �N;n .x/ � b�n .x/ � f .x/ :Since " > 0 and x 2 Œa; b� are arbitrary, we can conclude that

˚b�n� is anincreasing sequence of measurable simple functions that converges to f forall x 2 Œa; b�.

By the definition of the simple-Lebesgue integral, we see that

LZ b

a

f D s-LZ b

a

f D limn

s-LZ b

a

b�n D limn

LZ b

a

b�n.


The monotonicity of the Lebesgue integral implies that

LZ b

a

b�n � LZ b

a

fn �LZ b

a

f

so thatLZ b

a

f D limn

LZ b

a

fn

as claimed.

Theorem 37 is not directly useful in the context of Fourier series since thesequence of partial sums is certainly not monotone. But even when ffkg isneither increasing nor convergent, we still can say something.

Theorem 38 (Fatou’s lemma). Let ffkg be a sequence of nonnegative,measurable functions. Then

LZ b

a

limk

fk � limk

LZ b

a

fk :7

Proof. Define f n D infk�n fk . Thennfn

ois an increasing sequence of

measurable functions. Since f n � fk for all k � n,

LZ b

a

fn� infk�n

LZ b

a

fk .

Furthermore,ninfk�n L

R bafk

on

is an increasing sequence of extended real

numbers.8 Taking limits and invoking the monotone convergence theorem,

LZ b

a

limk

fk DLZ b

a

limnfnD lim

n

LZ b

a

fn� lim

ninfk�n

LZ b

a

fk D limk

LZ b

a

fk .

This brings us to the main theorem of this section, the dominated con-vergence theorem. It replaces the constant uniform bound in the boundedconvergence theorem with an integrable function.

7 limk

fk D limk infj�k fj . Since˚infj�k fj

�is a monotone increasing sequence, lim

k

fk

will always be defined in the extended real numbers. See Appendix A.3.8 The extended real numbers are NR D R[f�1;C1g.

Theorem 39 (Dominated convergence, Lebesgue, 1910). Suppose thatffkg is a sequence of measurable functions that converges to f a.e. on Œa; b�.If there is a measurable function g with L

R ba g <1 and for which jfkj � g

a.e. on Œa; b� for all k; then

LZ b

a

f D limk

LZ b

a

fk .

Proof. We may assume that ffkg converges to f and that jfkj � g on allof Œa; b�. Define the sequences

˚fk

�and

˚f k�

by fkD infj�k fj and

f k D supj�k fj . Then˚fk

�and

˚f k�

are monotone sequences of func-tions converging to f and satisfying

�g � fk� fk � f k � g.

Nowng C f

k

ois a nonnegative, increasing sequence of measurable func-

tions converging to g C f , so by the monotone convergence theorem,

LZ b

a

g C LZ b

a

f D LZ b

a

.g C f /

D limk

LZ b

a

g C f

k

�D lim

k

LZ b

a

g C LZ b

a

fk

!

D LZ b

a

g C limk

LZ b

a

fk

.

Since LR ba g is finite, limk

LR ba f k

D LR ba f . A similar proof shows that

limkLR ba f k D

LR ba f . By monotonicity of the Lebesgue integral,

LZ b

a

fk� L

Z b

a

fk �LZ b

a

f k

so that limkLR ba fk exists and is equal to L

R ba f as claimed.

Since the investigation of convergence properties of the various integralswas historically motivated by questions arising from trigonometric series, wewill close the loop and show how the Monotone and dominated convergencetheorems apply in the context of series of functions.

Theorem 40 (Interchange of summation and integral). Suppose ffkg is asequence of measurable functions defined on Œa; b�. Suppose further that theseries

Pk

LR ba jfkj converges. Then

Pk fk converges almost everywhere on

Œa; b� and

LZ b

a

Xk

fk DXk

LZ b

a

fk .

Proof. As a positive-term series,Pk jfk .x/j converges in the extended real

numbers for every x 2 Œa; b�. SincePk

LR ba jfkj < 1, we can conclude

thatPk jfk .x/j and so

Pk fk .x/ must converge to a finite real number for

almost every x 2 Œa; b�. Then by the monotone convergence theorem,

LZ b

a

Xk

jfkj DXk

LZ b

a

jfkj <1.

Moreover,

�

1XkD1

jfkj �

nXkD1

fk �

1XkD1

jfkj

for all n 2 N. The desired conclusion now follows from the dominated con-vergence theorem.

6.4 The fundamental theoremsWe begin our investigation of the fundamental theorem of calculus for theLebesgue integral with a statement of the evaluation form.

Theorem 41 (FTC-1). If F is a differentiable function with a boundedderivative in Œa; b�, then F 0 is Lebesgue integrable on Œa; b� and

LZ x

a

F 0 D F .x/ � F .a/

for all x 2 Œa; b�.

Before turning to the proof, compare this theorem to the correspondingtheorem for the Riemann integral. Both theorems require F to be differen-tiable on all of Œa; b�. FTC-1 (page 39) for the Riemann integral requiresthat F 0 be continuous whereas the Lebesgue version requires only the muchweaker assumption that F 0 be bounded. The bounded version of the theoremis false for the Riemann integral since the Volterra function has a bounded


derivative that is not Riemann integrable. We will see an additional weak-ening of the boundedness hypothesis for the Lebesgue integral later in thissection.

Proof. First note that F , being continuous, is measurable. Extend F toŒa; b C 1� by setting F .x/ D F .b/ C F 0 .b/ .x � b/ for x 2 Œb; b C 1�.Then

fn .x/ D n

�F

�x C

1

n

�� F .x/

; x 2 Œa; b�

is also measurable with ffng converging pointwise to F 0 on Œa; b�. Hence F 0

is measurable and, being bounded by assumption, is Lebesgue integrable.By the mean value theorem, given t 2 Œa; b� and n 2 N, there is a value

c 2�t; t C 1

n

�such that

n

�F

�t C

1

n

�� F .t/

D F 0 .c/ .

As we are assuming that F 0 is bounded, we see that ffng is uniformlybounded. Thus we can use the bounded convergence theorem, the continuityof F , and FTC-2 for the Riemann integral (page 40) to conclude that

LZ x

a

F 0 D LZ x

a

limnn

�F

�t C

1

n

�� F .t/

dt

D limnn � L

Z x

a

�F

�t C

1

n

�� F .t/

dt

D limnn �

�LZ x

a

F

�t C

1

n

�dt � L

Z x

a

F .t/ dt

D lim

nn �

"LZ xC 1n

aC 1n

F .t/ dt � LZ x

a

F .t/ dt

#

D limn

"n � L

Z xC 1n

x

F � n � LZ aC 1n

a

F

#

D limnn � R

Z xC 1n

x

F � limnn � R

Z aC 1n

a

F

D F .x/ � F .a/ :

Make sure that you thoroughly understand the preceding proof and theway it uses fn .x/ D n

�F�x C 1

n

�� F .x/

�. Closely related strategies will

be used multiple times in this section with the critical exchange of the limit

and the integral justified by the convergence results, including theorem 38,of the previous section.

The proof of the derivative form of the fundamental theorem of calculusfor the Lebesgue integral is more complex. We state the theorem now, butwe need to do quite a bit of preparatory work before we get to the proof.

Theorem 42 (FTC-2, Lebesgue, 1904). Let f be a Lebesgue integrablefunction on Œa; b�. Define F .x/ D L

R xa f on Œa; b�. Then F is absolutely

continuous on Œa; b� and F 0 D f a.e. on Œa; b�.

Absolute continuity is a strengthening of uniform continuity. It is the crit-ical condition for the Lebesgue integral.

Definition 20 (Absolute continuity). A function f is absolutely contin-uous on a set A if given any " > 0 there is a ı > 0 so thatPk jf .xk/ �f .yk/j<" for any finite choice of disjoint intervals f.xk ; yk/g

from A satisfyingPk jxk � ykj < ı.

Intuitively, continuity means that as x approaches x0, the value of f .x/approaches f .x0/. Absolute continuity requires that, in addition, f .x/ doesnot “wiggle around too much” as it approaches f .x0/. For example, thefunction

f .x/ D

�x sin �

x; x ¤ 0

0; x D 0

is uniformly continuous but not absolutely continuous on Œ0; 1�. (Seeexercise 84.)

Figure 6.2. f .x/ is uniformly but not absolutely continuous

Before turning to the proof of this theorem, notice how it differs fromFTC-2 for the Riemann integral (page 40). First, Riemann integrands mustbe bounded. This is no longer required for Lebesgue integrals. In addition,FTC-2 for the Riemann integral only concludes that F 0 .x/ D f .x/ forpoints where f is continuous. For some Lebesgue integrable functions suchas the Dirichlet function, the set of points of continuity is empty. Never-theless, F 0 D f a.e. on Œa; b�. Moreover, Lebesgue’s theorem draws thestronger conclusion that F is absolutely continuous.

The proof of FTC-2 for the Lebesgue integral depends on several proper-ties of the integral as well as some facts related to absolute continuity. Wewill prove the integration-related results here but will only state the resultsrelated to absolute continuity, deferring their proofs until Chapter 7:

Lemma 43. Suppose that f is Lebesgue integrable over Œa; b�. Then, givenany " > 0, there is a ı > 0 so that L

RE jf j < " for any measurable subset

E of Œa; b� with � .E/ < ı.

Proof. Define

fn .x/ D

�jf .x/j ; jf .x/j � n

n; jf .x/j > n:

Then ffng converges pointwise to jf j and, since 0 � jf j � fn � jf j, wecan apply the dominated convergence theorem to conclude that

limn

LZ b

a

.jf j � fn/ D 0.

Choose n so that LR ba.jf j � fn/ < "=2 and set ı D "

2n. Then for any

measurable subset E of Œa; b� with � .E/ < ı,

LZE

jf j D LZE

.jf j � fn/CLZE

fn

� LZ b

a

.jf j � fn/CLZE

fn < ":

Did you notice how variations of both the divide-and-conquer and trunca-tion techniques were used in the preceding proof?

Lemma 44. Suppose that f is Lebesgue integrable over Œa; b�. IfF .x/ D L

R xa f D 0 for all x 2 Œa; b� then f D 0 a.e. on Œa; b�.

Proof. Suppose that fx 2 Œa; b� W f .x/ > 0g has positive measure. Then forsome n 2 N, A D fx 2 Œa; b� W f .x/ � 1=ng also has positive measure.


Since A is measurable, theorem 22 (page 104) guarantees a closed set Econtained in A with � .E/ > 0. Since E is closed, .a; b/ nE is open and socan be written as the countable union of disjoint open intervals f.ak ; bk/g.(See exercise 74.) By the dominated convergence theorem,

LZ.a;b/nE

f D LZ b

a

1.a;b/nE � f

D LZ b

a

Xk

1.ak ;bk/ � f

DXk

LZ b

a

1.ak ;bk/ � f

DXk

LZ bk

ak

f

DXk

LZ bk

a

f � LZ ak

a

f

!:

This last sum is zero as all its terms are zero by assumption. But then

F .b/ D LZ b

a

f D LZE

f C LZ.a;b/nE

f

D LZE

f �1

n� .E/ > 0

contrary to our hypothesis. Thus f � 0 a.e. on Œa; b�. By applying the pre-ceding argument to �f , we can conclude that f D 0 a.e. on Œa; b�. 9

The following three results will be proved in Chapter 7, Section 8.

Lemma 45. Suppose that f is absolutely continuous on Œa; b�. Then f canbe expressed as f D f1 � f2 where f1 and f2 are increasing functions.(If convenient, we can require f1 and f2 to be strictly increasing and/orabsolutely continuous.)

Lemma 46 (Lebesgue, 1904). If f is increasing on Œa; b� then f is differ-entiable a.e. on Œa; b�.

Lemma 47. If f is absolutely continuous on Œa; b� and f 0 D 0 a.e. on Œa; b�,then f is constant on Œa; b�.

9 Exercise 39 outlines an alternate proof of Lemma 44.

With these tools, we are now prepared to prove the derivative form of thefundamental theorem of calculus for Lebesgue integrals which we restatehere.

Theorem 42 (FTC-2, Lebesgue, 1904). Let f be a Lebesgue integrablefunction on Œa; b� and define F .x/ D L

R xaf on Œa; b�. Then F is absolutely

continuous on Œa; b� and F 0 D f a.e. on Œa; b�.

Proof. To prove that F is absolutely continuous, let " > 0 be given: Sincef is Lebesgue integrable, lemma 43 tells us that there is a ı > 0 suchthat L

RA jf j < " whenever A is a measurable set with � .A/ < ı. Sup-

pose now that f.xk ; yk/g is a finite set of disjoint subintervals from Œa; b�

withPk jxk � ykj < ı. Setting A D [k .xk ; yk/ we have � .A/ DP

k jxk � ykj < ı so thatXk

jF .xk/ � F .yk/j DXk

ˇLZ yk

xk

f

ˇ

�Xk

LZ yk

xk

jf j

D LZA

jf j < ".

Hence F is absolutely continuous and therefore differentiable almost every-where.

We first prove that F 0 D f a.e. under the additional assumption thatjf j � B . ExtendF to Œa; b C 1� by takingF .x/ D F .b/ for x 2 Œb; b C 1�.Then limn n

�F�x C 1

n

�� F .x/

�D F 0 .x/ for those x 2 .a; b/ at which F

is differentiable. Sinceˇn

�F

�x C

1

n

�� F .x/

ˇD n

ˇˇ LZ xC 1n

x

f

ˇˇ � n � L

Z xC 1n

x

jf j � B ,

we can follow the proof of theorem 41 (page 141) using the bounded con-vergence theorem to conclude that

LZ x

a

F 0 D LZ x

a

limnn

�F

�t C

1

n

�� F .t/

D F .x/ � F .a/ D F .x/ .

This means that

LZ x

a

�F 0 � f

�D

�LZ x

a

F 0�� F .x/ D 0

for all x 2 Œa; b�. Hence, by lemma 44 (page 144), F 0 D f a.e. in Œa; b�.


If f is not bounded, assume temporarily that f � 0. Repeat the analysisabove using Fatou’s lemma (theorem 38, page 139) instead of the boundedconvergence theorem to conclude that, for those x 2 .a; b/ for which F isdifferentiable,

LZ x

a

F 0 D LZ x

a

limnn

�F

�t C

1

n

�� F .t/

� limn

LZ x

a

n

�F

�t C

1

n

�� F .t/

D F .x/ � F .a/ D F .x/ .

HenceLZ x

a

�F 0 � f

�D

�LZ x

a

F 0�� F .x/ � 0:

For the reverse inequality, define

fn .x/ D

�f .x/ ; f .x/ � n

n; f .x/ > n

and Fn .x/ D LR xa fn for x 2 Œa; b�. Then F � Fn is nonnegative and in-

creasing on Œa; b� so, by lemma 46 (page 145), .F � Fn/0 exists and is non-

negative almost everywhere in Œa; b�. Furthermore, our earlier work showsthat F 0n D fn a.e. on Œa; b�. Recalling that we are assuming that f , and sofn, is nonnegative,

0 � .F � Fn/0 D F 0 � F 0n D F

0 � fn � F0 a.e.

ThusLZ x

a

�F 0 � fn

�� 0:

Since the inequality holds for all values of n;

LZ x

a

�F 0 � f

�� 0

by the dominated convergence theorem.Thus when f is nonnegative, F 0 D f a.e. The general case now follows

by considering f D f C � f �.

Notice that, even though F and so n�F�t C 1

n

�� F .t/

�are continuous

functions and therefore Riemann integrable, the Riemann integral does not


allow the critical exchanges of the integral with the limit or with lim used inthe proofs that L

R xa F

0 D F .x/ and LR xa F

0 � F .x/.In light of FTC-2, it is tempting to conclude that as long as F is differ-

entiable a.e. on Œa; b�, then (subtracting F .a/ to make the function zero atx D a) we have

R xaF 0 D F .x/�F .a/. This would be an error. As a coun-

terexample, consider the Cantor function c. Its derivative is c0 D 0 a.e. onŒa; b� so that

0 D

Z 1

0

c0 ¤ c .1/ � c .0/ D 1.

Even when F is differentiable everywhere, the conclusion of the theoremmay not hold. (See exercise 89.) The conditions under which we can con-clude that

R ba F

0 D F .b/�F .a/ are given in the following theorem. Noticethat, since changing an integrand on a set of measure zero does not changethe integral and since antiderivatives may differ by a constant, the theoremis as close to a converse of FTC-2 as is possible.

Theorem 47 (FTC-1, Lebesgue, 1904). If F is absolutely continuous onŒa; b�, then F is differentiable a.e., F 0 is Lebesgue integrable on Œa; b�, and

LZ x

a

F 0 D F .x/ � F .a/

for all x 2 Œa; b�.

Proof. Since F is absolutely continuous, it can be written as F D F1 � F2where F1 and F2 are increasing functions. Since increasing functions aredifferentiable a.e., so is F and, as in theorem 41 (page 141), F 0 is measur-able. Moreover, jF 0j � F 01 C F

02 a.e. so that we can use theorem 38 (page

139) as in the proof of FTC-2 to conclude that

LZ b

a

ˇF 0ˇ� L

Z b

a

F 01 CLZ b

a

F 02

� F1 .b/ � F1 .a/C F2 .b/ � F2 .a/ .

Thus F 0 is Lebesgue integrable over Œa; b�.Now define G .x/ D L

R xa F

0. Then G; and so G � F; are absolutelycontinuous on Œa; b�. By theorem 42 (page 146), .G � F /0 D G0 � F 0 D 0

a.e. on Œa; b�, so that G � F is constant on Œa; b� by lemma 47 (page 145).Since G .a/ D 0, G .x/ � F .x/ D �F .a/, so that

LZ x

a

F 0 D G .x/ D F .x/ � F .a/ :

6.5. Exercises 149

The gauge integral, introduced in the next chapter, allows us to replace therequirement that F be absolutely continuous with simple continuity. On theother hand, F 0 must be differentiable except at a countable number of pointsrather than differentiable almost everywhere.

6.5 Exercises

6.1 Variations: filling the gaps1. Suppose that f W Œa; b�! .˛; ˇ/ is a measurable function. Prove that if

P DfŒyk�1; yk�gnkD1 is a partition of Œ˛; ˇ� and P� is a refinement of P ,then SL .f;P/ � SL .f;P�/ and SL .f;P/ � SL .f;P�/. (First showSL .f;P�/ � SL .f;P/ � 0 when a single subinterval is split into two.Then use induction.)

2. Suppose that f W Œa; b� ! .˛; ˇ/. Explain why SL .f;P/ � SL .f;Q/for any pair of measurable partitions P and Q of Œ˛; ˇ�. Use this fact toprove that supP SL .f;P/ � infP SL .f;P/ where the supremum andinfimum are taken over all partitions of Œ˛; ˇ�.

3. Explain why the definition of the Lebesgue integral does not mentionupper Lebesgue sums.

4. Why does the definition of the simple-Lebesgue integral require the se-quence of simple functions converging to f to be increasing?

5. Explain why LR ba jf j < 1 is equivalent to the condition that the inte-

grals of both f C and f � are finite.

6. Verify the standard integral properties for the Lebesgue-style integrals.Chose a different definition for each of the first three properties. (Thinkabout which definition will be most efficacious for proving each of theproperties.)

(a) Uniqueness. The value of the integral is unique (if it exists).(b) Linearity. If f and g are integrable over the interval Œa; b� and c 2

R, then so are f Cg and cf . Moreover,R ba.f C g/ D

R ba f C

R ba g

andR ba cf D c

R ba f:

(c) Monotonicity. If f and g are Lebesgue integrable over the intervalŒa; b� with f .x/ � g .x/ for x 2 Œa; b�, then

R ba f �

R ba g:

(d) Triangle inequality. If f and jf j are integrable over Œa; b�, thenˇR ba f

ˇ�R ba jf j.


6.1 Variations: deeper reflections7. Suppose that f W Œa; b� ! .˛; ˇ/. Even if we cannot use Lebesgue

measure because f is not measurable, we can still use Lebesgue outermeasure. Let P DfŒyk�1; yk �gnkD1 be a partition of Œ˛; ˇ� and setEk D f �1 ..yk�1; yk �/, k D 1; 2; : : : ; n. We can define lower and up-per sums by SL� .f;P/ D

PnkD1 yk�1 � �

� .Ek/ and SL� .f;P/ DPn

kD1 yk � �� .Ek/ where �� .A/ is the Lebesgue outer measure of the

set A. How do things go wrong when we try to define the Lebesgue(outer) integral of f by supP SL� .f;P/ or infP SL� .f;P/ where thesupremum/infimum is taken over all possible partitions of Œ˛; ˇ�?

8. Suppose that fEkgnkD1 are sets of real numbers and that � DPn

kD1 ˛k � 1Ek . How many distinct values could � take on?

9. If f is Lebesgue integrable over Œa; b� then so is jf j. Does the integra-bility of jf j imply the integrability of f ? Explain.

10. Let f be a nonnegative, measurable function on Œa; b�. Prove that fis Lebesgue integrable on Œa; b� if an only if

P1kD1 k� .Ek/ converges

where Ek D f �1 .Œk; k C 1//.

6.2 Reconciling the approaches: filling the gaps11. In the proof of theorem 28 (page 124), why is the set C measurable?

12. Prove that the two alternate statements in Definition 17 (page 125) areequivalent.

13. Explain how Definition 17 (page 125) relates to Definition 3 (page 30).

14. Suppose that f W Œa; b� ! .˛; ˇ/. Prove that if P DfEkg is a measur-able partition of Œa; b� and P� D

nE�j

ois a measurable refinement of P ,

then SL-Y .f;P�/ � SL-Y .f;P/. (See lemma 29 on page 126.)

15. Suppose that f W Œa; b� ! .˛; ˇ/. Use the previous exercise to provethat SL-Y .f;P/ � SL-Y .f;Q/ for any measurable partitions P and Qof Œa; b�. Conclude that L-Y

R ba f �

L-YR ba f .

16. Suppose that � and are measurable simple functions and that c is areal number. Prove that 1 D � C and 2 D c� are also measurablesimple functions.

17. Verify that LR ba � D

PnkD1 ˛k � � .Œa; b� \Ek/ for any simple function

� DPnkD1 ˛k � 1Ek .

6.5. Exercises 151


(a) Explain why � .Dk/ � ��Dk;n

�D L

R ba 1Dk;n �

LR ba k .�n � n/.

(b) Fill in the details to explain why f is measurable if � is measurable,� D a.e. and � � f � .

(c) Fill in the details at the end of the proof to show that L-YR ba f D

LR ba f .

19. Supply the justifications on page 128 to explain why

DZ b

a

f � L-YZ b

a

f � L-YZ b

a

f � D

Z b

a

f:

20. Suppose that � and are simple functions. Explain why j� � j is asimple function.

21. Suppose that � and are simple functions. Prove that max f�; g andmin f�; g are simple functions. Extend this conclusion to finite sets ofsimple functions.


(a) Why does the second half of the proof start withL < limk s-LR ba �k

and an n0 for which L < s-LR ba �n0 instead of beginning with " > 0

and an n0 for whichˇlimk

s-LR ba �k �

s-LR ba �n0

ˇ< "?

(b) Explain why˚��k

�and

˚ �k

�are increasing sequences of measurable

simple functions that converge to f �.

23. Suppose that f W Œa; b� ! R is Lebesgue integrable over Œa; b� and letE be a measurable subset of Œa; b�. L

RE f could be defined two ways:

(a) LREf D L

R baf � 1E or

(b) by modifying the standard definition to useEk D fx 2 E W yk�1 < f .x/ � ykg.

Prove that the two definitions are equivalent.

24. When proving that the Lebesgue and simple-Lebesgue integrals agreefor bounded functions, we used the monotonicity of the simple-Lebesgue integral. Prove this.

25. Suppose that f W Œa; b� ! R is Lebesgue integrable over Œa; b�.Prove that if E and F are disjoint measurable subsets of Œa; b�, thenLREf C L

RFf D L

RE[F f .

26. Fill in the details at the end of the proof of theorem 32 (page 131). Letf�ng be a sequence of simple functions that are bounded above by f andconverge uniformly to f and for which L

R ba f D limn

s-LR ba �n. Defineb�n D maxk�n �k . Prove that

nb�no is an increasing sequence of measur-

able simple functions bounded by f for which LR baf D limn

s-LR bab�n.

6.2 Reconciling the approaches: deeper reflections27. Use the "–ı definition of continuity to prove that if f W A ! R is

continuous andU is an open set of real numbers, then f �1 .U / D V \Afor some open subset V of R.

28. Let P DfŒ0; 0:2�; .0:2; 0:5/; Œ0:5; 0:8/; Œ0:8; 1�g andQ DfŒ0; 0:1/; Œ0:1; 0:5�; .0:5; 0:7/ ; Œ0:7; 1�g. What is P [Q?

29. Approximate f .x/ D x2; x 2 Œ0; 4� within 0:25 using a measurablesimple function � with � � f .

30. Suppose that f W Œa; b� ! .˛; ˇ/ and that P and Q are measurablepartitions of Œa; b� with Q being a refinement of P . Let � and be thelower simple functions corresponding to the partitions P and Q. In otherwords, � D

PE2P infE f �1E and D

PF 2Q infF f �1F . Prove that

� � . (Given x 2 Œa; b�, how is � .x/ computed?)

31. Give an example of a pair of sequences fakg and fbkg for whichlimk jak � bkj D 0 but for which limk ak ¤ limk bk since neither limitexists (even in the extended real numbers). Explain why this is not anissue in the proof of theorem 31 (page 128).

32. Give an example of a function f and a sequence of partitions fPng forwhich the naturally associated sequence f�ng of measurable lower sim-ple functions satisfies L

R baf D limn SL .f;Pn/ but is not increasing.

(See theorem 32 on page 131.)

33. Suppose that f is a measurable function on Œa; b� with 0 � f < ˇ. Usethe values yn;k D

k2n

to prove that there is an increasing sequence ofsimple functions f�ng that converges uniformly to f . (To simplify yourwork, you can assume that ˇ D N for some integer N .)

34. Use the sequence of simple functions of the previous exercise to verifythat any nonnegative, bounded, measurable function on Œa; b� has a finitesimple-Lebesgue integral that agrees with the Lebesgue integral.

6.5. Exercises 153

35. Use the previous exercise to show that any bounded, measurable (notnecessarily nonnegative) function has a finite simple-Lebesgue integralthat agrees with the Lebesgue integral.

36. Modify exercise 33 to prove if f is a nonnegative, measurable (not nec-essarily bounded) function on Œa; b�, there is an increasing sequenceof simple functions f�ng that converges pointwise to f . (Work withnyn;k D

k2n

on�2nkD0

.)

37. Outline the problems one would have to address in order to prove thatfor a Riemann integrable function f , L

R baf D R

R baf using the original

Riemann and Lebesgue definitions of the integral.

38. How would the proofs for simple-Lebesgue results change if we modi-fied the definition for nonnegative functions to be

s-LZ b

a

fDsup

(s-LZ b

a

� W� is a measurable simple function with ��f

)?

39. Suppose that f is Lebesgue integrable and that LRE f D 0 for every

measurable set E Œa; b�.

(a) Prove that f D 0 a.e. on Œa; b�. (Begin by showing that� .fx W f .x/ > 1=ng/ D 0.)

(b) Why is it sufficient to know that LR xaf D 0 for all x 2 Œa; b�? (Use

sigma algebras.)

6.3 Convergence theorems: filling the gaps40. Prove that if f D g a.e., then f C D gC a.e. and f � D g� a.e.

41. In the proof of theorem 34 (page 135), why are f C and gC used insteadof f and g?

42. In the proof of theorem 35 (page 135), why is An;k measurable?

43. Suppose that fAng is a sequence of measurable sets satisfying AnC1 An and � .\nAn/ D 0: Suppose further that � .A1/ < C1. Prove thatfor any " > 0, there is an integer n so that � .An/ < ". (Express A1 asthe disjoint union of measurable sets fBng such that An D

�[k�nBk

�[

.\nAn/.)


44. Suppose that ffkg is a sequence of functions that converges a.e. to fand ffkg is uniformly bounded by B . Define f � by

f � .x/ D

�f .x/ ; ffk .x/g converges to f .x/0; otherwise.

(a) Prove that jf �j � B .(b) Give an example where jf j � B fails.

45. The proof of theorem 37 (page 138) begins by redefining f and fk .Provide a specific procedure for this redefinition. Why is the redefinitionpermissible?

46. In the proof of theorem 37 (page 138), why is max1�k�n �k;nC1 �max1�k�n �k;n?


(a) Why isnfn

oan increasing sequence of measurable functions?

(b) Why is LR ba f n

� infk�n LR ba fk?

(c) Why isninfk�n L

R ba fk

on

an increasing sequence?

48. Suppose that fakg and fbkg are increasing sequences of extended realnumbers with ak � bk for all k. Prove that limk ak � limk bk .

49. In theorem 39 (page 140)

(a) Why is f integrable on Œa; b� ?(b) Why can we assume that ffkg converges to f and that jfkj � g on

all of Œa; b� ?

(c) Why isnfk

oincreasing and why is

nf k

odecreasing?

(d) Why donfk

oand

nf k

oconverge to f ?

(e) Why does limkLR ba f k

exist?

(f) Fill in the details to show that LR ba f D limk

LR ba f k .

50. Prove that if ��˚x 2 Œa; b� W

Pk jfk .x/j D 1

��D m > 0 thenP

kLR ba jfkj D 1. (Use exercise 58 to show that for any M , we can

find an N so that � nx W

PNkD1 jfk .x/j �M

o�� m=2. Explain how

this forcesPk

LR ba jfkj D 1.)

6.5. Exercises 155

51. Explain how the Monotone and dominated convergence theorems guar-antee that L

R ba

Pk fk D

Pk

LR ba fk in theorem 40 (page 141).

6.3 Convergence theorems: deeper reflections52. Give an alternate proof of theorem 33 (page 135) based on the defini-

tion of the Lebesgue (rather than Lebesgue-Young) integral under theadditional assumption that f � 0.

53. Give an alternate proof of theorem 33 (page 135) based on the definitionof the simple-Lebesgue (rather than Lebesgue-Young) integral.

54. Prove that if ffkg is a sequence of bounded functions that convergesuniformly to a function f , then ffkg is uniformly bounded. In otherwords, there is a real number B such that jfkj < B for all values of k.

55. Suppose that f is measurable and that g D f a.e. Supply an alternateproof that g is measurable by considering the function f � g.

56. Give a counterexample to exercise 43 when � .A1/ D C1.

57. Suppose that fAng is a sequence of measurable sets. Prove that if� .[nAn/ > 0 then � .An/ > 0 for some n 2 N.

58. Suppose that A1 A2 � � � An AnC1 � � � is a nested sequenceof measurable sets with � .[nAn/ < 1. Prove that there is an integerN for which � .AN / � 1

2� .[nAn/. Under what circumstances can you

conclude � .AN / > 12� .[nAn/?

59. Give an example of a sequence of continuous functions ffkg that con-verges to f on all of Œ0; 1� but for which limk

LR 10 fk ¤

LR 10 f .

60. State and prove a version of theorem 37 (page 138) for deceasing se-quences of functions.

61. Define f on Œ0; 1� by

f .x/ D

(1px; x ¤ 0

0; x D 0:

Use an increasing sequence of functions to evaluate LR 10 f .


f .x/ D

(1

.x�0:5/2; x ¤ 0:5

0; x D 0:5::

Use an increasing sequence of functions to show that LR 10f D C1.

f .x/ D

(1

.x�0:5/3; x ¤ 0:5

0; x D 0:5:

Show that LR 10 f is not defined.

64. The statement that LR ba g C

LR ba f D

LR ba.g C f / in the proof of

Theorem 39 (page 140) relies on the assumption that 0 � LR bag < 1.

Give an example of a pair of measurable functions f and g for whichLR ba.g C f / D 0 but for which L

R bag C L

R baf is not defined.

65. Compute the following (provide justifications)

(a)P1nD0

LR 10

x˛

.1Cxˇ/n d� .x/ where ˛ � ˇ.

(b)P1nD1

LR 10

x.1C.n�1/x/.1Cnx/

d� .x/ :

(c) limnLR 10

nx1Cn2x2

d� .x/ :

(d) limnLR 2�1

x4n

1Cx4nd� .x/ :

(e) LR 10

P1nD1

xn

n

�d� .x/ :

(f) limnLR 10 nxe

�nx2 d� .x/ :

(g) LR 10 g where g is defined on Œ0; 1� by

g .x/ D

�0; x is in the Cantor setn; x is in a removed interval of length 1=3n.

66. Give an example of a sequence of functions ffng on Œ0; 1� for whichlimn

LR 10 jfnj D 0 but ffn .x/g does not converge to zero for any x 2

.0; 1/. (Try using characteristic functions.)

6.4 The fundamental theorems: filling the gaps67. Suppose that F is differentiable almost everywhere on Œa; b�. Why is F 0

measurable? (Consider limn n�F�t C 1

n

�� F .t/

�.)

68. Without using F.T.C., prove that limn n �LR xC 1nx f D f .x/ whenever

f is continuous at x.

69. Prove that if f is bounded and measurable on Œa; b C 1� then

LZ b

a

f

�t C

1

n

�D L

Z bC 1n

aC 1n

f .t/ .

6.5. Exercises 157

70. Prove that if f is nonnegative and A is a measurable subset of Œa; b�,then L

RA f �

LR ba f .

71. In the proof of lemma 43 (page 144), why is LRE fn < "=2?

72. Suppose that f is a measurable function for which fx 2 Œa; b� W

f .x/ > 0g has positive measure. Why must fx 2 Œa; b� W f .x/ � 1=ngalso have positive measure for some n?

73. Suppose that A is measurable with � .A/ > 0. Why must A contain aclosed set E with � .E/ > 0?

74. Prove that any bounded open set E can be written as a countableunion of disjoint open intervals. (The hard part is to make the inter-vals disjoint. Work inductively. Having chosen I1; I2; : : : ; In, let En DEn[n

kD1I n where I n is the closure of In. Given x 2 En, compute ˛x D

inf ft 2 Œa; b� W .t; x/ Eg and ˇx D sup fs 2 Œa; b� W .x; s/ Eg.Take InC1 D .˛x ; ˇx/ for some x 2 En. Make sure to choose x insuch a way that [nIn D E.)

75. Extend the result of the previous exercise to all open sets.

76. Why can the sum and integrals be interchanged in the proof of lemma 44(page 144)? What is the dominating function?

77. Supply a proof that f � 0 a.e. on Œa; b� for lemma 44 (page 144). (Youdo not need to mimic what was done to prove f � 0 a.e. on Œa; b�.)

78. Prove that if f is a nonnegative function on Œa; b�, then F .x/ D LR xaf

is increasing.


(a) Fill in the details to show that LR xaF 0 D F .x/ � F .a/ D F .x/

when f is bounded.(b) Why is F � Fn nonnegative and increasing on Œa; b�?(c) Why is F 0n D fn a.e. on Œa; b�?(d) Fill in the details near the middle of the proof to show that L

R xa F

0 �

F .x/ when f � 0 but unbounded.(e) Justify the statement L

R xa.F 0 � fn/ � 0, near the end of the proof.

(f) How is the dominated convergence theorem used to conclude thatLR xa.F 0 � f / � 0? What is the dominating function?

(g) Complete the proof by showing how f D f C � f � can be used toprove the theorem when f is unbounded but not nonnegative.


80. Supply the details needed to conclude that LR ba jF

0j � F1 .b/�F1 .a/C

F2 .b/ � F2 .a/ in theorem 47.

6.4 The fundamental theorems: deeper reflections81. Suppose that f is Riemann integrable on Œa; b�. Prove that F .x/ D

RR xa f is absolutely continuous on Œa; b�.

82. Prove that if f and g are absolutely continuous on a set A and c is ascalar, then f C g and cf are also absolutely continuous.

83. Prove that any absolutely continuous function is uniformly continuous.

84. Define

f .x/ D

�x cos �

x; x ¤ 0

0; x D 0:

(a) Prove that f is uniformly continuous on Œ0; 1�.(b) Prove that f is not absolutely continuous on Œ0; 1�. (Use .xk ; yk/ D�

1kC1

; 1k

�.)

85. Prove that the Cantor function is not absolutely continuous.

86. Show that continuity is not sufficient in lemma 47 (page 145). In otherwords, give an example of a nonconstant, continuous function on Œ0; 1�for which f 0 D 0 a.e. Is uniform continuity sufficient?

87. Suppose that ffng is a sequence of measurable functions and that

limn

LZ b

a

jfn � f j D 0

for some integrable function f . Prove that if ffng converges a.e. onŒa; b�, then ffng converges a.e. to f on Œa; b�. (See exercise 59.)

88. Let f .x/ Dpx, x 2 Œ0; 1�.

(a) Is f .x/ absolutely continuous on Œ0; 1�?(b) Is f 0 Lebesgue integrable on Œ0; 1�?(c) Is L

R x0 f

0 D f .x/ for x 2 Œ0; 1�?(d) Why are we not worried about the fact that f 0 does not exist at

x D 0?

89. Let

f .x/ D

�x2 cos �

x2; x ¤ 0

0; x D 0:

6.6. References 159

(a) Prove that f is differentiable on all of Œ0; 1�.(b) Is f absolutely continuous? (Consider intervals of the form

1pnC1

; 1pn

�.)

(c) Show that LR x0 f

0 ¤ f .x/ for x 2 Œ0; 1� since LR x0 f

0 does not

exist. (Again, consider intervals of the form

1pnC1

; 1pn

�.)

90. In some situations, the Riemann integral can be extended to handle iso-lated points near which a function is unbounded. In particular, when afunction f is unbounded at 0, we can attempt to define R

R x0f by

RZ x

0

f D lim˛!0C

RZ x

˛

f .

Show that the extension works for the function f 0 in exercise 88 andthat in this extended sense

RZ x

0

f 0 D f .x/ � f .0/

for all x 2 .0; 1/.

91. Give an example of an unbounded function that is Lebesgue integrablebut for which it is impossible to use the technique of exercise 90 toassign a value to the Riemann integral.



America.Daniell, P.J. (1918), A general form of integral, Annals of Mathematics, Sec-

ond Series 19 (4): 279–294. JSTOR 1967495.Hewitt, E. and K. Stromberg (1975). Real and Abstract Analysis. Springer.Taylor, A.E. (2010). General Theory of Functions and Integration. Dover.

CHAPTER 7

The Gauge Integral

As we have seen, the Lebesgue integral has far more powerful convergenceproperties than the Riemann integral. The Lebesgue integral also has signif-icantly more flexible fundamental theorems of calculus. Why?

From the Darboux-integral point of view, expanding the set of partitions ofŒa; b� to include measurable partitions in addition to partitions by intervalscreates much larger sets of upper and lower sums. These larger sets allowthe infimum of the upper sums to meet the supremum of the lower sumsmore often, thus creating a much larger set of integrable functions. Morespecifically, the use of measurable partitions makes it possible to choose setson which the supremum and infimum of the integrand f are nearly identical.A bound-and-telescope argument shows that whenever we can ensure thatsupE f � infE f < "= .b � a/ for each set E in a partition of Œa; b�, theupper and lower Lebesgue-Young sums for the partition will be within " ofeach other.

The connection between the values of the integrand and the elements ofa partition is even stronger in the original conception of the Lebesgue inte-gral. The use of inverse images of small intervals forces the upper and lowerbounds on the Lebesgue integral together. Fundamentally, this forcing is pos-sible because measurable sets allow us to guarantee that whenever x and ybelong to the same set in the partition of Œa; b�, f .x/ and f .y/ are nearlythe same.

While the Lebesgue integral is based on a strong connection between thevalues of the integrand f and the sets used in the partition, the Riemannintegral does not support any such connection. At its core, the definition ofthe Riemann integral is uniform. A single ı > 0 is selected to control thebehavior of the entire partition with no accommodation available to accountfor the local behavior of f . From this point of view, it is not particularly

161

162 CHAPTER 7. The Gauge Integral

surprising that convergence theorems for the Riemann integral rely on uni-form convergence.

What would happen if, instead of employing a single value of ı > 0 forthe entire interval of integration, the value of ı is allowed to vary from pointto point in a fashion analogous to the way that ı in the "-ı definition ofcontinuity (as opposed to uniform continuity) depends on both the value of" and a particular point? Such an approach would move us away from anessentially uniform definition of integration and support a tie between thevalues of f and a partition of Œa; b�. Although allowing the value of ı todepend on x 2 Œa; b� cannot preclude the integrand taking on significantlydifferent values within a subinterval, as we shall see, the contribution suchsubintervals make to the sum can be controlled by restricting their lengths.

Independently, Ralph Henstock (1923–2007) and Jaroslav Kurzweil (b.1926) took up this line of inquiry around 1960. The result is an integralthat avoids the complications of measure theory while extending the set ofintegrable functions. In particular, any derivative is integrable in this newsetting. (Compare theorem 71 on page 191 with example 47 on page 192.)

We will refer to this type of integral as a gauge integral. The gauge integralmaintains the Riemann integral’s focus on the integrand’s domain, uses onlyintervals in the partition, and is based on Riemann sums. The power (andcomplexity) of the gauge integral stems from the localized control a gaugeexerts on the size of the intervals in a partition.

7.1 Definition and basic examplesRecall that a function f W Œa; b� ! R is Riemann integrable over Œa; b� ifthere is a real number A such that for any " > 0 we can find a ı > 0 so that

jSR.f;P/ � Aj DˇˇnXkD1

f .tk/ .xk � xk�1/ � A

ˇˇ < "

whenever P is a tagged partition of Œa; b� with jxk � xk�1j < ı for all k D1; 2; : : : ; n. The gauge integral modifies this definition slightly by replacingı with ı .tk/.

Definition 21 (ı gauge). Let E be a subset of R. A gauge on E is a positivefunction ı W E ! RC.

Think of a gauge ı .t/ as comparable to creating a function by piecingtogether the required values of ı in the "-ı definition of continuity. In the

7.1. Definition and basic examples 163

case of the gauge, however, we are creating a function to verify integrabilityrather than continuity.

Definition 22 (ı-fine). Given a gauge ı, a tagged partitionP D


��nkD1

is said to be ı-fine if for all k D 1; 2; : : : ; n wehave jxk � xk�1j < ı .tk/.1

Definition 23 (Gauge integrable). A function f W Œa; b� ! R is gaugeintegrable over Œa; b� if there is a real number A satisfying the followingcondition: For any " > 0 there is a gauge ı such that

jSR.f;P/ � Aj DˇˇnXkD1

f .tk/ .xk � xk�1/ � A

ˇˇ < "

for any ı-fine partition P of Œa; b�. In this case, the value of the integral isgR baf D A.

Notice the strong relationship between the gauge integral and the Riemannintegral. In particular, we are once again working with Riemann sums. Alsotake note of the fact that if f is a Riemann integrable function, then f canbe shown to be gauge integrable using the uniform gauge ı .t/ D ı0 whereı0 > 0 is taken from the definition of the Riemann integral. We will ignorethis fact in the initial examples in order to investigate the gauge integralbased on its definition.

Just as continuity has an alternative characterization that uses open sets,there is an alternative and equivalent (see exercise 1) definition of gauge interms of open intervals.2

Definition 24 (� gauge). LetE be a subset of R. A gauge onE is a function� from E to the set of open intervals in R such that for any t 2 E, we havet 2 � .t/ :

Definition 25 (� -fine). Given a gauge � , a tagged partition P Df.tk ; Ik/gis said to be � -fine if for all k D 1; 2; : : : ; n we have Ik � � .tk/.

While the second definition may appear to be more awkward, in practice itis often simpler and more transparent to use. It is typically (but not always)easier to use the first type of gauge when working with specific functions

1 Some texts require instead that Œxk�1; xk� � .tk � ı .tk/ ; tk C ı .tk//. (See Exercise 7.)2 One could more closely mimic the open set definition of continuity and take � .t/ to be an

open set containing t . Since we eventually want to use partitions consisting of intervals, thereis nothing to be gained from this generality.

and more efficient to use the second type when proving general results. Thisstate of affairs should not surprise us as it is similar to the situation for the"-ı versus the open-set characterizations of continuity.

We will consistently use ı or � to indicate which type of gauge we areusing.

Example 35 (Constant functions). Let f .x/ D c on Œa; b�. Fix " > 0 andset � .t/ D .a � 1; b C 1/. If P Df.tk ; Ik/g is a � -fine tagged partition ofŒa; b�, then

SR.f;P/ DXPc ��xk D c .b � a/

so thatjSR.f;P/ � c .b � a/j D 0 < ".

We conclude that the constant function f .x/ D c is gauge integrable overŒa; b� with

gZ b

a

f D c .b � a/ :


d .x/ D

�1; x 2 Q

0; x 62 Q

is gauge integrable over Œa; b�.To see why, suppose that " > 0 and let frkg be an enumeration of the

rational numbers in Œa; b�. Define a gauge ı on Œa; b� by

ı .t/ D

� "

2k; t D rk

1; t 62 Q.

If P is a ı-fine tagged partition of Œa; b�, we can separate P into Pr , thosetagged intervals with rational tags, and Pi , those with irrational tags. Then

jSR.f;P/ � 0j DXPd .tk/ ��xk

DXPr

1 ��xk CXPi

0 ��xk

<Xn

"

2nD ".

Thus the Dirichlet function is gauge integrable with gR ba d D 0.

7.1. Definition and basic examples 165

Example 37 (Identity function). Let f .x/ D x: We know that gR 20 f D 2

since RR 20 f D 2, but we will verify this result directly from the definition

of the gauge integral.Let " > 0 be given. Define a gauge ı by ı .t/ D "=2 and suppose that

P D˚�tk; Œxk�1; xk �

��nkD1

is a ı-fine tagged partition of Œ0; 2�. Because both

tk and xkCxk�12

fall in the interval Œxk�1; xk�,ˇtk �

xk C xk�1

2

ˇ� xk � xk�1 < ı .tk/ D "=2:

Hence

jSR.f;Pn/ � 2j DˇˇnXkD1

tk .xk � xk�1/ �1

2

nXkD1

�x2k � x

2k�1

�ˇˇ�

nXkD1

ˇ�tk �

xk C xk�1

2

�ˇ.xk � xk�1/

<

nXkD1

"

2.xk � xk�1/

D"

2� 2 D ".

We conclude that the identity function is gauge integrable over Œ0; 2� withgR 20 f D 2.

The first and third examples used gauges whose widths do not dependon x 2 Œa; b�. The use of a constant-width gauge reflects the fact that theconstant and identity functions are Riemann integrable. This is not the casefor the Dirichlet function nor for the next example.

Also note the appearance of the telescoping sum 12

PnkD1

�x2k� x2

k�1

�D

2 in the last example. When proving that a function is integrable, we willoften replace the value of the integral by such a telescoping sum.


f .x/ D

(1px; x > 0

0; x � 0

is not Riemann integrable over Œ0; 1� since f is unbounded. However, f isgauge integrable over Œ0; 1� with g

R 10f D 2.

To verify this, let 0 < " � 1 be given. We will use a gauge � on Œ0; 1� ofthe form

� .t/ D

( t � a2 .t/ ; t

a2.t/

�; t > 0�

�"2; "2�; t D 0

where 0 < a .t/ < 1 is yet to be defined. Suppose that P D˚�tk; Œxk�1; xk�

��nkD1

is a � -fine tagged partition of Œ0; 1�. Then, for k > 1,2p

xkCpxk�1

and 1ptk

both fall in the intervalh

1pxk; 1p

xk�1

iso thatˇ

2pxk C

pxk�1

�1ptk

ˇ<

1pxk�1

�1pxk

<1p

a2 .tk/ � tk�

1ptk=a2 .tk/

D1ptk

�1

a .tk/� a .tk/

�.

If we take a .t/ D 1 � "pt

4, then (recalling that 0 < "; t � 1)

1

a .t/� a .t/ D

2"pt � 1

4"2t

4 � "pt

<2

3"pt :

Hence ˇ2

pxk C

pxk�1

�1ptk

ˇ< ".

Due to the way � is defined, the tagged interval .t0; Œx0; x1�/ cannot havex0 D 0 unless t0 D 0. Thus when k D 1, f .t1/ D f .0/ D 0 and 0 < x1 <"2. Hence

jSR.f;Pn/ � 2j DˇˇnXkD2

1ptk.xk � xk�1/ � 2

nXkD1

�pxk �

pxk�1

�ˇˇ�

ˇˇnXkD2

�1ptk�

2pxk C

pxk�1

�.xk � xk�1/

ˇˇC 2px1

<

nXkD2

" � .xk � xk�1/C 2"

< 3".

Thus f is gauge integrable with gR 10f D 2.

7.2. The art of constructing gauges 167

Notice the structure of the last two examples. In both cases, a specialvalue sk is chosen for the subinterval Œxk�1; xk� in such a way that, whensk � .xk � xk�1/ is expanded, the sum

PnkD1 sk � .xk � xk�1/ simpli-

fies by telescoping. The gauge is selected to control jf .tk/ � skj so thatwe can also bound and telescope

Pk jf .tk/ � skj .xk � xk�1/ : Though it

need not always be the case, jf .tk/ � skj often is bounded by controllingjf .xk/ � f .xk�1/j :

Before turning to more general results, we need to verify that, given agauge � , a � -fine partition exists. If we could construct a gauge � for whichthere were no � -fine tagged partitions, we would be in much the same sit-uation as if the real-valued ı used to bound the mesh of a partition in theRiemann integral were allowed to be negative. The condition for gauge in-tegrability would be satisfied vacuously. In this case, any value of A wouldsatisfy the conditions of the definition.

Theorem 48 (Cousin, 1895). If � is a gauge on Œa; b�, then there is a � -finetagged partition of Œa; b�.3

Proof. Let S D fs 2 Œa; b� W there exists a � -fine tagged partition of Œa; s�gand let ˇ be the supremum of the bounded and nonempty set S . We willprove that ˇ 2 S and that ˇ D b. Hence b 2 S which means that there is a� -fine tagged partition of Œa; b�.ˇ 2 S : By the definition of supS , the open interval � .ˇ/ \ .a; ˇ/ must

contain an element c of S . Since c 2 S , there is a � -fine partition P of Œa; c�.Add .ˇ; Œc; ˇ�/ to P to create a � -fine tagged partition of Œa; ˇ� thus showingthat ˇ 2 S .ˇ D b: Since b is an upper bound for S , ˇ � b. Suppose that ˇ < b.

Because ˇ 2 S , there is a � -fine tagged partition P of Œa; ˇ�. Choose d 2� .ˇ/\ .ˇ; b/ and add .ˇ; Œˇ; d �/ to P . The result is a � -fine tagged partitionof Œa; d �. Since d > ˇ, we have contradicted the fact that ˇ D supS . Henceˇ D b.

7.2 The art of constructing gaugesWhile the general role of a gauge � is to ensure that the Riemann sum ofany � -fine tagged partition of Œa; b� is close to the posited value of the inte-gral, it is not yet clear how such a task is accomplished. This general objec-tive is usually achieved by accomplishing one or more specific jobs: force a

3 In fact, there are infinitely many � -fine tagged partitions of Œa; b�.

particular point to be a tag, ensure that the contribution made by a particularpoint is small, guarantee that the value of the function throughout a subin-terval is close to the functions’s value at the interval’s tag. In the course of aproof or when working with a particular integral, one often wants to achievemore than one such objective. This is easily accomplished using the inter-section of gauges. If �1 and �2 are two gauges on Œa; b� and � is defined by� .t/ D �1 .t/\�2 .t/, then � is also a gauge on Œa; b� and any � -fine partitionis also �1-fine and �2-fine. Thus any � -fine partition will have the propertiesthat both �1 and �2 were designed to guarantee. The point is that one candeal with individual conditions separately and then easily create a gauge thathandles all the conditions at once using intersection. The same goals can beaccomplished with delta gauges using ı .t/ D min fı1 .t/ ; ı2 .t/g.

The following examples illustrate some of the ways that gauges can beused.

7.2.1 Forced tags and tag splittingExample 39 (Forced tags). Design a gauge � on Œ0; 1� that will force 0; 1

2,

and 1 to be tags of any � -fine partition.Take

� .t/ D

8<ˆ:

�0; 12

�; t 2

�0; 12

��12; 1�; t 2

�12; 1�

�t � 1

2; t C 1

2

�; t D 0; 1

2; 1:

A tagged interval in a � -fine partition can only contain 12

if its tag is 12

. Thesame is true for 0 and 1.

In a similar fashion, we can use a gauge � to force any finite set to beincluded among the tags of any � -fine partition. Sometimes it is convenientto assume that these tags are also division points of the partition. This can beaccomplished by tag splitting: if tj is not already a division point, replace.tj ; Œxj�1; xj �/ with .tj ; Œxj�1; tj �/ and .tj ; Œtj ; xj �/. The resulting partitionis still � -fine and has the same Riemann sum as the original partition. Tagsplitting can be combined with an appropriately chosen gauge � to force� -fine tagged partitions to be refinements of a given partition.

7.2.2 Controlling small setsThe next two examples illustrate how gauges can be used to control the con-tribution of a small set.

Example 40 (Finite set). Let f be a function on Œa; b�. Given a finite setS D fckg

nkD1 from Œa; b� and " > 0, construct a gauge � so that if P is a � -

fine tagged partition of Œa; b�, then the total contribution to SR .f;P/ madeby tagged intervals whose tags come from S is bounded by ".

Let

� .t/ D

8<: t � "

2n.jf .t/jC1/; t C "

2n.jf .t/jC1/

�; t 2 S

.t � 1; t C 1/ ; t 62 S:

If ti D ck is a tag in the � -fine partition P , then the contribution to SR .f;P/made by that tagged interval4 satisfies

jf .ti /�xi j < jf .ck/j2"

2n .jf .ck/j C 1/<"

n.

Hence the maximal contribution to SR .f;P/ made by tagged intervals withtags in S is less than ".

Example 41 (Sets of measure zero). Suppose that f is a function on Œa; b�with jf j � B . Given a set Z of measure zero and " > 0, construct a gauge� so that for any � -fine tagged partition of Œa; b�, the contribution to theRiemann sum made by terms with tags in Z is at most ".

Since Z has measure zero, it can be covered by a countable set of openintervals fIkg such that

Pk l .Ik/ < "=B . If t 2 Z, then t 2 Ik for some k.

Let k .t/ be the smallest such index and define

� .t/ D

�Ik.t/; t 2 Z

.t � 1; t C 1/ ; t 62 Z:

Suppose now that P D˚�tk;Jk

��nkD1

is a � -fine tagged partition of Œa; b�.If several tags satisfy k

�ti1�D k

�ti2�D � � � D k .tim/ D k0, then the

corresponding subintervals will all fit inside Ik0 . Since the subintervals donot overlap, the total length of the intervals corresponding to those tags canbe at most that of Ik0 . Consequently, the total contribution of all the termswith tags in Z will be at mostˇ

ˇXti2Z

f .ti / l .Ji /

ˇˇ �X

k

B � l .Ik/ < ".

4 If a tag is used for two adjoining intervals, we can use tag joining to treat them as a singleinterval.

7.2.3 Controlling function differencesAs suggested by our previous work, it is often helpful to boundjf .tk/ � f .wk/j where tk is the given tag for the interval Œxk�1; xk � andwk is the tag that we wish we had. Often gauges can be used to bound thisdifference so that

Pk jf .tk/ � f .wk/j .xk � xk�1/ can be telescoped.

Example 42. Let

f .x/ D

(1x; x > 0

0; x D 0

and suppose that 0 < " < 1. Find a gauge ı on Œ0; 1� so that forany ı-fine tagged partition P D

˚�tk; Œxk�1; xk�

��nkD1

of Œ0; 1�, we havejf .xk�1/ � f .tk/j < " when tk > 0.5

Set

ı .t/ D

("t2

1C"t; t > 0

1; t D 0:

Then when tk > 0, we have xk�1 > tk �"t2k

1C"tkso that

0 � f .xk�1/ � f .tk/

<1

tk �"t2k

1C"tk

�1

tk

D1C "tk

tk�1

tkD ".

Of course the point of all these gauge constructions is to gain the capacityto create a gauge to verify that a function is integrable. We close this sectionwith such an example.

Example 43. Show that the unbounded function

f .x/ D

(.�1/k k; x 2

�1kC1

; 1k

�0; x D 0;

is gauge integrable over Œ0; 1� with

gZ 1

0

f D

1XkD1

.�1/k

k C 1D ln 2 � 1:

5 Note that this bound does not prove that f is gauge integrable since we have not shown thatPk fk .xk�1/ .xk � xk�1/ telescopes or otherwise simplifies.

Begin by noting that f is constant on the interval�1kC1

; 1k

�and that

gR 1k1kC1

f D .�1/k k�1k� 1kC1

�D .�1/k

kC1(see exercise 13). With this fact in

mind, we want to construct a gauge so that

1. the union of the intervals with tags in�1kC1

; 1k

�approximates

�1kC1

; 1k

�and

2. a sufficient number of the intervals�1kC1

; 1k

�are considered to get a

good approximation toP1kD1

.�1/k

kC1.

To this end, let " > 0 and define our gauge by

� .t/ D

( 1kC1

; 1kC "

2k�2k

�; t 2

�1kC1

; 1k

�;�

� "2; "2

�; t D 0:

Now suppose that P D˚�tk; Œxk�1; xk �

��nkD1

is a � -fine partition of Œ0; 1�.Let m be the natural number satisfying 1

mC1< x1 �

1m

. Note that the tagfor Œ0; x1� must be t1 D 0 because no other value can tag an interval with aleft endpoint of 0. If we express the union of the intervals from P with tagsin�1kC1

; 1k

�as ŒzkC1; zk�, then SR.f;P/ can be expressed as

SR.f;P/ DmXkD1

.�1/k k .zk � zkC1/ .

Since z1 D 1 and 1kC1

< zkC1 for 1 � k � m,

1

k� zk <

1

kC

"

2k � 2kfor 1 � k � mC 1.

Thus for 1 � k � m we can use the fact that k�1k� 1kC1

�D 1

kC1to see that

1

k C 1�"

2

1

2k< k

�1

k�

�1

k C 1C

"

2 .k C 1/ � 2kC1

��< k .zk � zkC1/

< k

�1

kC

"

2k � 2k�

1

k C 1

�D

1

k C 1C"

2

1

2k.

Hence ˇˇSR.f;P/ �

mXkD1

.�1/k

k C 1

ˇˇ �

mXkD1

ˇk .zk � zkC1/ �

1

k C 1

ˇ

<

mXkD1

"

2

1

2k<1

2".

As 1mC1

< x1 <"2

, the properties of alternating series tell us thatˇˇ1XkD1

.�1/k

k C 1�

mXkD1

.�1/k

k C 1

ˇˇ < "

2.

Hence

jSR.f;P/ � .ln 2 � 1/j �ˇˇSR.f;P/ �

mXkD1

.�1/k

k C 1

ˇˇ

C

ˇˇmXkD1

.�1/k

k C 1�

1XkD1

.�1/k

k C 1

ˇˇ < ".

We conclude that f is gauge integrable over Œ0; 1� with

gZ 1

0

f D

1XkD1

.�1/k

k C 1D ln 2 � 1:

Note that f , being unbounded, is not Riemann integrable.

7.3 Basic integrability resultsWe begin by formally recording our previously noted connection betweenthe Riemann and gauge integrals.

Theorem 49 (Riemann H) gauge). Any Riemann integrable function isgauge integrable and the integrals agree.

Proof. Exercise 2.

In many contexts we do not need to identify the value of an integral. Wesimply want to verify the integrability of a function. As with the Darboux in-tegral there is a form of the Cauchy criterion that guarantees a gauge integralexists without requiring us to identify a specific value for the integral.

Theorem 50 (Cauchy criterion). Let f be a function on Œa; b�. Then f isgauge integrable over Œa; b� if and only if for each " > 0 there is a gauge �on Œa; b� such that

jSR.f;P1/ � SR.f;P2/j < "

for any pair of � -fine tagged partitions P1 and P2 of Œa; b�.

7.3. Basic integrability results 173

Proof. The necessity of the condition is left as an exercise (exercise 37).To prove sufficiency, let f�ng be a sequence of gauges on Œa; b� such that

jSR.f;P1/ � SR.f;P2/j <1

n

whenever P1 and P2 are two �n-fine tagged partitions of Œa; b�. Define asecond sequence of gauges byb�n .t/ D \niD1�i .t/ and let fPng be a corre-sponding sequence ofb�n-fine tagged partitions of Œa; b�. By construction, ifm > n then Pm is alsob�n-fine. Hence

jSR.f;Pm/ � SR.f;Pn/j <1

N

whenever n;m > N . Thus fSR.f;Pn/g is a Cauchy sequence and so con-verges to some value A.

Now let " > 0 and select a natural number n0 so that 1n0

< "2

andˇSR

�f;Pn0

�� A

ˇ< "=2. If P is ab�n0 -fine tagged partition, then

jSR.f;P/ � Aj �ˇSR.f;P/ � SR.f;Pn0/

ˇCˇSR.f;Pn0/ � A

ˇ< ".

Hence f is gauge integrable over Œa; b� with gR ba f D A:

The next theorem is an example of the type of circumstance in which theexistence but not the value of a gauge integral is important.

Theorem 51 (Subintervals 1). Suppose that the function f is gauge inte-grable over Œa; b� and Œc; d � � Œa; b�. Then f is gauge integrable over Œc; d �.

Proof. For definiteness, consider the case where a < c < d < b, the caseswhen a D c or d D b being simpler.

Let " > 0 and let � be a gauge on Œa; b� such that


whenever P1 and P2 are two � -fine tagged partitions of Œa; b�. Let b� bethe restriction of � to Œc; d � and suppose that Q1 and Q2 are two b�-finepartitions of Œc; d �. Letb�a andb�b be the restrictions of � to Œa; c� and Œd; b�respectively and let Ra and Rb be b�a-fine and b�b-fine tagged partitionsof Œa; c� and Œd; b� respectively. Then S1 D Ra [ Q1 [ Rb and S2 DRa [Q2 [Rb are � -fine tagged partitions of Œa; b�.6 Hence

jSR.f;Q1/ � SR.f;Q2/j D jSR.f;S1/ � SR.f;S2/j < ".

6 Note that, since the tagged partitions are on non-overlapping intervals, [ in this case is theusual union of sets. If we were working with tagged partitions of a common interval, then [would refer to a refinement.

We conclude that f is gauge integrable over Œc; d � by the Cauchy criterion.

The next theorem serves as a converse:

Theorem 52 (Subintervals 2). Suppose that the function f is gauge in-tegrable on Œa; c� and Œc; b�. Then f is gauge integrable on Œa; b� with

gR ba f D

gR ca f C

gR bc f .

Proof. Exercise 39.

In the future, rather than assigning a new symbol when a gauge � is re-stricted to a subinterval, we will continue to use � . This slight abuse of no-tation will cause no harm and will make our notation cleaner. Following thisconvention, we would refer to Ra, Rb , and Q1 in the above proof as � -finetagged partitions of Œa; b�, Œd; b�, and Œc; d � respectively.

As with the Lebesgue integral, we can safely ignore changes to a functionwhen they occur on sets of measure zero.

Theorem 53 (Zero a.e.). If f D 0 a.e. on Œa; b�, then f is gauge integrablewith g

R baf D 0.

Proof. Let " > 0 be given and define

En D fx 2 Œa; b� W n � 1 < jf .x/j � ng; n D 0; 1; 2; : : : :

When n > 0, � .En/ D 0 so (using the union of a collection of open inter-vals) there is an open set Gn with En Gn and � .Gn/ < "=n2n . As Gnis an open set, for each x 2 En there is an rx > 0 so that y 2 Gn wheneverjy � xj < rx . Set E D [nEn and define a gauge ı by

ı .t/ D

�rt ; t 2 E

1; t 62 E:

Now suppose that P Df.ti ; Ii /g is a ı-fine tagged partition of Œa; b�. LetPn be the set of intervals with tags in En. Then for n > 0, [PnIi Gn and,since the intervals in Pn do not overlap,X

Pn

l .Ii / D � .[PnIi / � � .Gn/ < "=n2n:

Thus

jSR .f;P/j D

ˇˇ 1XnD0

XPn

f .ti / l .Ii /

ˇˇ � 1X

nD1

nXPn

l .Ii / <

1XnD1

"

2nD ".

Hence f is gauge integrable with gR ba f D 0.

7.3. Basic integrability results 175

Corollary 54 (Equal a.e.). If f is gauge integrable on Œa; b� and f D g

a.e. on Œa; b�, then g is gauge integrable on Œa; b� with gR ba f D

gR ba g.

Proof. Exercise.

The proof of theorem 53 uses sums taken over subsets of a tagged par-tition. We have used similar tactics before, but this technique plays a morecentral role in proofs related to the gauge integral.

On a seemingly unrelated note, one of the standard techniques when work-ing with Riemann integrals is to create a common refinement. When this isdone, however, the tags are lost. We can use a gauge and tag splitting tomake sure that the tags are preserved, but then the intervals may change andthe new intervals will have additional tags. Consequently, the common re-finement approach generally is not effective when working with the gaugeintegral.

Both of these issues are addressed by relating partial Riemann sums tosums of integrals over subintervals. The key observation is that sums of inte-grals over subintervals do not depend on a choice of tags. The next definitionand the two theorems that follow provide us with the needed vocabulary andtools to make effective use of this strategy.

Definition 26 (Partial tagged partition). A partial tagged partition ofŒa; b� is a finite set of tagged, closed intervals f.ti ; Ii /g where

1. ti 2 Ii Œa; b�, and

2. Iiand Ij are non-overlapping when i ¤ j .

A partial tagged partition is like a tagged partition of Œa; b� except that theunion of the intervals in a partial tagged partition need not (but may) coverall Œa; b�. Typically a partial tagged partition arises from considering a subsetof a tagged partition.

Lemma 55 (Henstock). Let f be a gauge integrable function on Œa; b�. Fix" > 0 and suppose further that � is a gauge on Œa; b� for which any � -finepartition P satisfies ˇ

ˇSR.f;P/ � gZ b

a

f

ˇˇ < ":

If bP D f.ti ; Ii /g is a � -fine partial tagged partition of Œa; b�, thenˇˇXbP

�f .ti / ��xi �

gZIi

f

ˇˇ � ":

Alternatively, we can express the last inequality asˇˇSR.f;bP/ �XbP

gZIi

f

ˇˇ � ".

Proof. The set Œa; b� n [i Ii consists of a finite number (perhaps zero) ofdisjoint intervals. Add their endpoints and label the resulting closed in-tervals as J1; J2; : : : ; Jm. By theorem 51, f is gauge integrable over Jk ,k D 1; 2; : : : ; m. Thus for any > 0 and for each k D 1; 2; : : : ; m, we canfind a � -fine partition Pk of Jk for whichˇ

SR.f;Pk/ � gZJk

f

ˇ<

m:

Putting the partial tagged partitions together, Q DbP[P1[P2[� � �[Pmforms a � -fine tagged partition of Œa; b�. Henceˇ

ˇSR.f;bP/ �XbPgZIi

f

ˇˇ

D

ˇˇSR.f;Q/ � g

Z b

a

f �

mXkD1

�SR.f; Pk/ �

gZJk

f

�ˇˇ

�

ˇˇSR.f;Q/ � g

Z b

a

f

ˇˇC

mXkD1

ˇSR.f; Pk/ �

gZJk

f

ˇ< "C .

Since > 0 was arbitrary, we conclude thatˇˇSR.f;bP/ �XbP

gZIi

f

ˇˇ � ":

Corollary 56. Under the hypotheses of Henstock’s lemma,XbPˇf .ti / ��xi �

gZIi

f

ˇ� 2"

and ˇˇXbP

�jf .ti /j ��xi �

ˇgZIi

f

ˇ�ˇˇ � 2"

7.4. Absolute integrability 177

or, equivalently, ˇˇSR.jf j ;bP/ �XbP

ˇgZIi

f

ˇˇˇ � 2":Proof. Let bPC be those intervals from bP for which f .ti / ��xi� g

RIif � 0.

Define bP� analogously. Then both bPC and bP� are partial tagged partitionssatisfying the conditions of Henstock’s lemma. ThusX

bPˇf .ti / ��xi �

gZIi

f

ˇ

D

ˇˇXbPC

�f .ti / ��xi �

gZIi

f

ˇˇCˇˇXbP�

�f .ti / ��xi �

gZIi

f

ˇˇ� 2".

The second inequality is a consequence of the fact that for a; b 2 R,jjaj � jbjj � ja � bj so thatˇˇXbP

�jf .ti /j ��xi �

ˇgZIi

f

ˇ�ˇˇ �XbPˇjf .ti /j ��xi �

ˇgZIi

f

ˇˇ

�XbPˇf .ti / ��xi �

gZIi

f

ˇ� 2":

7.4 Absolute integrabilityAt the end of Section 7.2, we showed that the function

f .x/ D

(.�1/k k; x 2

�1kC1

; 1k

�0; x D 0

is gauge integrable. The function jf j is not gauge integrable, however. Tosee why, suppose that jf j is gauge integrable over Œ0; 1�. Then by theorems51 and 52 (page 173),

gZ 1

0

jf j D gZ 1

n

0

jf j C gZ 1

1n

jf j > gZ 1

1n

jf j .

But jf j is constant on each of the intervals�1kC1

; 1k

�, 1 � k < n, so that

gZ 1

1n

jf j D

n�1XkD1

gZ 1

k

1kC1

jf j D

n�1XkD1

1

k C 1.

Thus gR 10 jf j cannot be finite and jf j is not gauge integrable.

We introduce the terms absolutely integrable to describe those func-tions f for which f and jf j are integrable and conditionally integrable todescribe those cases when f is integrable but jf j is not. For the Lebesgueintegral, the integrability of a measurable function f is equivalent to the inte-

grability of jf j andˇ

LR ba f

ˇ� L

R ba jf j. As just illustrated, this equivalence

is not true for the gauge integral. A function can be gauge integrable withoutbeing absolutely gauge integrable because a gauge can force a Riemann sumto consider both positive and negative contributions in a balanced way.

Initially, the fact that some functions are conditionally gauge integrableappears to be interesting but not particularly consequential. In fact, the dis-tinction between integrable and absolutely integrable complicates the devel-opment of the theory of convergence for the gauge integral. For example,many of the convergence proofs for the Lebesgue integral rely on the factthat measurable functions are closed under taking linear combinations, max-imums, minimums, supremums, and infimums. Functions constructed withthose operations can be used freely in Lebesgue integral proofs. This is notthe case for the gauge integral. Given a pair of gauge integrable functions fand g and a scalar c, f Cg and cf are gauge integrable, but h1 D max ff; ggand h2 D min ff; ggmay not be. However, as we shall see, h1 D max ff; ggand h2 D min ff; gg are gauge integrable when f and g are absolutelygauge integrable.

So when is a gauge integrable function f absolutely integrable over Œa; b�?It is relatively straightforward to prove that a gauge integrable function f isabsolutely integrable if and only if both f C and f � are gauge integrable(see exercise 51). This characterization suffices for some, but certainly notall, of our purposes. For example, it is tempting to try to prove that the max-imum of two absolutely gauge integrable functions is gauge integrable us-ing max ff; gg D 1

2.f C g C jf � gj/, but we cannot easily prove that the

sum or difference of two absolutely gauge integrable functions is absolutelygauge integrable since .f C g/C is not f C C gC.

For a measurable function, the only requirement for absolute Lebesgueintegrability is that the Lebesgue sums be bounded. We will eventually provea related but rather more subtle result for the gauge integral. However, unlike


the situation for the Lebesgue integral, the gauge theorem does not followdirectly or easily from the definition of the gauge integral. The pathway runsthrough the concept of bounded variation.

Definition 27 (Variation). Let f be a function on Œa; b�. Given a partitionP DfŒxk�1; xk�g of Œa; b�, the variation of f with respect to P is

V .f;P/ DXk

jf .xk/ � f .xk�1/j .

The variation of f over Œa; b� is

V ba f D supP

V .f;P/

where the supremum is taken over all partitions P of Œa; b�. If V ba f is fi-nite, then f is said to be of bounded variation on Œa; b�. The set of all suchfunctions is denoted by BV .Œa; b�/.

Example 44. The gauge integrable function

f .x/ D

(.�1/k k; x 2

�1kC1

; 1k

�0; x D 0

(see example 43, page 170) does not belong to BV .Œ0; 1�/.To see why, use the partition PnD

˚�0; 1n

�;�1n; 1��

. Then

V .f;Pn/ D2XkD1

jf .xk/ � f .xk�1/j

Dˇ.�1/n n � 0

ˇCˇ1 � .�1/n n

ˇ� 2n � 1.

Since 2n � 1 is not bounded, f 62 BV .Œ0; 1�/.

Theorem 57 (Monotone functions). If f is a monotone function on Œa; b�,then f 2 BV .Œa; b�/ with V ba f D jf .b/ � f .a/j.

Proof. For definiteness, assume that f is monotone increasing. LetP DfŒxk�1; xk�g be a partition of Œa; b�. Then

V .f;P/ DXk

jf .xk/ � f .xk�1/j

DXk

.f .xk/ � f .xk�1// D f .b/ � f .a/ .

Hence V ba f D jf .b/ � f .a/j.The verification for monotone decreasing functions is similar.

Step functions form another class of functions with bounded variation.Step functions have appeared in the exercises (see exercises 13, 14, and 31)and will play an important role in Section 7. However, we have not yet for-mally defined them. We do so now.

Definition 28 (Step function). A function f defined on Œa; b� is a step func-tion if there is a partition P D fŒxi�1; xi �g of Œa; b� such that f is constanton each open interval .xi�1; xi /.

At first glance, step functions appear deceptively similar to measurablesimple functions. In a sense, step functions are the gauge integral’s analogto the Lebesgue integral’s measurable simple functions. Step functions arebased on partitions (by intervals) and simple functions are based on mea-surable partitions. Despite their apparent similarities, you should not as-sume that step functions and measurable simple functions have the sameproperties. For example, the Dirichlet function is a measurable simple func-tion whose variation is definitely not bounded while all step functions havebounded variation. Both simple and step functions take on a finite number ofvalues, but while simple functions can have an infinite number of points ofdiscontinuity, step functions have only finitely many discontinuities.


f .x/ D

8ˆ<ˆˆ:

1; 0 � x < 12

3; x D 12

2; 12< x < 3

4

5; x D 34

4; 34< x < 1

0; x D 1

is a step function with three points of discontinuity: x D 12

, x D 34

, andx D 1.

Theorem 58 (Step functions). If f is a step function on Œa; b�, then f hasbounded variation.

Proof. Since a step function f can only take on a finite number of values, fis bounded by some value B . Suppose that f has m points of discontinuity.If P DfŒxk�1; xk �g is an arbitrary partition of Œa; b�, then at most 2m of theintervals of P can produce different values for f .xk�1/ and f .xk/ and the

difference can be at most 2B . Thus V .f;P/ DPk jf .xk/ � f .xk�1/j is

bounded by 4mB .

Before developing the relationship between functions of bounded varia-tion and functions that are absolutely gauge integrable, we investigate theproperties of BV .Œa; b�/ and V ba f .

Theorem 59 (BV.Œa; b�) is a vector space). Suppose that f; g2BV.Œa; b�/and that c 2 R. Then f C g and cf are both in BV Œ.a; b/�.

Proof. Exercise.

Theorem 60 (Properties of V ba f ). Suppose that f is a function on Œa; b�.Then

1. V .f;P/ � V .f;Q/ whenever P and Q are partitions of Œa; b� and Qis a refinement of P .If, in addition, f 2 BV .Œa; b�/, then

2. V ba f D Vca f C V

bc f for any c 2 .a; b/ and

3. V xa f and V xa f � f .x/ are increasing functions of x on Œa; b�.

Proof. We leave (1) as an exercise.To prove (2), suppose that f 2 BV .Œa; b�/ and c 2 .a; b/. Then f 2

BV .Œa; c�/ and f 2 BV .Œc; b�/. Fix " > 0 and choose partitions Pa andPb of Œa; c� and Œc; b� such that

V ca f �"

2< V .f;Pa/ � V ca f

andV bc f �

"

2< V .f;Pb/ � V bc f .

Then P D Pa [ Pb is a partition of Œa; b� for which

V ca f C Vbc f � " < V .f;Pa/C V .f;Pb/

D V .f;P/� V ba f .

HenceV ca f C V

bc f � V

ba f .

For the reverse inequality, select a partition Q of Œa; b� satisfying

V ba f � " < V .f;Q/ � V ba f .

Create a new partition Q� by inserting c as a division point and let Qa andQb consist of the intervals from Q� to the left and right of c respectively.Then by (1),

V ba f � " < V .f;Q/� V

�f;Q�

�D V .f;Qa/C V .f;Qb/� V ca f C V

bc f .

Hence

V ba f � Vca f C V

bc f

and equality follows.To prove (3), suppose that a � x < y � b. From (2),

V ya f � Vxa f D V

yx f � 0:

Thus V xa f is increasing. Moreover,

V ya f � Vxa f D V

yx f � jf .y/ � f .x/j � f .y/ � f .x/ .

Hence V xa f � f .x/ is also increasing.

The previous theorem has a corollary that characterizes functions ofbounded variation.

Corollary 61 (BVD difference of increasing functions). Let f be a func-tion defined on Œa; b�. Then f 2 BV .Œa; b�/ if and only if f can be writtenas the difference of a pair of increasing functions on Œa; b�.

Proof. Since monotone functions on Œa; b� have bounded variation andBV Œ.a; b/� is a vector space, any function on Œa; b� that can be expressedas the difference of increasing functions is in BV Œ.a; b/�.

For the converse, note that f .x/ D V xa f ��V xa f � f .x/

�:

By writing f .x/ D�V xa f C x

��V xa f � f .x/ � x

�we can replace

“increasing” with “strictly increasing” in the corollary.We now come to the key theorem that connects BV .Œa; b�/ with abso-

lutely integrable functions on Œa; b�. In the proof, watch for the use of agauge and tag cutting to force a tagged partition to be a refinement of agiven partition.

Theorem 62 (Absolute integrability and BV.Œa; b�/). Suppose that f is agauge integrable function on Œa; b�. Then f is absolutely gauge integrableon Œa; b� if and only if F .x/ D g

R xa f has bounded variation on Œa; b�. In

this case,

V ba F DgZ b

a

jf j .

Proof. Begin by noting that if P DfŒxk�1; xk�g is a partition of Œa; b�, then

V .F;P/ DXk

jF .xk/ � F .xk�1/j DXk

ˇgZ xk

xk�1

f

ˇ. (7.1)

Suppose that f is absolutely integrable and let P DfŒxk�1; xk�g be a par-tition of Œa; b�. Then

V .F;P/ DXk

ˇgZ xk

xk�1

f

ˇ�Xk

gZ xk

xk�1

jf j D gZ b

a

jf j .

Hence V ba F �gR ba jf j.

For the converse, suppose that F 2 BV .Œa; b�/. Given " > 0, choose apartition P0 D fŒxk�1; xk�gnkD1 such that

V ba F � "=2 < V .f;P0/ � V ba F .

Set x�1 D a � 1 and xnC1 D b C 1 and define �0 on Œa; b� by

�0 .t/ D

�.xk�1; xk/ ; t 2 .xk�1; xk/

.xk�1; xkC1/ ; t D xk :

Then any �0-fine tagged partition P will include fx0; x1; : : : ; xng among thetags so that, after tag cutting, P is a refinement of P0. Hence, by theorem 60(page 181),

V ba F � "=2 < V .f;P0/ � V .f;P/ � V ba F .

Because f is integrable, we can find a gauge �1 such that any �1-finetagged partition P D

˚�tj ;�yj�1; yj

��satisfiesˇ

ˇSR .f;P/ � gZ b

a

f

ˇˇ < "

4.

Using (7.1) and Corollary 56 of Henstock’s lemma (page 176) applied to allof P we conclude that

jSR .jf j ;P/ � V .F;P/j DˇˇSR .jf j ;P/ �X

k

ˇgZ xk

xk�1

f

ˇˇˇ � "

2.

Set � .t/ D �0 .t/\�1 .t/ and suppose that P is any � -fine tagged partitionof Œa; b�. Thenˇ

SR .jf j ;P/ � V ba Fˇ

� jSR .jf j ;P/ � V .F;P/j CˇV .F;P/ � V ba F

ˇ< ".

Hence jf j is gauge integrable with

gZ b

a

jf j D V ba F

as claimed.

While theorem 62 provides a complete characterization of absolutelygauge integrable functions, it is rather awkward to apply. The next threeresults (the goals of this section) provide tools that are much more straight-forward to use.

Theorem 63 (Comparison test). Let f and g be gauge integrable functionson Œa; b� with jf j � g. Then f is absolutely integrable withˇ

ˇ gZ b

a

f

ˇˇ � g

Z b

a

jf j � gZ b

a

g.

Proof. Set F .x/ D gR xa f , x 2 Œa; b�. If P DfŒxk�1; xk �g is a partition of

Œa; b�, then

V .F;P/ DXk

jF .xk/ � F .xk�1/j

DXk

ˇgZ xk

xk�1

f

ˇ�Xk

gZ xk

xk�1

g D gZ b

a

g.

Thus

V ba F �gZ b

a

g


so that f is absolutely integrable with

gZ b

a

jf j D V ba F �gZ b

a

g:

Corollary 64 (Absolutely integrable functions form a vector space). Letf and g be absolutely gauge integrable functions on Œa; b� and let c be ascalar. Then f C g and cf are also absolutely gauge integrable.

Proof. Exercise.

Theorem 65 (Max and min of absolutely integrable functions). Supposethat f and g are absolutely gauge integrable functions. Then

1. f C and f � are gauge integrable and

2. max ff; gg and min ff; gg are gauge integrable.

Proof. 1. Since f and jf j are gauge integrable, so are f C D 12.jf j C f /

and f � D 12.jf j � f /.

2. Since f � g, jf j, and jgj are gauge integrable and jf � gj �jf j C jgj, jf � gj is gauge integrable by the comparison test.Hence so are min ff; gg D 1

2.f C g � jf � gj/ and max ff; gg D

12.f C g C jf � gj/.

The condition that f and g are absolutely gauge integrable can be relaxeda bit.

Theorem 66 (Max and min of dominated functions). Suppose that f; g,and h are gauge integrable functions on Œa; b�.

1. If f � h and g � h then max ff; gg and min ff; gg are gauge inte-grable.

2. If f � h and g � h then max ff; gg and min ff; gg are gauge inte-grable.

Proof. 1. Suppose that f; g � h. Observe that

jf � gj D 2max ff; gg � f � g � 2h � f � g:

By the comparison test, jf � gj is gauge integrable. Consequently, so are

max ff; gg D1

2.f C g C jf � gj/


and

min ff; gg D1

2.f C g � jf � gj/ .

Negate the functions to prove 2.

This result will be extended to apply to supremums and infimums in thenext section once we have established some convergence results.

7.5 Convergence theoremsThe convergence results for the gauge integral are similar to those for theLebesgue integral. However, the proofs have a rather different flavor.

Theorem 67 (Monotone convergence). Suppose that ffkg is a monotonesequence of gauge integrable functions that converges pointwise to f on

Œa; b�. Then f is gauge integrable if and only ifn

gR ba fk

ois bounded. In this

case,

limk

gZ b

a

fk DgZ b

a

f .

Before turning to the proof, consider how it will need to differ from theproof of the corresponding result for the Lebesgue integral. The measurabil-ity of the limit function was never an issue. All we needed to worry aboutwas the value of the integral. We proved the monotone convergence theoremfor the Lebesgue integral (theorem 37 on page 138) by approximating eachof the functions fk with an increasing sequence of measurable simple func-tions. A new increasing sequence of measurable simple functions converg-ing to f was constructed using maximums and the conclusion of the mono-tone convergence theorem followed from the monotonicity of the Lebesgueintegral.

We know that the gauge integral is monotone and that, with some care,maximums of gauge integrable functions are again gauge integrable. How-ever, we don’t know that measurable simple functions are gauge integrablenor do we know that a gauge integrable function is the limit of a monotonesequence of measurable simple functions. Moreover, proving that the limitof the gauge integrals of a such a sequence is the gauge integral of the limitfunction requires the theorem we are trying to prove. We need to work fromscratch here and construct an appropriate gauge.

Since we are dealing with pointwise convergence, different values of xwill require different values of k in order to force jf .x/ � fk .x/j to beappropriately small. Thus the gauge we seek will be defined in terms of

a gauge associated with fk where k depends on the particular value of x.Henstock’s lemma (page 175) will be used to provide the critical connectionfor the triangle inequality.

Proof. For definiteness, assume that ffkg is monotone increasing.

If f is gauge integrable, then the increasing sequencen

gR bafk

ois

bounded below by gR baf1 and above by g

R baf .

For the converse, suppose thatn

gR ba fk

ois bounded above. Sincen

gR ba fk

ois increasing, the sequence converges to a real value A. We will

show that f is gauge integrable with gR ba f D A.

To that end, let " > 0 be given. Then we can find a natural number N sothat ˇ

ˇ gZ b

a

fk � A

ˇˇ < "=3

for all k � N . Since ffkg converges pointwise to f , we can associate to eachx in Œa; b� a natural number n .x/ � N such that jf .x/ � fk .x/j <

"3.b�a/

for all k � n .x/. Finally, we can use the gauge integrability of fk to find agauge �k on Œa; b� such thatˇ

ˇSR.fk ;P/ � gZ b

a

fk

ˇˇ < "

3 � 2k

for any �k-fine tagged partition P of Œa; b�.Now define a new gauge on Œa; b� by � .x/ D �n.x/ .x/ and suppose that

P D f.tk ; Ik/gnkD1 is a � -fine tagged partition of Œa; b�. By the triangleinequality,

jSR.f;P/ � Aj �ˇˇnXkD1

f .tk/�xk �

nXkD1

fn.tk/ .tk/�xk

ˇˇ

C

ˇˇnXkD1

fn.tk/ .tk/�xk �

nXkD1

gZIk

fn.tk/

ˇˇ

C

ˇˇnXkD1

gZIk

fn.tk/ � A

ˇˇ .

We will show that each of the terms on the right side is less than "=3.

For the first term, note that n .x/ was chosen so that jf .tk/�fn.tk/ .tk/ j <

"3.b�a/

. ThusˇˇnXkD1

f .tk/�xk �

nXkD1

fn.tk/ .tk/�xk

ˇˇ �

nXkD1

ˇf .tk/ � fn.tk/ .tk/

ˇ�xk

<"

3 .b � a/

nXkD1

�xk D"

3.

For the second term, let Pj be the partial tagged partition consist-ing of those tagged intervals from P for which n .tk/ D j . Set M D

max1�k�n n .tk/. Using Henstock’s lemma (page 175) with the �j -fine par-tial tagged partition Pj , we conclude thatˇ

ˇnXkD1

fn.tk/ .tk/�xk �

nXkD1

gZIk

fn.tk/

ˇˇ

�

MXjD1

ˇˇXPj

�fj .tk/�xk �

gZIk

fj

ˇˇ�

MXjD1

"

3 � 2j<"

3.

Keeping the same value of M for the third term, use the facts that ffkg isan increasing sequence and that N � n .tk/ �M to see that

gZ b

a

fN D

nXkD1

gZIk

fN �

nXkD1

gZIk

fn.tk/ �

nXkD1

gZIk

fM DgZ b

a

fM � A.

Hence ˇˇnXkD1

gZIk

fn.tk/ � A

ˇˇ �

ˇˇ gZ b

a

fN � A

ˇˇ < "

3.

Putting the three terms together, we find that jSR.f;P/ � Aj < " for any� -fine tagged partition P of Œa; b�. Thus f is gauge integrable with

gZ b

a

f D A D limk

gZ b

a

fk :

We are now in a position to extend theorem 66 to apply to infimums andsupremums. We need this result for the proof of the dominated convergencetheorem.


Theorem 68 (Dominated supremums). Suppose that ffkgis a sequence ofgauge integrable functions on Œa; b� and that g is a gauge integrable functionon Œa; b�.

1. If fk � g for all k 2 N, then supk fk is gauge integrable.

2. If fk � g for all k 2 N, then infk fk is gauge integrable.

Proof. Suppose that fk � g for all k 2 N. If we define gn D max1�k�n fk ,then theorem 66 tells us that gn is a gauge integrable function. By its defini-tion, fgng is an increasing sequence satisfying

gZ b

a

g1 �gZ b

a

gn �gZ b

a

g.

Whence supk fk D limn maxk�n fk D limn gn is gauge integrable.Negate the functions to establish (2).

With infimums and supremums now available, the proof of the gauge inte-gral form of the dominated convergence theorem is very similar to the proofin the Lebesgue context.

Theorem 69 (Dominated convergence). Suppose that ffkg is a sequenceof gauge integrable functions that converges pointwise to f on Œa; b�. If thereexist gauge integrable functions g1 and g2 on Œa; b� satisfying g1 � fk � g2for all k 2 N, then f is gauge integrable and

gZ b

a

f D limk

gZ b

a

fk .

Proof. Define the sequences˚fk

�and

˚f k�

by fkD infj�k fj and

f k D supj�k fj . By construction,˚fk

�and

˚f k�

are monotone sequencesconverging to f and satisfying

g1 � f k� f k � g2.

By theorem 68, fk

and f k are gauge integrable for k 2 N. The monotonic-ity of the gauge integral allows us to bound their integrals by

gZ b

a

g1 �gZ b

a

fk� g

Z b

a

f k �gZ b

a

g2.

By the monotone convergence theorem, f is gauge integrable with

limk

gZ b

a

fkD g

Z b

a

f D limk

gZ b

a

f k .

Moreover,

fk� fk � f k

so that

gZ b

a

fk� gZ b

a

fk �gZ b

a

f k .

Hence

gZ b

a

f D limk

gZ b

a

fk

as claimed.

Example 46. Use the dominated convergence theorem to show that

f .x/ D

(.�1/k ; x 2

�1kC1

; 1k

�0; x D 0

is gauge integrable over Œ0; 1� with gR 10f D 2

P1kD2

.�1/k

k�1 D 1�2 ln 2.

Define

fn .x/ D

8<:.�1/k ; x 2

�1kC1

; 1k

�; k < n

0; x 2�0; 1n

�:

Then ffng converges to f on Œ0; 1� with �1 � fn � 1. Hence, by thedominated convergence theorem,

gZ 1

0

f D limn

gZ 1

0

fn D limn

n�1XkD1

gZ 1

k

1kC1

.�1/k

D limn

n�1XkD1

.�1/k�1

k�

1

k C 1

�D 2

1XkD2

.�1/k

k� 1 D 1 � 2 ln 2:

Compare this example to example 43 on page 170.

7.6 The fundamental theoremsGiven the close relationship between the definitions of the gauge and Rie-mann integrals, we should expect that the proofs of the fundamental theo-rems for the gauge integral should have a structure similar to those for theRiemann integral. Review the proof of FTC-1 for the Riemann integral (the-orem 9, page 39) and you will find the key to the proof: Given a partition

of Œa; b�, use the mean value theorem to select the tag for each subintervalso that the resulting Riemann sum telescopes. This is not possible for thegauge integral since a subinterval and its tag are not selected independently.So the critical step in the proof of FTC-1 for the gauge integral is to showthat there is a gauge � that will guarantee that, given any � -fine tagged parti-tion of Œa; b�, the term f 0 .tk/ .xk � xk�1/ in the Riemann sum and the termf .xk/ � f .xk�1/ in the telescoping sum are essentially the same. This re-sult is known as the straddle lemma since the endpoints of the subintervalstraddle the tag. In contrast to the mean value theorem which starts with aninterval and selects a point in that interval, the straddle lemma begins with apoint and identifies a containing interval.

Lemma 70 (Straddle). Let f be defined on Œa; b� and differentiable at z 2Œa; b�. Then for each " > 0 there is an open interval Iz containing z so thatˇ

f .v/ � f .u/ � f 0 .z/ .v � u/ˇ< " .v � u/

for all u; v satisfying z 2 Œu; v� Œa; b� \ Iz .

Proof. Fix " > 0. Since f is differentiable at z, there is a ı > 0 such thatfor all x 2 Œa; b� \ .z � ı; z C ı/ˇ

f .x/ � f .z/

x � z� f 0 .z/

ˇ< ".

Equivalently, ˇf .x/ � f .z/ � f 0 .z/ .x � z/

ˇ< " jx � zj .

Set Iz D .z � ı; z C ı/ and suppose that z 2 Œu; v� Œa; b� \ Iz . If z D u

or z D v, the conclusion follows. So suppose that u < z < v. Thenˇf .v/ � f .u/ � f 0 .z/ .v � u/

ˇ�ˇf .v/ � f .z/ � f 0 .z/ .v � z/

ˇCˇf .z/ � f .u/ � f 0 .z/ .z � u/

ˇ< " .v � z/C " .z � u/ D " .v � u/

as claimed.

Theorem 71 (FTC-1). Suppose that F is differentiable on Œa; b�. Then F 0

is gauge integrable on Œa; b� and

gZ y

a

F 0 D F .y/ � F .a/

for all y 2 Œa; b�.

Proof. Let " > 0 be given and define a gauge � on Œa; b� by taking � .t/to be the open interval It guaranteed by the straddle lemma when appliedto F and "

.b�a/> 0. Suppose that P Df.ti ; Œxi�1; xi �/g is a � -fine tagged

partition of Œa; y�. ThenˇSR

�F 0;P

�� ŒF .y/ � F .a/�

ˇD

ˇˇX

PF 0 .ti / .xi � xi�1/ �

XPŒF .xi / � F .xi�1/�

ˇˇ

�XP

ˇF 0 .ti / .xi � xi�1/ � ŒF .xi / � F .xi�1/�

ˇ<XP

"

b � a.xi � xi�1/ D ":

Compare theorem 71 to the statements of FTC-1 for the Riemann (theo-rem 9, page 39) and Lebesgue (theorem 41, page 141) integrals. Where theRiemann integral requires the derivative to be continuous (at least a.e.) andthe Lebesgue integral requires the derivative to be bounded a.e., the gaugeintegral has no constraints on the derivative.

The gauge-integral version of FTC-1 (theorem 71) points us toward func-tions that are gauge integrable but not Lebesgue integrable.

Example 47. Let

F .x/ D

(x2 cos �

x2; x ¤ 0

0; x D 0

and set

f .x/ D F 0 .x/ D

(2x cos �

x2C 1

x2� sin �

x2; x ¤ 0

0; x D 0:

FTC-1 for the gauge integral (theorem 71) tells us that f is gauge integrable.However, L

R x0f does not exist when x > 0 since both L

R x0f C and L

R x0f �

are infinite.To see why, let In D Œ 1p

nC1; 1p

n�, choose m so that 1p

m< x, and define

gn D�Pn

kDm 1I2k�� f . Since f is bounded on Ik , we can apply FTC-1 for

the Lebesgue integral to conclude that

LZ x

0

gn D

nXkDm

LZ 1p

2k

1p2kC1

f D

nXkDm

�F

�1p2k

�� F

�1

p2k C 1

�

D

nXkDm

�1

2kC

1

2k C 1

.

But gn � f C so that

nXkDm

�1

2kC

1

2k C 1

� L

Z x

0

f C.

As n increases, the summation will grow without bound forcing LR x0 f

C D

C1.The verification that L

R x0 f

� D C1 is similar.

Similarly to the way that FTC-1 for the Lebesgue integral can be extendedto allow F to be differentiable almost everywhere as long as F is also ab-solutely continuous, FTC-1 for the gauge integral can be extended to allowF to be nondifferentiable at a countable number of points. For the gaugeintegral, ordinary continuity, the minimal condition on F , still suffices.

Theorem 72 (FTC-1). Suppose that F is a continuous function on Œa; b�that is differentiable except at a countable number of points. Then F 0 isgauge integrable on Œa; b� with

gZ y

a

F 0 D F .y/ � F .a/

for all y 2 Œa; b�.

Proof. Denote the countable exceptional set on which F is not differentiableby fzkg. By Corollary 54 (page 175), we can set F 0 .zk/ D 0 for all k.

Given " > 0; define a gauge � on Œa; b� in the following manner. If Fis differentiable at t , take � .t/ to be the open interval It guaranteed by thestraddle lemma when applied to F and "

2.b�a/. When t D zk , use the conti-

nuity of F to choose an open interval � .zk/ that contains zk and such that

jF .x/ � F .zk/j <"

2kC2

for all x 2 Œa; b� \ � .zk/.

Let P Df.ti ; Œxi�1; xi �/gniD1 be a � -fine tagged partition of Œa; y�. Sup-pose that ti D zk and compare the term associated with zk in the telescopingsum

Pi .F .xi / � F .xi�1//with the corresponding term from the Riemann

sumPi F0 .ti / .xi � xi�1/.ˇ

F .xi / � F .xi�1/ � F0 .ti / .xi � xi�1/

ˇ� jF .xi / � F .zk/j C jF .zk/ � F .xi�1/j C

ˇF 0 .zk/ .xi � xi�1/

ˇ< 2

"

2kC2C 0 D

"

2kC1.

It could be that zk is the tag for two adjacent subintervals (zk D xj D tj DtjC1). In this case, an additional term in the second line will be zero and thetotal difference between the corresponding terms in the telescoping sum andthe Riemann sum is still bounded by "

2kC1.

Now separate P into two partial tagged partitions Pe and Pr correspond-ing to the intervals whose tags belong to the exceptional set fzkg and theintervals whose tags do not. Thenˇ�

F .y/ � F .a/�� SR .f;P/

ˇ�

ˇˇXPe

�F .xi / � F .xi�1/ � F

0 .ti / .xi � xi�1/�ˇˇ

C

ˇˇXPr

�F .xi / � F .xi�1/ � F

0 .ti / .xi � xi�1/�ˇˇ

<

1XkD1

"

2kC1C

"

2 .b � a/

nXiD1

.xi � xi�1/ D ".

Hence F 0 is gauge integrable with

gZ y

a

F 0 D F .y/ � F .a/ .

You of course recognized the by now familiar divide-and-conquer movein the proof.

For the derivative form of the fundamental theorem of calculus, we beginwith a simple form that is closely related to the corresponding theorem forthe Riemann integral.

Theorem 73 (FTC-2). Suppose that f is a gauge integrable function onŒa; b�. Then the function F .x/ D g

R xaf is continuous on Œa; b� and differ-

entiable with F 0 D f at those points where f is continuous.

Proof. We will show that F is continuous at x0 2 .a; b/, the analysis at theendpoints being similar. Fix " > 0 and choose a gauge ı on Œa; b� so that anyı-fine tagged partition P satisfiesˇ

ˇSR.f;P/ � gZ b

a

f

ˇˇ < "

2:

Let ı0 be the positive value ı0 D minnı .x0/ ;

"2.jf .x0/jC1/

oand suppose

that x 2 Œa; b� \ .x0; x0 C ı0/. Then f.x0; Œx0; x�/g is a ı-fine partial taggedpartition of Œa; b�. Using the triangle inequality followed by an applicationof Henstock’s lemma, we find that

jF .x/ � F .x0/j �

ˇgZ x

x0

f � f .x0/ .x � x0/

ˇC jf .x0/ .x � x0/j

<"

2C

ˇf .x0/

"

2 .jf .x0/j C 1/

ˇ< ".

The analysis when x0 � ı0 < x < x0 is similar. Thus F is continuous at x0.The proof that F 0 .x0/ D f .x0/ when f is continuous at x0 closely

follows that for the Riemann integral and is left as an exercise.

As with FTC-2 for the Lebesgue integral, we can remove the conditionthat f is continuous at x0 and still conclude that F is differentiable al-most everywhere. For the gauge integral, the Vitali covering theorem (be-low) plays a role analogous to the roles of lemmas 45, 46, and 47 (page 145)in the Lebesgue context. Not coincidentally, the Vitali covering theorem willalso do a lot of heavy lifting in Section 8 where we provide the deferredproofs of these lemmas.

Definition 29 (Vitali covering). Let E be a bounded set. A Vitali coveringofE is a collection V of nondegenerate, closed intervals such that, given anyx 2 E and " > 0, there is an interval I 2 V with x 2 I and l .I / < ". Wedo not assume that V is countable.

Example 48. Let C be the Cantor set. Then V D˚ �x � 1

n; x�W x 2 C;

n 2 N�

is a Vitali covering of C . Note that V� D˚�x � 1

2; x�W x 2 C

�is

not a Vitali covering of C because there is no interval I in V� for which132 I with l .I / < 1

4.

The Vitali covering theorem, next, captures an idea similar to the con-cept of compactness. Any open cover of a compact set has a finite subcover.

Given a Vitali covering, there is a finite subset of disjoint sets that is “almost”a subcover.

Theorem 74 (Vitali Covering Theorem). Let E Œa; b� and let V be aVitali covering of E. Then given any " > 0, there is a finite set of disjointintervals fIkg

nkD1 from V with ��

�En [n

kD1Ik�< ".

The following proof will use a slightly unusual construction. Given a non-degenerate interval I D Œ˛; ˇ�, let OI D Œ˛ � 2l; ˇ C 2l� where l D ˇ � ˛ isthe length of I . The construction extends the interval I by twice its lengthon each end. The important features of OI are the length of OI is 5 times thatof I and any point within 2l of I is an element of OI .

Proof. If all the intervals of length greater than 1 are removed from V , theresult is still a Vitali covering of E. Hence we can assume that the intervalsin V have length at most 1.

Choose any interval from V to be I1. Suppose that I1; : : : ; Im have beenchosen. If E [m

kD1Ik , then we are done. Otherwise, let Vm be the set of

intervals in V that are disjoint from [mkD1

Ik and that contain at least oneelement of E. Let �m be the supremum of the lengths of the intervals in Vmand choose an interval from Vm with length greater than �m=2 to be ImC1.

Suppose that the process does not terminate with a finite number of in-tervals and that x 2 En

�[1kD1

Ik�. Since the intervals fIkg

1kD1 are disjoint

and contained in Œa � 1; b C 1�,P1kD1 l .Ik/ converges. Hence we can se-

lectN 2 N so thatP1kDNC1 l .Ik/ < "=5. Pick any interval J from VN that

contains x and let � be the length of J . Since �m � � as long as J 2 Vmand since limn �n D 0, we see that J 62 Vm for sufficiently large values ofm. Let M be the smallest integer for which J 62 VM . As J is disjoint from[k<M Ik , J must intersect IM which implies that x is within � of IM . Sincethe length of IM is at least �M=2 � �=2, we know that x 2 OIM .

Because x 2 En�[1kD1

Ik�

was arbitrary, we can conclude that

En�[1kD1

Ik� [1

kDNC1OIk . Consequently,

��En [nkD1 Ik

��

[1kDNC1

OIk

��

1XkDNC1

l OIk

�D 5

1XkDNC1

l .Ik/ < "

as claimed.

In the proof of the next theorem, the Vitali covering theorem is used toprove that a set ZC has measure zero. By constructing a Vitali covering ofZC, the theorem allows us “essentially” to cover ZC with a union of a finiteset fIkg

nkD1 of disjoint intervals. Henstock’s lemma is used to prove that the

size of [nkD1

Ik is arbitrarily small. This divide-and-conquer use of the Vitalicovering theorem is typical.

Theorem 75 (FTC-2). Suppose that f is a gauge integrable function onŒa; b�. Then the function F .x/ D g

R xa f is continuous on Œa; b� and, except

on a set of measure zero, F is differentiable with F 0 D f .

Proof. A proof of the continuity of F appears in theorem 73 (page 194).To prove that F 0 exists and is equal to f almost everywhere, let ZC be

the set of points t where

F 0C .t/ D limx!tC

F .x/ � F .t/

x � t

fails to exist or is not equal to f .t/. Z� is defined analogously. If t 2 ZCthen, negating the definition of a limit, there is some "t > 0 such that forevery s > 0 there is an xt;s 2 Œa; b� \ .t; t C s/ for whichˇ

F .xt;s/ � F .t/

xt;s � t� f .t/

ˇ> "t :

Equivalently,

jF .xt;s/ � F .t/ � f .t/ .xt;s � t /j > "t .xt;s � t / . (7.2)

Set En D˚t 2 ZC W "t �

1n

�and let " > 0 be given. Since f is gauge

integrable, there is a gauge ı such thatˇˇSR .f;P/ � g

Z b

a

f

ˇˇ < "

3n

for any ı-fine tagged partition P of Œa; b�.Observe that Vn D fŒt; xt;s� W t 2 En, 0 < s < ı .t/g is a Vitali covering

ofEn. By the Vitali covering theorem we can find a finite set of disjoint inter-vals fIkg D fŒtk ; xk�g from Vn for which �� .Enn [k Ik/ < "=3. Since theelements of fIkg are disjoint, P� D f.tk ; Œtk ; xk�/g is a ı-fine partial taggedpartition of Œa; b�. Using the corollary to Henstock’s lemma (page 176) and

then inequality (7.2), we can conclude that

2"

3n�XP�

ˇf .tk/ � .xk � tk/ �

gZ xk

tk

f

ˇDXP�jf .tk/ .xk � tk/ � ŒF .tk/ � F .xk/�j

>XP�

"t .xk � tk/

�XP�

1

n.xk � tk/ .

Hence

�� .[kIk/ DXP�

.xk � tk/ �2

3".

Thus

�� .En/ � �� .[kIk/C �

� .Enn [k Ik/ < ".

Because " > 0 was arbitrary, � .En/ D 0 and since ZC D [nEn, weconclude that ZC is a set of measure zero. A similar argument shows that� .Z�/ D 0 so ZC[Z�, the set of points where F fails to have a derivativeequal to f , has measure zero.

We observe that the conclusion of theorem 75 is somewhat weakerthan the corresponding theorem for the Lebesgue and hence the Riemann-Darboux integrals. For Lebesgue integrable functions, the functionF .x/ D L

R xaf is absolutely continuous. This need not be true for the gauge

integral. To see why not, consider the functions of example 47 (page 192)and the intervals fŒ 1p

2kC1; 1p

2k�gnkDm

. Given any ı > 0, we can choose mand n so that the total length of the intervals is

nXkDm

�1p2k�

1p2k C 1

�<

1p2m

< ı

while

nXkDm

�F

�1p2k

�� F

�1

p2k C 1

�D

nXkDm

�1

2kC

1

2k C 1

> 1:

7.7. Integral relationships 199

7.7 Integral relationshipsWe have already seen (exercise 2) that any Riemann integrable function isgauge integrable and that the two integrals agree. We have also encountered afunction that is gauge but not Lebesgue integrable (example 47 on page 192).Our purpose in this section is to more closely examine the relationship be-tween Lebesgue and gauge integrability. The nature of the Lebesgue integraldictates that the discussion must focus on the measurability of functions and,consequently, on measurable sets.

We begin our investigation by considering characteristic functions. Firstnote that when I is an interval then 1I is a step function and so is gaugeintegrable over Œa; b� with

gZ b

a

1I D l .I \ Œa; b�/ D � .I \ Œa; b�/ DLZ b

a

1I .

Let G be an open set. Then G can be expressed as a countable union ofdisjoint, open intervals, G D [kIk . If the union is finite, then the linearityof the Lebesgue and gauge integrals implies that 1G is gauge integrable overŒa; b� with

gZ b

a

1G DgZ b

a

Xk

1Ik DXk

gZ b

a

1Ik DXk

LZ b

a

1Ik DLZ b

a

1G .

The result for countably infinite unions follows from the monotone conver-gence theorems for the gauge and Lebesgue integrals. If F is a closed set,then G D F c is open and 1F D 1 � 1G . Hence 1F is gauge integrable overŒa; b� with

gZ b

a

1F DLZ b

a

1F D � .F \ Œa; b�/ .

In fact, the conclusion extends to all measurable sets.

Theorem 76 (Measurable sets have gauge integrable characteristic func-tions). Suppose that E is a (Lebesgue) measurable subset of R. Then 1E isgauge integrable over Œa; b� with

gZ b

a

1E DLZ b

a

1E D � .E \ Œa; b�/ .

Proof. Since E is measurable, we can use theorem 22 (page 104) to finda sequence of closed sets fFng contained in E such that � .EnFn/ < 1

n.

Replacing Fn by [nkD1

Fk , we may assume that the sequence of closed set

is increasing (Fn FnC1). Set E� D [nFn. Again applying the monotoneconvergence theorems, we see that 1E� is gauge integrable over Œa; b� with

gZ b

a

1E� DLZ b

a

1E� D ��E� \ Œa; b�

�.

Now for any n 2 N, we have Fn E� E so � .EnE�/ � � .EnFn/ <1n

. Hence EnE� has measure zero so that 1E D 1E� a.e. Therefore 1E isgauge integrable with

gZ b

a

1E DLZ b

a

1E D � .E \ Œa; b�/ .

This is a satisfying result that begs the question: Can we use�g .E/ D

gR ba1E to define a gauge measure that extends Lebesgue measure

to be defined on a larger collection of subsets of R? One suspects not, butthe question should be investigated. In any case, the previous theorem pro-vides us with exactly the tools we need to verify that any Lebesgue integrablefunction is gauge integrable.

Theorem 77 (Lebesgue H) gauge). Suppose that f is a Lebesgue inte-grable function over Œa; b�. Then f is also gauge integrable over Œa; b� andthe two integrals agree.

Proof. First assume that f � 0. Then there is an increasing sequence ofmeasurable simple functions f�ng that converges to f . By linearity of theintegrals, the previous theorem implies that each �n is gauge integrable overŒa; b� with

gZ b

a

�n DLZ b

a

�n:

The conclusion now follows from the monotone convergence theorems forthe Lebesgue and gauge integrals. When f is not nonnegative, considerf D f C � f �.

We already know from example 47 (page 192) that the converse of theo-rem 77 is false. Can we say anything about those functions that are gauge butnot Lebesgue integrable? Indeed we can. And the proof of the result featuresa cameo appearance of a familiar sequence of functions.

Theorem 78 (Gauge H) measurable). Let f be a gauge integrable func-tion on Œa; b�. Then f is measurable.

7.8. Loose ends and Dini derivatives 201

Proof. Define F .x/ D gR xa f for x 2 Œa; b�. Then by the FTC-2 (theorem

75 on page 197), F is continuous on Œa; b� and F 0 D f a.e. on Œa; b�. ExtendF to Œa; b C 1� by taking F .x/ D F .b/ for x 2 Œb; b C 1� and definegn on Œa; b� by gn .x/ D n

�F�x C 1

n

�� F .x/

�. Then fgng is a sequence

of continuous and therefore measurable functions that converges a.e. to f .Hence f is measurable.

Note that we have settled the question of whether or not the gauge integralcan be used to extend the Lebesgue measure. If E is a set for which 1E isgauge integrable, thenE is Lebesgue measurable. The gauge integral cannotbe used to extend Lebesgue measure.

So if a gauge integrable function f is measurable, how can it fail to beLebesgue integrable? Only if at least one of L

R ba f

C or LR ba f

� is infi-nite. This observation allows us to completely characterize the relationshipbetween the gauge and Lebesgue integrals.

Theorem 79 (LebesgueD absolutely gauge). Let f be a real-valued func-tion on Œa; b�.

1. If f � 0, then f is Lebesgue integrable over Œa; b� if and only if f isgauge integrable over Œa; b�.

2. In general, f is Lebesgue integrable over Œa; b� if and only if f is ab-solutely gauge integrable over Œa; b�.

Proof. Exercise.

7.8 Loose ends and Dini derivatives

The techniques we have developed for the gauge integral can be used toprovide the proofs that were deferred from the previous chapter. Specifically,bounded variation and the Vitali covering theorem are the appropriate toolsto prove lemmas 45, 46, and 47 (page 145).

You no doubt noticed the similarity between the definitions of boundedvariation and of absolute continuity. Absolute continuity appears to be astrengthening of the condition of bounded variation. This is indeed the case.

Theorem 80 (Absolutely continuous H) BV). If f is an absolutely con-tinuous function on Œa; b�, then f 2 BV .Œa; b�/.

Proof. Since f is absolutely continuous, we can find a ı > 0 so that when-ever f.xk ; yk/g is a finite set of disjoint intervals satisfying

Pk jyk � xkj <

ı; we havePk jf .yk/ � f .xk/j < 1. Choose n so that b�a

n< ı and set

zk D a C k.b�an/, k D 0; 1; 2; : : : ; n. Suppose that P is a partition of

Œzk�1; zk �. Then the total length of the intervals in P is b�an

< ı so thatV .f;P/ < 1. This inequality holds for all partitions of Œzk�1; zk� so thatVzkzk�1f � 1. By theorem 60 (page 181) we conclude that

V ba f D

nXkD1

V zkzk�1f � n:

Hence f 2 BV .Œa; b�/.

The previous theorem when combined with Corollary 61 (page 182) pro-vides the deferred proof of lemma 45 on page 145:

Lemma 45. Suppose that f is absolutely continuous on Œa; b�. Then f canbe expressed as f D f1 � f2 where f1 and f2 are increasing functions. (Ifconvenient, we can require f1 and f2 to be strictly increasing or absolutelycontinuous.)

Proof. If f is absolutely continuous, then f 2 BV .Œa; b�/. By theorem 60(page 60), V xa f and V xa f � f are increasing functions. Thus we can ex-press f .x/ D V xa f �

�V xa � f .x/

�as the difference of increasing func-

tions. To make the functions strictly increasing write f .x/ D V xa f C x ��V xa � f .x/C x

�: Since V xa f is absolutely continuous, so are all the func-

tions in the decompositions.To see that V xa f is absolutely continuous, let " > 0 be given. By the

absolute continuity of f , there is a ı > 0 so that whenever f.xk ; yk/gis a finite set of disjoint intervals satisfying

Pk jyk � xkj < ı; we haveP

k jf .yk/ � f .xk/j < ". Let f.xk ; yk/gnkD1 be such a set of intervals and

suppose that for each k, Pk D˚�xk;j�1; xk;j

��nkjD1

is a partition of Œxk ; yk�for which

V ykxk f �"

n< V .f;Pk/ .

Then˚�xk;j�1; xk;j

��k;j

is a finite set of disjoint intervals satisfying

nXkD1

nkXjD1

ˇxk;j � xk;j�1

ˇD

nXkD1

jyk � xkj < ı

so thatnXkD1

V .f;Pk/ DnXkD1

nkXjD1

ˇf�xk;j

�� f

�xk;j�1

�ˇ< ":

Thus Xk

jV yka f � V xka f j DXk

V ykxk f <Xk

V .f;Pk/C " < 2".

We conclude that V xa f is absolutely continuous.

The proofs of the other two lemmas make heavy use of the Vitali cover-ing theorem. The next proof uses the Vitali covering theorem to divide-and-conquer. We split .a; b/ into two parts: a collection of intervals on whichf cannot change much because the derivative is zero at one endpoint and acollection of intervals with very small total length on which the absolutelycontinuous function f can change very little.

Lemma 47. If f is absolutely continuous on Œa; b� and f 0 D 0 a.e. on Œa; b�,then f is constant on Œa; b�.

Proof. Using lemma 45, it is sufficient to assume that f is also monotoneincreasing and prove that f .a/ D f .b/.

Fix " > 0 and let ı be the corresponding ı of absolute continuity. SetE D fx 2 .a; b/ W f 0 .x/ D 0g. Then if x 2 E,

limt!x

f .t/ � f .x/

t � xD 0

so that

V DfŒx; t � W x 2 E, t 2 .x; b/ , jf .t/ � f .x/j < " .t � x/g

is a Vitali covering of E. The Vitali covering theorem implies that there is afinite set of disjoint, closed intervals fIkg D fŒxk ; yk �g

nkD1 from V such that

�� .En [k Ik/ < ı and jf .yk/ � f .xk/j < " jyk � xkj.Set y0 D a and xnC1 D b. Then .a; b/ is the disjoint union of the 2nC 1

intervals fŒxk ; yk �gnkD1 and f.yk ; xkC1/g

nkD0. Since

En [k Ik .a; b/ n [k Ik ..a; b/ nE/ [ .En [k Ik/

and .a; b/ nE has measure zero, we conclude that

nXkD0

jxkC1 � ykj D �� ..a; b/ n [k Ik/ D �

� .En [k Ik/ < ı

so thatnXkD0

jf .xkC1/ � f .yk/j < ".

Thus

jf .b/ � f .a/j D

ˇˇnXkD1

.f .yk/ � f .xk//C

nXkD0

.f .xkC1/ � f .yk//

ˇˇ

�

nXkD1

jf .yk/ � f .xk/j C

nXkD0

jf .xkC1/ � f .yk/j

< "

nXkD1

jyk � xkj C " � " .b � aC 1/ .

As " > 0 was arbitrary, we conclude that f .b/ D f .a/. Hence f , beingmonotone increasing, is constant on Œa; b�.

To prove lemma 46, we employ a more general notion of the derivative.A function f is not differentiable at x D c when limx!x0

f .x/�f .c/x�c

fails toexist. However, there are four related limits that always exist in the extendedreal numbers. These are the Dini derivatives. A function is differentiable atc exactly when its four Dini derivatives at c are finite and equal.

Definition 30 (Dini derivatives). Let f be a function defined on a neigh-borhood of c. The four Dini derivatives of f at c are

D�f .c/ D limx!c�

f .x/�f .c/x�c

, DCf .c/ D limx!cC

f .x/�f .c/x�c

,

D�f .c/ D limx!c�

f .x/�f .c/x�c

, DCf .c/ D limx!cC

f .x/�f .c/x�c

.

The limits may take values in the extended real numbers.

The positions of the four Dini derivatives in the definition reflect the rela-tive positions of rays having the corresponding slopes. Figure 7.1 displays afunction with its four Dini derivatives. In this case, the four Dini derivativesare all different.

Example 49. The four Dini derivatives of the absolute value functionf .x/ D jxj at zero are

D�f .0/ D �1, DCf .0/ D 1,D�f .0/ D �1, DCf .0/ D 1.


D f– (0)

D f– (0) D f+ (0)

D f+ (0)

Figure 7.1. f with rays for the four Dini derivatives at x D 0

Example 50. The four Dini derivatives of

f .x/ D

�x sin 1

x; x ¤ 0

0; x D 0

at zero areD�f .0/ D �1, DCf .0/ D 1,

D�f .0/ D 1, DCf .0/ D �1.

From the definition of the Dini derivatives, we see that DCf � DCf

and D�f � D�f . Moreover, when f is an increasing function, all fourderivatives are nonnegative. Our goal is to prove that when f is an increasingfunction on Œa; b�, then f is differentiable a.e. on Œa; b�. This will involveshowing that the four Dini derivatives are equal and finite a.e. on Œa; b�. Webegin by showing that DCf is finite a.e. on Œa; b�.

Theorem 81. Let f be an increasing function on Œa; b�. ThenDCf is finitea.e. on Œa; b�.

Proof. Let E D˚x 2 .a; b/ W DCf .x/ D1

�and set m D �� .E/. If

x 2 E, then

limt!xC

f .t/ � f .x/

t � xD C1

so that, for any constant B ,

V DfŒx; t � W x 2 E; t 2 .x; b/ , f .t/ � f .x/ > B .t � x/g

is a Vitali covering of E. By the Vitali covering theorem, there is a fi-nite, disjoint collection of intervals fIkg D fŒxk ; tk �g from V so that�� .En [k Ik/ <

m2

and hencePk .tk � xk/ D �� .[kIk/ >

m2

. Becausef is increasing,

f .b/ � f .a/ �Xk

.f .tk/ � f .xk// > BXk

.tk � xk/ > Bm

2.

Since B is arbitrary, we must have �� .E/ D m D 0 as claimed.

By reflecting f , we can interchange the Dini derivatives.

Lemma 82. Suppose that f is a function on Œa; b�. Define h on Œ�b;�a� byh .x/ D �f .�x/. Then for any c 2 .a; b/

1. DCf .c/ D D�h .�c/,

2. D�f .c/ D DCh .�c/, and

3. if f is increasing, so is h.

Proof. Exercise.

We are now prepared to provide the last of the deferred proofs. The proofof lemma 46 proceeds by carefully approximating the set of points where fis not differentiable in three stages: by an open set, by a Vitali approxima-tion7 associated with left limits, and a second Vitali approximation relatedto right limits. The open set contains the set of points where f is not differ-entiable and each subsequent approximating set will fit inside the previousone.

Lemma 46 (Lebesgue, 1904). If f is increasing on Œa; b� then f is differ-entiable a.e. on Œa; b�.

Proof. Since we have already established that the DCf is finite a.e. whenf is an increasing function, our task is to show the Dini derivatives are equala.e. on Œa; b�. If we can show that D�f � DCf and DCf � D�f a.e. onŒa; b�, then our conclusion will follow from the chain of inequalities

D�f � DCf � DCf � D

�f � D�f .

In fact, it suffices to show thatD�f � DCf for any increasing function f ,since lemma 82 then can be used to conclude that DCf � D�f .

7 The union of the finite set of closed disjoint intervals guaranteed by the Vitali coveringtheorem.

Suppose then that f is an increasing function on Œa; b�. Let

E D˚x 2 .a; b/ W D�f .x/ < D

Cf .x/�

and note that E D [p<q2QCEqp where

Eqp D˚x 2 .a; b/ W D�f .x/ 0, there

is an open set G with E G .a; b/ and �� .G/ < mC ".If x 2 Eqp , then

limt!x�

f .t/ � f .x/

t � x< p

so that

V D˚Œt; x� W x 2 Eqp , t < x, Œt; x� � G, f .x/ � f .t/ < p .x � t /

�is a Vitali covering of Eqp . Use the Vitali covering theorem to find a finite setof disjoint, closed intervals fIkg D fŒxk ; yk�g such that��

�Eqpn [k Ik

�< "

and Xk

.f .yk/ � f .xk// < pXk

.yk � xk/

< p�� .G/ < p .mC "/ . (7.3)

Without affecting the outer measure of Eqp , remove any endpoints of thefIkg from E

qp and consider the set F qp D E

qp \ [k .xk ; yk/. Note that our

modified Eqp is the disjoint union of F qp and Eqpn [k Ik . Thus ��Fqp

�>

m � ".If x 2 F qp then x 2 Eqp and x belongs to exactly one interval Ikx . Since

x 2 Eqp;

limt!xC

f .t/ � f .x/

t � x> q

so that

V D˚Œx; t � W x 2 F qp , t > x, Œx; t � � Ikx , f .t/ � f .x/ > q .t � x/

�is a Vitali covering of F qp . Again, the Vitali covering theorem asserts theexistence of a finite number of disjoint closed intervals fJig D fŒui ; vi �gsuch that ��

�Fqp n [i Ji

�< " andX

i

.f .vi / � f .ui // > qXi

.vi � ui / . (7.4)

Since F qp is contained in the disjoint union of F qp n [i Ji and [iJi , we seethat

�� .[iJi / > ��F qp�� " > m � 2". (7.5)

As f is increasing and every Ji is contained in some Ik ,Xi

.f .vi / � f .ui // �Xk

.f .yk/ � f .xk// . (7.6)

Hence from (7.5), (7.4), (7.6), and (7.3),

m < �� .[iJi /C 2"

DXi

.vi � ui /C 2"

<1

q

Xi

.f .vi / � f .ui //C 2"

�1

q

Xk

.f .yk/ � f .xk//C 2"

 0 was arbitrary, we conclude that ��

�Eqp

�D m D

0: Hence E D [p<q2QCEqp , the countable union of sets of measure zero, is

itself a set of measure zero. Since D�f � DCf a.e. on Œa; b�, we concludethat the increasing function f is differentiable a.e. on Œa; b�.

Since functions of bounded variation can be expressed as the difference ofincreasing functions, we have actually proved a slightly stronger result.

Theorem 83 (BV H) differentiable a.e). If f 2 BV .Œa; b�/ then f is dif-ferentiable a.e. on Œa; b�.

7.9 Some reflections on the gauge integralIt generally takes students a while to become comfortable working with thegauge integral. Once the basic ideas are well grasped, however, the devel-opment of the theory of the gauge integral seems to proceed more smoothlythan that of the Lebesgue integral which requires the development of the the-ory of Lebesgue-measurable sets and functions. Our proof of FTC-2 for theLebesgue integral required five lemmas, three of which were postponed until

7.10. Exercises 209

Section 7.8 in the current chapter on the gauge integral. Since the three post-poned proofs use ideas related to the gauge integral (bounded variation andthe Vitali covering theorem) they have been placed at the end of the chap-ter on the gauge integral. In terms of logical complexity, the gauge integralcomes out a winner.

In addition to having a simpler development, the gauge integral strictlyextends the Lebesgue integral and has a more satisfying version of FTC-1. When a function is differentiable on an interval, its derivative is alwaysgauge integrable over any subinterval. Why then is the gauge integral rel-atively unknown compared to the Lebesgue integral? Let me suggest threereasons.

First, Lebesgue published his integral in 1902 while the gauge integralwas not introduced until around 1960. Consequently, the Lebesgue integralhad over 50 years to become established in the structure of mathematics andto be extended before the gauge integral was developed. In some sense, theecological niche was already occupied by the time the gauge integral wasdeveloped. Second, the mathematical community took note of the Lebesgueintegral since it solved a problem that mathematicians were struggling withat the time. In 1960, the goal of extending FTC-1 to all derivatives did nothave the same kind of priority. Third and most enduring, the Lebesgue in-tegral can more naturally be extended to nonscalar-valued integrals of func-tions over non-Euclidean domains. We will see a hint of these extensions inthe next chapter.

7.10 Exercises

7.1 Definition and basic examples: filling the gaps1. Prove that the two definitions of gauge on page 162 are equivalent in

the following sense.

(a) Given any gauge ı of the first type on Œa; b�, there is a gauge � ofthe second type so that any � -fine tagged partition of Œa; b� is alsoı-fine.

(b) Given any gauge � of the second type on Œa; b�, there is a gauge ıof the first type such that any ı-fine tagged partition of Œa; b� is also� -fine.

2. Prove that any Riemann integrable function is gauge integrable and thatthe integrals agree.

3. Prove that if f is gauge integrable over Œa; b� then for each " > 0 thereis a gauge � on Œa; b� such that



4. In example 38 (page 165)

(a) What is it about the calculations related to the function

f .x/ D

(1px; x > 0

0; x � 0

that would lead one to think that using a gauge of the form

� .t/ D

8<: t � a2 .t/ ; t

a2.t/

�; t > 0�

�"2; "2�; t D 0

might be an effective line of attack?(b) Why was 2p

xkCpxk�1

used?

(c) Explain how a .t/ D 1 � "pt

4was probably derived.

5. Explain what is meant by the phrase “satisfied vacuously” on page 167.

6. In the proof of Cousin’s theorem (page 167),

(a) Why is S nonempty and bounded?(b) Why must there be a c 2 S with c 2 � .ˇ/ \ .a; ˇ/?(c) Why is P[f.ˇ; Œc; ˇ�/g � -fine?

7.1 Definition and basic examples: deeper reflections7. Some texts use an alternative condition for a tagged partition

P D˚�tk;Ik

��to be ı-fine: tk 2 Ik � .tk � ı .tk/ ; tk C ı .tk// for

all tagged intervals. Prove that this condition is equivalent to the condi-tion given in Definition 21 (page 162) in the sense that given a gauge ı1using one of the conditions, there is a second gauge ı2 using the othercondition such that any partition that is ı2-fine is ı1-fine.

8. How would you change � in example 38 (page 165) if the interval ofintegration was Œ0; 3�?

9. In examples 37 and 38 (page 165), the size of jf .tk/ � f .sk/j wascontrolled by bounding jf .xk/ � f .xk�1/j : Under what circum-stances can this approach be effective?

7.10. Exercises 211

10. Let frkg be the enumeration of the rational numbers in Œ0; 1� given by˚0; 1; 1

2; 13; 23; 14; 34; 15; 25; 35; : : :

�and define a gauge ı on Œ0; 1� by

ı .t/ D

(1

2k; t D rk

0:1; t 62 Q.

Construct two ı-fine tagged partitions of Œ0; 1�.

11. Let

� .t/ D

( �14t; 2t

�; t > 0

.�0:1; 0:1/ ; t D 0:

Construct two � -fine tagged partitions of Œ0; 1�.

12. Using the definition of the gauge integral, prove that f .x/ D x2 isgauge integrable over Œ0; 2� and that g

R 20 f D

83

. (Note that x2i�1 <x2iCxixi�1Cx

2i�1

3< x2i .)

13. Prove from the definition that

f .x/ D

8<:ˇ1; x D a

˛; a < x < b

ˇ2; x D b

is gauge integrable on Œa; b� and find the value of its integral.

14. Prove from the definition that

f .x/ D

8<:˛1; a � x < c

ˇ; x D c

˛2; c < x � b

is gauge integrable on Œa; b� and find the value of its integral.

15. Explain why Cousin’s theorem (page 167) also implies the existence ofa ı-fine partition.

16. Construct an alternative proof of Cousin’s theorem (page 167) basedon nested subintervals. Suppose that there is no � -fine tagged partitionof Œa; b�.

(a) Let c be the midpoint of Œa; b�. Can both Œa; c� and Œc; b� have � -finetagged partitions?

(b) Create a nested sequence of closed subintervals whose lengths goto zero. Explain how the intersection point can be used to generatea contradiction.


17. Since f� .t/gt2Œa;b� is an open cover of the compact set Œa; b�, it istempting to try to prove Cousin’s theorem (page 167) by construct-ing a tagged partition based on a finite subcover f� .tk/g

nkD1 of Œa; b�.

What goes wrong when this approach is attempted?

7.2 The art of constructing gauges: filling the gaps18. Suppose that ı1 and ı2 are two gauges on Œa; b� with ı1 � ı2. Prove

that any ı1-fine tagged partition is also ı2-fine:

19. Suppose that �1 and �2 are two gauges on Œa; b� with �1 .t/ �2 .t/.Prove that any �1-fine tagged partition is also �2-fine.

20. Given a finite set of points A D fa1; a2; : : : ; amg from Œa; b�, explainhow a gauge � on Œa; b� can be constructed so that the elements of Amust be tags of any � -fine partition. (You may find it convenient to addghost points a0 D a � 1 and anC1 D b C 1.)

21. In example 40 (page 169), why was "2n.jf .t/jC1/

used instead of"

2n.jf .t/j/?

22. Suppose that ı1 and ı2 are two gauges on Œa; b�. Prove that ı3 Dmin fı1; ı2g is also a gauge on Œa; b� and that any ı3-fine partition isboth ı1-fine and ı2-fine.

23. Suppose that �1 and �2 are two gauges on Œa; b�. Prove that �3 definedby �3 .t/ D �1 .t/ \ �2 .t/ is also a gauge on Œa; b� and that any any�3-fine partition is both �1-fine and �2-fine.

24. Fill in the details to explain why we can assume that the tags of anypartition are also division points of the partition without changing theassociated Riemann sum.

25. Given a gauge � on Œa; b� and a � -fine partition P of Œa; b�, explainhow a second gauge � 0 can be combined with tag splitting to forcesubsequent � 0-fine tagged partitions of Œa; b� to be � -fine refinementsof P .

7.2 The art of constructing gauges: deeper reflections26. Design a gauge ı on Œ0; 1� that that will force 0; 1

2, and 1 to be tags of

any ı-fine partition.

27. Let P D f.0:2; Œ0; 0:25�/ ; .0:3; Œ0:25; 0:5�/ ; .0:6; Œ0:5; 0:75�/ ; .0:8;

Œ0:75; 1�/g.

7.10. Exercises 213

Identify an alternative tagged partition P� so that all the tags of Pare division points of P� and SR .f;P�/ D SR .f;P/ for any functionf defined on Œ0; 1�.

28. Explain how ı probably was derived in example 42 (page 170).

29. Explain why it is impossible to construct a gauge � that will prevent apredetermined point from being a tag of a � -fine partition.

30. Let f be defined on Œa; b� and let " > 0.

(a) Design a gauge ı on Œa; b� so that for any ı-fine tagged partitionP D

˚�tk;Jk

��nkD1

of Œa; b�, the contribution of any one term in theRiemann sum will be less than ":

(b) Part (a) may appear to bound the value of SR.f;P/. Explain whythis is not the case.

31. Find a gauge that shows that any step function of the form

f .x/ D

8ˆˆ<ˆˆ:

˛1; a � x < c1ˇ1; x D c1˛2; c1 < x < c2ˇ2; x D c2˛3; c2 < x < c3ˇ3; x D c3˛4; c3 < x < c4

� � �

˛n; cn�1 < x � b

is gauge integrable on Œa; b� : What is the value of gR ba f ?

32. Let

f .x/ D

�n; x D m

nwithn;m 2 N and m

nin lowest terms,

0; x 62 Q,

and let " > 0 be given. Design a gauge � so that for any � -fine partitionP of Œ0; 1�, SR.f;P/ < ". What can you conclude?

33. Let f be defined on Œa; b� and let " > 0. Given a countable set S ,design a gauge � on Œa; b� so that for any � -fine tagged partition P ofŒa; b�, the total contribution to SR .f;P/ by tagged intervals with tagsin S is at most ".

34. Suppose that f is gauge integrable over Œa; b� and that g D f exceptat a countable set of points. Prove that g is gauge integrable and thatgR ba g D

gR ba f:

35. Suppose that f is gauge integrable over Œa; b� and that g D f excepton a setZ of measure zero. Suppose further that the difference betweenf and g is bounded.

(a) Prove that g is gauge integrable and that gR bag D g

R baf:

(b) What goes wrong with your proof when you try to extend it byremoving the condition that the difference between f and g isbounded? Can you find a way around this difficulty?

7.3 Basic integrability results: filling the gaps36. Verify the standard integral properties for the gauge integral.

(a) Uniqueness. The value of the gauge integral is unique (if itexists).

(b) Linearity. Let c 2 R. If f and g are gauge integrableover the interval Œa; b�, then so are f C g and cf . Moreover,gR ba.f C g/ D g

R ba f C

gR ba g and g

R ba cf D c

gR ba f:

(c) Monotonicity. If f and g are gauge integrable over the intervalŒa; b� with f .x/ � g .x/ for x 2 Œa; b�, then g

R baf � g

R bag:

(d) Triangle inequality. If f and jf j are gauge integrable over Œa; b�,

thenˇ

gR bafˇ� gR ba jf j.

37. Supply a proof for the forward direction of the Cauchy criterion (theo-rem 50 on page 172): If f is gauge integrable over Œa; b� then for each" > 0 there is a gauge � on Œa; b� such that




(a) Why is jSR.f;Q1/ � SR.f;Q2/j D jSR.f;S1/ � SR.f;S2/j?(b) How should the proof be modified when a D c < d < b?

39. Prove that if the function f is gauge integrable on Œa; c� and Œc; b� thenf is gauge integrable on Œa; b� and that g

R ba f D

gR ca f C

gR bc f .

7.10. Exercises 215

(Using gauges �a and �b on Œa; c� and Œc; b�, define

� .t/ D

8<:�a .t/ \ .�1; c/ ; t < c

�a .t/ \ �b .t/ ; t D c

�b .t/ \ .c;C1/ ; t > c

and split the tag at t D c.)

40. Let � be a gauge on Œa; b� and suppose that P1 is a � -fine tagged par-tition of Œa; c� where a < c < b. Prove that there is another � -finetagged partition P2 of Œc; b� so that P1 [ P2 is a � -fine partition ofŒa; b�.

41. Use theorem 53 (page 174) to prove Corollary 54 (page 175):

42. In the proof of Henstock’s lemma (page 175)

(a) The gauge integrability of f over Jk ensures that there is a gauge�k on Jk such that any �k-fine tagged partition P of Jk satisfiesˇ

SR.f;P/ � gZJk

f

ˇ<

m:

So why is there a � -fine (as opposed to �k-fine) tagged partition Pksuch that ˇ

SR.f;Pk/ � gZJk

f

ˇ<

m?

(b) Why isˇˇSR.f;bP/ �X

i

gZIi

f

ˇˇ

D

ˇˇSR.f;Q/ � g

Z b

a

f �

mXkD1

�SR.f; Pk/ �

gZJk

f

�ˇˇ ?

7.3 Basic integrability results: deeper reflections43. The Darboux and gauge integrals both have a Cauchy criterion. Though

not presented in this text, Cauchy first developed the criterion for theRiemann integral. Of all the integrals we have considered, only theLebesgue integral does not have a Cauchy criterion. Why not?

44. (Alternative approach for exercise 31) Use exercises 39 and 13 to provethat any step function (see Definition 28, page 180) is gauge integrable.Describe the value of the integral.

45. Show that

f .x/ D

(.�1/k ; x 2

�1kC1

; 1k

�0; x D 0

is gauge integrable.

46. Prove that

f .x/ D

(k; x 2

�1kC1

; 1k

�0; x D 0

is not gauge integrable.

47. Define f by

f .x/ D

8<:1; x 2 Œ0; 1�

�1; x 2 .1; 2�

2; x 2 .2; 13�

Define a gauge � on Œ0; 3� by

� .t/ D

8<ˆ:

.�1; 1/ ; t 2 Œ0; 1/

.1 � "; 1C "/ ; t D 1

.1; 2/ ; t 2 .1; 2/

.2 � "; 2C "/ ; t D 2

.2; 3/ t 2 .2; 3�

and suppose that P Df.ti ; Œxi�1; xi �/gniD1 is a � -fine tagged partitionof Œa; b�.

(a) ComputeˇSR.f;P/ � g

R ba f

ˇand give a bound in terms of ".

(b) Why must 1 be a tag in the partition P?Let be k the minimal integer for which tk D 1 and set P0 Df.ti ; Œxi�1; xi �/g

kiD1 and P3 D f.ti ; Œxi�1; xi �/gniDkC1.

(c) ComputeˇSR.f;P0/ �

PkiD1

gR xixi�1

fˇ

and identify a bound in

terms of ".(d) Compute and bound

ˇSR.f;P3/ �

PniDkC1

gR xixi�1

fˇ.

(e) Compute and boundPi

ˇf .ti / ��xi �

gRIifˇ.

(f) Compute and bound SR.jf j ;P3/ �Pi

ˇgRIifˇ:

(g) Relate the previous parts of this exercise to Henstock’s lemma andits corollary.

48. Suppose that f is defined on Œa; b� and gauge integrable over Œc; b� forall c 2 .a; b/. Prove that f is gauge integrable over Œa; b� if and only

7.10. Exercises 217

if limc!aCR bc f exists. In this case,

gZ b

a

f D limc!aC

gZ b

c

f .

(For one direction, define a gauge on Œa; b� by patching together gaugeson subintervals like

�aC 1

n; b�. For the other direction, use theorem 51

(page 173) and Henstock’s lemma (page 175) applied to a single inter-val with tag t D a.)

49. Use exercise 48 to give an alternative proof that the function of ex-ample 43 (page 170) is gauge integrable over Œ0; 1� with g

R 10 f DP1

kD1.�1/k

kC1D ln 2 � 1. (First show that limn!1

gR 11nf D ln 2 � 1.

Then use the triangle inequality to show that lima!0CgR 1a f D

ln 2 � 1.)

50. Suppose that f is a gauge integrable function on Œa; b� and that s > 0.Define fs on Œsa; sb� by fs .x/ D f .x=s/. Prove that fs is gaugeintegrable on Œsa; sb� with

gZ sb

sa

fs D sgZ b

a

f .

7.4 Absolute integrability: filling the gaps51. Let f be a gauge integrable function on Œa; b�. Prove that f is abso-

lutely gauge integrable over Œa; b� if and only if f C and f � are bothgauge integrable over Œa; b�. (Express f C as a linear combination of fand jf j.)

52. Prove that the Dirichlet function does not have bounded variation.

53. Suppose that P and Q are partitions of Œa; b� and that Q is a refinementof P . Prove that

V .f;P/ � V .f;Q/ .

54. Suppose that f; g 2 BV .Œa; b�/ and that c 2 R. Prove that f C g andcf are both in BV .Œa; b�/.

55. Suppose that f 2 BV .Œa; b�/ and that c 2 R. Explain why f 2BV .Œa; c�/ and f 2 BV .Œc; b�/.

56. Why is V ba f � jf .b/ � f .a/j?

57. Near the end of the proof of theorem 60 (page 181), why does V yx f �f .y/ � f .x/ imply that V xa f � f .x/ is increasing?

58. When proving the first half of theorem 62 (page 183), we only provedV ba F �

gR ba jf j rather than V ba F D

gR ba jf j. Why is this OK?

59. Why isPk

ˇgR xkxk�1

fˇ�Pk

gR xkxk�1

g in the proof of theorem 63

(page 184)?

60. Prove that if f and g are absolutely gauge integrable functions on Œa; b�and c is a scalar, then f C g and cf are also absolutely gauge inte-grable.

61. Suppose that f; g, and h are gauge integrable functions with f � h

and g � h. Prove that max ff; gg and min ff; gg are gauge integrable.

7.4 Absolute integrability: deeper reflectionsFor the next two exercises, take

f .x/ D

(.�1/k k; x 2

�1kC1

; 1k

�0; x D 0:

62. Show that given any partition P of Œ0; 1� and B > 0, it is possibleto assign one set of tags so that SR.f;PC/ > B and another set oftags so that SR.f;P�/ < �B . What does this say about the Riemannintegrability of f ?

63. Show that s-LR 10 f

C and s-LR 10 f

� are both infinite so that f is notLebesgue integrable.

64. Prove that

f .x/ D

(.�1/k ; x 2

�1kC1

; 1k

�0; x D 0

is not in BV .Œ0; 1�/.

65. Is

f .x/ D

8<: .�1/k

k; x 2

�1kC1

; 1k

�0; x D 0

in BV .Œ0; 1�/?

66. Is

f .x/ D

(sin��x

�; x 2 .0; 1�

0; x D 0

in BV .Œ0; 1�/?

7.10. Exercises 219

67. Is

f .x/ D

(x2 sin

��x

�; x 2 .0; 1�

0; x D 0

in BV .Œ0; 1�/?

68. Prove that if f 2 BV .Œa; b�/ then f is bounded on Œa; b�. Give anexample to show that the converse is not true.

69. Give two examples of functions to show that continuous functions neednot have bounded variation and that functions of bounded variationneed not be continuous.

70. Give an example of a pair of gauge integrable functions f and g forwhich h1 D max ff; gg and h2 D min ff; gg are not gauge integrable.

71. Give an example of a step function f on Œ0; 1� that has m points ofdiscontinuity, satisfies jf j � B; and attains the maximum value ofV ba f D 4mB . (See the proof of theorem 58 on page 180.)

72. Prove that if f 2 BV .Œa; b�/, then jf j 2 BV .Œa; b�/.

73. Prove that if f; g 2 BV .Œa; b�/, then max ff; gg and min ff; gg havebounded variation on Œa; b�. (Use exercise 72.)

74. Prove that if f is differentiable with a bounded derivative on Œa; b�,then f 2 BV .Œa; b�/.

75. Express sin x and cos2 x on Œ0; 2�� as the difference of two increasingfunctions.

76. Suppose that f 2 BV .Œa; b�/.

(a) Show that f can be written as the difference of two increasingfunctions in an infinite number of ways.

(b) Define g0 D 12

�V xa f C f .x/

�and h0 .x/ D 1

2

�V xa f � f .x/

�for x 2 Œa; b�. Prove that g0 and h0 are increasing functions withf D g0 � h0 and V xa f D g0 C h0.

(c) Prove that g0 and h0 are most efficient in the sense that any decom-position of f D g � h, where g and h are increasing functions,satisfies g0 .b/ � g0 .a/ � g .b/ � g .a/ and h0 .b/ � h0 .a/ �h .b/ � h .a/.

7.5 Convergence theorems: filling the gaps77. In the proof of the monotone convergence theorem (page 67), we used

Henstock’s lemma with a sum of the form

MXjD1

ˇˇ8<:X

Pj

�fj .tk/�xk �

gZIk

fj

9=;ˇˇ :

Explain why we could not just use

nXkD1

ˇfn.tk/ .tk/�xk �

gZIk

fn.tk/

ˇ�

nXjD1

"

3 � 2j<"

3.

78. Prove theorem 67 (page 186) for monotone decreasing sequences offunctions.

79. Prove part (2) of theorem 68 (page 189): If ffkgis a sequence of gaugeintegrable functions on Œa; b� and g is a gauge integrable function onŒa; b� satisfying fk � g, then infk fk is gauge integrable.

80. Explain why ffkg and ff kg in the proof of theorem 69 (page 189) are

monotone sequences converging to f:

7.5 Convergence theorems: deeper reflections81. Suppose that ffkg is a sequence of gauge integrable functions that con-

verges uniformly to f on Œa; b�. Without using the theorems of Section5, prove that f is gauge integrable with

gZ b

a

f D limk

gZ b

a

fk .

82. Let f be a continuous function on Œ0; 1�. Prove that limkgR 10 f

�xk�D

f .0/.

83. Suppose that ffkg is a sequence of nonnegative functions on Œ0; 1�.

(a) Prove that if ffkg is monotone; then limkgR 10 fk D 0 if and only

if ffkg converges to 0 a.e. on Œ0; 1�.(b) Find an example of a sequence of nonnegative functions ffkg on

Œ0; 1� such that limkgR 10 fk D 0 but ffk .x/g does not converge to

0 for any x 2 Œ0; 1�. (Work with intervals likehj

2i; jC12i

i.)

84. Suppose that ffkg is a sequence of gauge integrable functions on Œa; b�

with n � gR bafn and that f is a positive function on Œa; b� satisfying

fk � f for all k. Prove that f is not gauge integrable.

7.10. Exercises 221

85. State and prove a version of Fatou’s lemma (theorem 38, page 139) forthe gauge integral.

86. Explain how the hypotheses in the convergence theorems can be mod-ified with the phrase “almost everywhere.”

7.6 Fundamental theorems: filling the gaps87. In the proof of the straddle lemma (page 191), explain why the result

holds if z D u or z D v.

88. Let

f .x/ D

(x2 cos �

x2; x ¤ 0

0; x D 0:

Prove that

f 0 .x/ D

(2x cos �

x2C 1

x2� sin �

x2; x ¤ 0

0; x D 0:

In particular, verify that f 0 .0/ D 0.

89. Let

f .x/ D

(2x cos �

x2C 1

x2� sin �

x2; x ¤ 0

0; x D 0:

Prove that for any x > 0, LR x0 f

� D C1.

90. Let f be a function on Œa; b� and let E be a subset of Œa; b�. Prove that1E � f � f

C and �1E � f � f �.

91. In the proof of FTC-1 (theorem 72 on page 193)

(a) Explain why we can set F 0 .zk/ D 0 for all k.(b) Explain why the sum of the differences between the terms asso-

ciated with zk in the telescoping sum and the Riemann sum isbounded by "

2kC1when zk D xi D tiC1 D ti .

92. Modify the proof of theorem 10 on page 40 to complete the proof ofFTC-2 (page 194).

93. Let I be an interval of length l . Explain why

(a) the length of OI is 5 times that of I and(b) any point within 2l of I is an element of OI .


94. In the proof of the Vitali covering theorem (page 74),

(a) Why is �m finite?(b) Why does limn �n D 0?(c) Explain why IM intersects J and the length of IM is at least �=2.

95. Explain how the proof of theorem 75 (page 197) can be modifiedto show that the set of points x0 where the left derivative of F ,limx!x�

0

F .x/�F .x0/x�x0

, does not exist or is not equal to f .x0/ has mea-sure zero.

7.6 Fundamental theorems: deeper reflections96. Use an example to explain the statement that the continuity of F is “the

minimal condition on F ” on page 191.

97. Let f be a gauge integrable function on Œa; b� and defineF .x/ D g

R xa f for x 2 Œa; b�. Suppose that f is bounded on an open

interval containing c 2 .a; b/.

(a) Prove that F is continuous at c.(b) Give an example to show that F need not be differentiable at c.

98. Suppose that F and G are two continuous functions on Œa; b� such thatF 0 D G0 except at a countable number of points. Prove that F � G isconstant.

99. Since F in FTC-1 (page 193) is continuous on the compact set Œa; b�,F is uniformly continuous on Œa; b�. One can use the uniform conti-nuity of F together with a tight, countable, open-interval cover of theset where F is not differentiable to create a gauge to play the role of� in the proof. It seems like this proof should work for F that are dif-ferentiable except on a set of measure zero. What goes wrong with thisapproach? Under what circumstances will this tactic succeed?

100. Prove that V D˚�r � 1

n; r C 1

n

�W r 2 Œa; b� \Q,n 2 N

�is a Vitali

covering of Œa; b�.

101. Suppose that V is a Vitali covering of E and that E � I where I isan open interval. Prove that VI D fJ 2 V W J � I g is also a Vitalicovering of E.

102. The Vitali covering of the Cantor set in example 48 (page 195) is un-countable. Find a countable Vitali covering of the Cantor set.

7.10. Exercises 223

103. Given E and fIkg as constructed in the proof of the Vitali coveringtheorem, prove that En [1

kD1Ik has measure zero.

104. Give an example of a construction of fIkg as in the proof of the Vitalicovering theorem where E 6 [ 1

kD1Ik : (E D

˚1nW n 2 N

�[f0g is one

possible place to start.)

105. Give an example of a function f defined on Œ0; 1� that is not continuousat x D 1

2but for which F .x/ D g

R x0f is differentiable at x D 1

2.

106. FTC-1 (page 193) tells us that when F is a continuous function onŒa; b� and has at most a countable set of points of nondifferentiability,then F 0 is gauge integrable with g

R ya F

0 D F .y/ � F .a/ for y 2Œa; b�. FTC-2 (page 197) tells us that when f is a gauge integrablefunction on Œa; b�, the function F .x/ D g

R xa f is differentiable with

F 0 D f a.e. on Œa; b�.

(a) Give an example to show that we cannot extend FTC-1 (page 193)by allowing the set of points where F is not differentiable to havemeasure zero instead of being countable.

(b) Give an example to show that we cannot strengthen FTC-2(page 197) to conclude that F.x/ D g

R xa f has only a countable

number of points where F 0 fails to exist or does not equal f .

7.7 Integral relationships: filling the gaps107. How are the monotone convergence theorems used to prove that the

characteristic function of any open setG is gauge integrable over Œa; b�with g

R ba1G D

LR ba1G?


(a) How are the monotone convergence theorems used to concludethat

gZ b

a

1E� DLZ b

a

1E� D ��E� \ Œa; b�

�?

(b) Fill in the details to conclude that 1E is gauge integrable with

gZ b

a

1E DLZ b

a

1E D � .E \ Œa; b�/ .



(a) How does linearity imply that

gZ b

a

�n DLZ b

a

�n

for all n 2 N?(b) How is the monotone convergence theorem applied to conclude

that f is gauge integrable over Œa; b� and

gZ b

a

f D LZ b

a

f ?

110. Prove theorem 79 (page 201).

7.7 Integral Relationships: deeper reflections111. Prove directly that the collection of sets whose characteristic function

is gauge integrable over every closed and bounded interval Œa; b� isa sigma algebra. Explain why every Borel set has a gauge integrablecharacteristic function.

112. Give an example of a function that is gauge integrable on Œa; b� and aset E � Œa; b� for which

g .x/ D

�f .x/ ; x 2 E

0; x 62 E

is not gauge integrable. What hypotheses can you add so that g is gaugeintegrable?

113. Give an example of two functions that are not gauge integrable overŒ0; 1� but whose sum is.

114. What are the advantages and disadvantages of the gauge integral whencompared to the Lebesgue integral?

7.8 Loose ends: filling the gaps115. In the proof of lemma 47 (page 145),

(a) Why it is sufficient to assume that f is also monotone increasingand prove that f .a/ D f .b/?

(b) Is is possible to have degenerate intervals in f.yk ; xkC1/g? Explain.

7.10. Exercises 225


(a) Explain why .a; b/ n [k Ik ..a; b/ nE/ [ .En [k Ik/ :(b) Explain why V DfŒx; t � W t 2 .x; b/ ,f .t/ � f .x/ > B .t � x/g

is a Vitali covering of E.(c) Why is

Pk .tk � xk/ D �

� .[kIk/ >m2

?(d) Justify each of the inequalities in

f .b/�f .a/ �Xk

.f .tk/ � f .xk// > BXk

.tk � xk/ > Bm

2:

117. Suppose that f is a function on Œa; b�. Define h on Œ�b;�a� by h .x/ D�f .�x/. Prove that for any c 2 .a; b/

(a) DCf .c/ D D�h .�c/,(b) D�f .c/ D DCh .�c/, and(c) if f is increasing, so is h.

118. Suppose we know that D�f � DCf for any for increasing functionf . Explain how lemma 82 (page 206) can be used to show thatDCf �D�f as well.

119. Given a set E .a; b/ and " > 0, explain why there is an open set Gwith E G .a; b/ and �� .G/ < �� .E/C ".

120. In the proof of lemma 46 (page 145)

(a) Explain why E D [p;q2QCEqp:

(b) Explain why G is introduced into the proof.(c) Explain why

VD˚Œt; x� W x 2 Eqp , t < x, Œt; x� � G,f .x/ � f .t/ < p .x � t /

�is a Vitali covering of Eqp .

(d) Justify each of the inequalities inXk

.f .yk/ � f .xk// < pXk

.yk � xk/ < p�� .G/ m � "?

7.8 Loose ends: deeper reflections121. Compute the Dini derivatives of f .x/ D x sin 1

xC 2 .x � jxj/

at x D 0.

122. Given a; b; c; d 2 R with a < b and c < d , define a functionf for which D�f .0/ D a; D�f .0/ D b; DCf .0/ D c, andDCf .0/ D d .

7.11 ReferencesGordon, R.A. (1994). The integrals of Lebesgue, Denjoy, Perron, and Hen-

stock. Graduate Studies in Mathematics 4. American MathematicalSociety.

Thomson, B.S. (2012). Theory of the Integral. ClassicalRealAnalysis.com.Burk, F.E. (2007). A Garden of Integrals. Mathematical Association of

America.DePree, J. and C. Swartz. (1988). Introduction to Real Analysis. John Wiley

& Sons.Henstock, R. (1988). Lectures on the Theory of Integration. Series in Real

Analysis 1. World Scientific Publishing Company.Kurzweil, J. (2000). Henstock-Kurzweil Integration: Its Relation to Topolog-

ical Vector Spaces. Series in Real Analysis 7. World Scientific Publish-ing Company.

Vitali, G. (1908). Sui gruppi di punti e sulle funzioni di variabili reali,Atti dell’Accademia delle Scienze di Torino (in Italian) 43: 229–246.archive.org/stream/attidellarealeac43real#page/229.

CHAPTER 8

Stieltjes-type Integralsand Extensions

Thus far, all the integrals we have studied have treated intervals accordingto their length. In 1894, Thomas Joannes Stieltjes published the definitionof a new type of integral that did not treat all intervals of the same lengthequally. Stieltjes’ motivation came from the problem of computing momentsof inertia. If a massm lies at a distance l from a fulcrum, then it exerts a forceof glm on the lever where g is the constant for the acceleration of gravity. Ofcourse, objects typically do not place all their mass at a single point. Rather,an object’s mass is distributed along a range of distances from the fulcrum.The total force exerted on the lever comes from the cumulative contributionof bits of mass at these various distances. In a similar fashion, while a pointmass’s resistance to rotation is l2m where l is the distance from the centerof rotation, the resistance of a typical object must be computed from bits ofmass spread across an interval.

Instead of focusing on the mass at point x, Stieltjes represented the massto the left of a point x by g .x/. Then g .xi / � g .xi�1/ is the mass betweenxi�1 and xi . Since we will want to allow point masses (weights concentratedat a single point), we need to be a bit more precise. Typically, the functiong .x/ represents the mass in the interval .�1; x� so that g .xi / � g .xi�1/represents the mass in .xi�1; xi �.

Definition 31 (Riemann-Stieltjes Integral). Let f and g be functions onŒa; b� and let P Df.ti ; Œxi�1; xi �/g be a tagged partition of Œa; b�. Then theRiemann-Stieltjes sum of f with respect to g and P is

SR-S .f; g;P/ DnXiD1

f .ti / Œg .xi / � g .xi�1/� DXPf �g.

227

228 CHAPTER 8. Stieltjes-type Integrals and Extensions

The function f is Riemann-Stieltjes integrable with respect to g over Œa; b�if there is a real value A such that for every " > 0 there is an associatedı > 0 so that

jSR-S .f; g;P/ � Aj < "

whenever P is a tagged partition of Œa; b� with kPk < ı. In this case, wewrite R-S

R ba f dg D A.

Note that when g .x/ D x, the Riemann-Stieltjes integral becomes the fa-miliar Riemann integral. In Stieltjes’ applications referenced above, f wouldbe either f .x/ D gx or f .x/ D x2. However, we need not and will notlimit ourselves to these options.

8.1 Examples and counterexamples

Example 51 (Constant functions I). Let f .x/ D c on Œa; b� and letg be an arbitrary function on Œa; b�. Then for any tagged partitionP Df.ti ; Œxi�1; xi �/g of Œa; b�,


c �g D c .g .b/ � g .a// :

Hence the constant function f .x/ D c is Riemann-Stieltjes integrable withrespect to g over Œa; b� and

R-SZ b

a

f dg D c .g .b/ � g .a// :

Example 52 (Constant functions II). Let f be an arbitrary function onŒa; b� and let g .x/ D c. If P Df.ti ; Œxi�1; xi �/g is any tagged partition ofŒa; b�, then


f .ti / Œc � c� D 0:

Thus the function f is Riemann-Stieltjes integrable with respect to the con-stant function g over Œa; b� and

R-SZ b

a

f dg D 0:

8.1. Examples and counterexamples 229

We can interpret the preceding examples in terms of Stieltjes’ originalmotivating problem. When f is constant, it acts as a change of units factorand R-S

R ba f dg reflects the mass over .a; b� expressed in the selected units.

If g is constant, then there is no mass in the interval Œa; b�. Hence the mass,force, and moment of inertia will all be zero.

Example 53 (Dirichlet). The Dirichlet function d is Riemann-Stieltjes in-tegrable with respect to g over Œa; b� only if g is constant.

To see why, take any ˛ < ˇ 2 Œa; b� and let P DfŒxi�1; xi �g be a par-tition of Œa; b� that includes ˛ and ˇ as division points, say xj D ˛ andxk D ˇ. Create two tagged partitions from P . Create P1 by choosing ratio-nal numbers for the tags of the subintervals between ˛ and ˇ and irrationaltags for the remaining intervals. Construct P2 using all irrational tags. Then,irrespective of the mesh of P ,

SR-S .d; g;P1/ DkXiDj

1 Œg .xi / � g .xi�1/� D g .˛/ � g .ˇ/

and

SR-S .d; g;P2/ DnXiD1

0 Œg .xi / � g .xi�1/� D 0:

If d .x/ is integrable, then we can make SR-S .d; g;P1/ be arbitrarily closeto SR-S .d; g;P2/ by choosing P of sufficiently small mesh. But this meansthat g .˛/ � g .ˇ/ D 0. Thus d is Riemann-Stieltjes integrable with respectto g over Œa; b� only if g is constant.

Thus far, the examples have not exhibited any behaviors that are contraryto what we would expect from a Riemann integral. The next three examplesbreak this pattern.

Example 54 (Step functions). Let

f .x/ D

�1; 0 � x � 1

0; 1 < x � 2

and

g .x/ D

�0; 0 � x < 1

1; 1 � x � 2:

By examples 51 and 52, R-SR 10f dg D 1 and R-S

R 21f dg D 0. However, f

is not Riemann-Stieltjes integrable with respect to g over Œ0; 2�.

To see why, let P DfŒxi�1; xi �g be a partition of Œ0; 2� that does not in-clude 1 as a division point. Then there is an integer k such that xk�1 < 1 <xk . Let P1 be a tagged partition based on P where tk D xk�1 and let P2be a tagged partition based on P with tk D xk . Since �g D 0 for all butthe kth interval, SR-S .f; g;P1/ D 1 and SR-S .f; g;P2/ D 0. As the meshof P can be arbitrarily small, f cannot be Riemann-Stieltjes integrable withrespect to g over Œ0; 2�.

Contrary to the situation with the Riemann integral, Riemann-Stieltjesintegrability over Œa; b� and Œb; c� is not sufficient to conclude Riemann-Stieltjes integrability over Œa; c�. Moreover, while bounded functions are Rie-mann integrable as long as the set of discontinuities has measure zero, exam-ple 54 shows that discontinuity at even a single point can render a functionnon-Riemann-Stieltjes integrable. Even monotonicity can fail for Riemann-Stieltjes integrals.

Example 55 (Not monotone). Let f1 .x/ D 1, f2 .x/ D 2, and

g .x/ D

�1; 0 � x � 1

0; 1 < x � 2:

Then f1 < f2; but R-SR 20 f1 dg D �1 while R-S

R 20 f2 dg D �2.

One would not necessarily expect the fundamental theorem of calculus toapply to Riemann-Stieltjes integrals, but even more basic properties fail. Thefunction F .x/ D R-S

R xa f dg need not be continuous.

Example 56 (F not continuous). Let f .x/ D 1 and

g .x/ D

�0; 0 � x � 1

1; 1 < x � 2:

Then by example 51,

F .x/ D R-SZ x

0

f dg D g .x/

for x 2 Œ0; 2�. Thus F is not continuous at x D 1.

8.2 Basic integrability theoremsIn light of the previous examples, it seems that we are more than ready forsome general positive results. We begin with a pair of theorems that shouldseem very familiar.

8.2. Basic integrability theorems 231

Theorem 84 (Cauchy criterion). Given functions f and g on Œa; b�, f isRiemann-Stieltjes integrable with respect to g over Œa; b� if and only if givenany " > 0 there is a ı > 0 such that

jSR-S .f; g;P1/ � SR-S .f; g;P2/j < "

whenever P1 and P2 are tagged partitions of Œa; b� whose meshes are bothless than ı.

Proof. The proof of this theorem follows the same lines as previous Cauchycriteria and is left as an exercise.

The Cauchy criterion allows us to prove integrability over subintervalseven though example 54 (page 229) shows that the converse is false.

Theorem 85 (Subintervals). If f and g are functions on Œa; b� with fRiemann-Stieltjes integrable with respect to g on Œa; b�, then f is Riemann-Stieltjes integrable with respect to g over Œc; d � for any Œc; d � � Œa; b�.

Proof. This is a straightforward exercise in the application of the Cauchycriterion.

We can also prove a fairly general existence result for the Riemann-Stieltjes integral.

Theorem 86 (Integrability conditions). Let f be a continuous function onŒa; b� and suppose that g is increasing on Œa; b�. Then f is Riemann-Stieltjesintegrable with respect to g over Œa; b�.

Proof. Fix " > 0 and set B D g .b/ � g .a/ C 1. Since f is uniformlycontinuous on Œa; b� we can find a ı > 0 so that

jf .x/ � f .y/j <"

2B

for all x; y 2 Œa; b� satisfying jx � yj < ı. Suppose that P1 Df.ti ; Œxi�1; xi �/g and P2 are two tagged partitions of Œa; b� with mesh lessthan ı. Then P3 D P1[P2 is a common refinement to which we assign mid-point tags.1 Define the step functions �P1 D f .a/�1fagC

PP inf.xi�1;xi � f �

1.xi�1;xi � and �P1 D f .a/ � 1fag CP

P sup.xi�1;xi � f � 1.xi�1;xi �.Because g is increasing,

SR-S .�P1 ; g;P1/ � SR-S .f; g;P1/ � SR-S��P1; g;P1

�1 The particular choice of tags does not really matter.

and

SR-S��P1 ; g;P1

�D SR-S

��P1 ; g;P3

�� SR-S .f; g;P3/� SR-S

��P1 ; g;P1

�.

Moreover,

SR-S��P1 ; g;P1

�� SR-S

��P1 ; g;P1

�DXP1

��P1 .ti / � �P1 .ti /

�.g .xi / � g .xi�1//

�XP1

"

2B.g .xi / � g .xi�1//

D"

2B.g .b/ � g .a// <

"

2.

HencejSR-S .f; g;P1/ � SR-S .f; g;P3/j <

"

2.

Similarly,

jSR-S .f; g;P2/ � SR-S .f; g;P3/j <"

2.

ThusjSR-S .f; g;P1/ � SR-S .f; g;P2/j < ":

Integrability follows from the Cauchy criterion.

By linearity, theorem 86 can immediately be extended to the case where gis of bounded variation. This extension is quite useful in the context of theo-rem 89 (page 236) below. On the other hand, example 54 (page 229) servesas a warning that the restriction that f be continuous cannot be removed eas-ily. The restriction can be loosened, but to do so requires additional theorythat we have not yet developed. By the time we have developed the theory,the effort to extend theorem 86 will be superseded by Lebesgue-like results.

8.3 Evaluation theoremsWhile the theorems of the previous section allow us to determine that aRiemann-Stieltjes integral exists, they provide little help in evaluating anintegral. For the Riemann integral, FTC-1 is the basic evaluation tool. Thereis no such theorem for the Riemann-Stieltjes integral.

8.3. Evaluation theorems 233

One typically encounters two types of Riemann-Stieltjes integrals cor-responding to the two basic types of probability distributions: continuousand discrete. The following two theorems provide useful tools for theirevaluation.

Theorem 87 (Continuous distributions). Suppose that f and g are func-tions on Œa; b� where f is continuous and g is differentiable with g0 beingRiemann integrable over Œa; b�. Then f is Riemann-Stieltjes integrable withrespect to g over Œa; b� and

R-SZ b

a

f dg D RZ b

a

fg0.

Proof. Since g0 is Riemann integrable, g0 and so fg0 are bounded and con-tinuous a.e. Thus fg0 is Riemann integrable. Since g0 is bounded we canselect a real value B so that jg0j 0 and choose ı > 0 so that jf .x/ � f .y/j < "B.b�a/

when-ever x; y 2 Œa; b� with jx � yj < ı. Let P Df.ti ; Ii /g be a tagged parti-tion of Œa; b� with mesh less than ı. By FTC-1 for the Riemann integral,�gi D

RRIig0 so thatˇ

ˇSR-S .f; g;P/ � RZ b

a

fg0

ˇˇ D

ˇˇX

Pf �g �

XP

RZIi

fg0

ˇˇ

D

ˇˇX

P

RZIi

Œf .ti / � f � g0

ˇˇ

<XP

RZIi

"

B .b � a/B

D"

b � a

XP�xi D ".

Hence f is Riemann-Stieltjes integrable with respect to g over Œa; b� and

R-SZ b

a

f dg D RZ b

a

fg0:

Theorem 87 can be strengthened by allowing g to have points of non-differentiability. Assuming g0 is differentiable a.e. and Riemann integrablewhen set to zero at the points of nondifferentiability, the theorem remainsvalid. (See exercise 25.)

Before stating the evaluation theorem for the discrete case, we introducesome additional notation. Let f be defined on Œa; b� and let c 2 Œa; b�. Then

f�cC�D limx!cC

f .x/ and f .c�/ D limx!c�

f .x/

where it is understood that

f .a�/ D f .a/ and f�bC�D f .b/ .

These values are always defined for monotone functions.

Theorem 88 (Discrete distributions). Let g be a step function on Œa; b�with discontinuities at fcig

nkD1 and suppose that f is a continuous func-

tion on Œa; b�. Then f is Riemann-Stieltjes integrable with respect to g overŒa; b� and

R-SZ b

a

f dg DnXiD1

f .ci /�g�cCi�� g

�c�i��

.

Proof. Fix " > 0 and setM D maxx2Œa;b� jg .x/j. By the uniform continuityof f , there is a ı0 > 0 so that jf .x/ � f .y/j < "

4nMwhenever x; y 2

Œa; b� with jx � yj < ı0. Take ı1 D min1�i;j�n˚ˇci � cj

ˇ�and set ı D

min fı0; ı1=2g. Note that g .x/ D g�c�i�

for x 2 Œa; b� \ .ci � ı; ci / andg .x/ D g

�cCi�

for x 2 Œa; b� \ .ci ; ci C ı/.Now suppose that P D f.ti ; Œxi�1; xi �/g is a tagged partition of Œa; b� with

mesh less than ı. Since the mesh of P is less than ı1, any interval�xj�1; xj

�can contain at most one point where g is discontinuous and any point ofdiscontinuity is contained in at most two intervals from P . If

�xj�1; xj

�contains no point of discontinuity, then f

�tj� �g�xj�� g

�xj�1

��D 0. If

xj�1 < ck < xj thenˇf�tj� �g�xj�� g

�xj�1

�� f .ck/

�g�cCk

�� g

�c�k�� ˇ

Dˇf�tj� �g�cCk

�� g

�c�k�� f .ck/

�g�cCk

�� g

�c�k�� ˇ

Dˇf�tj�� f .ck/

ˇ ˇg�cCk

�� g

�c�k�ˇ

<"

4nM2M D

"

2n.

8.3. Evaluation theorems 235

If ck is a division point with xj�1 < ck D xj < xjC1, thenˇf�tj��g�xj�� g

�xj�1

��C f

�tjC1

��g�xjC1

�� g

�xj��

� f�ck��g�cCk

�� g

�c�k��ˇ

Dˇf�tj��g�ck�� g

�c�k��

C f�tjC1

��g�cCk

�� g

�ck��

� f�ck��g�cCk

�� g

�c�k��ˇ

�ˇf�tj�� f

�ck�ˇˇg�ck�� g

�c�k�ˇ

Cˇf�tjC1

�� f

�ck�ˇˇg�cCk

�� g

�ck�ˇ

<"

4nM2M C

"

4nM2M D

"

n:

The cases where ck D x0 or ck D xn are similar but simpler. HenceˇˇSR-S .f; g;P/ �

nXiD1


�c�i��ˇˇ < "

so that f is Riemann-Stieltjes integrable with respect to g over Œa; b� with

R-SZ b

a

f dg DnXiD1


�c�i��:

Theorem 92 (page 239) in the next section can be used to extend theorem88 to conclude that

R-SZ b

a

f dg DXk

f .ck/�g�cCk

�� g

�c�k��

when g has a countable number of discontinuities fckg satisfyingPk

ˇg�cCk

�� g

�c�k

�ˇ< 1. This situation is quite common when work-

ing with discrete probability distributions.Linearity allows us to apply the previous theorems in mixed situations.

Example 57. Compute R-SR 20 x

2 dg where

g .x/ D

8<ˆ:

0; x D 0

3x C 1; 0 < x < 1

1; x D 1

x3; 1 < x < 2

9; x D 2:

First note that we can write g D g1 C g2 where

g1 .x/ D

�3x; 0 � x � 1

x3 C 2; 1 < x � 2

and

g2 .x/ D

8<:0; x D 0

1; 0 < x < 1

�2; 1 � x < 2

�1; x D 2:

Then, by theorems 86 and 87 (pages 231 and 233),

R-SZ 2

0

x2 dg1 DRZ 1

0

x2 3 dx C RZ 2

1

x2 3x2 dx D 1C93

5D98

5

and, by theorem 88,

R-SZ 2

0

x2 dg2 D 0 � 1C 1 � .�3/C 4 � 1 D 1.

So by linearity, R-SR 20 x

2 dg D 1035

.

On occasion, it is simpler to consider the integral R-SR ba g df instead of

the integral R-SR ba f dg. The following theorem (which should have a very

familiar appearance) allows us to interchange the roles of the two functionsin a Riemann-Stieltjes integral.

Theorem 89 (Integration by parts). Suppose that f and g are boundedfunctions on Œa; b� with no common discontinuities. If f is Riemann-Stieltjesintegrable with respect to g over Œa; b� then g is Riemann-Stieltjes integrablewith respect to f over Œa; b� and

R-SZ b

a

g df D f .b/ g .b/ � f .a/ g .a/ � R-SZ b

a

f dg.

Proof. Let " > 0 be given. By assumption, there is a ı > 0 so thatˇˇSR-S .f; g;Q/ � R-S

Z b

a

f dg

ˇˇ < "

for any tagged partition Q of Œa; b� with mesh less than ı. Now suppose thatP Df.ti ; Œxi�1; xi �/gniD1 is a tagged partition of Œa; b� with mesh less than

ı=2. Define t0 D a and tnC1 D b. Then P� D f.xk�1; Œtk�1; tk �/gnC1kD1 is atagged partition of Œa; b� with mesh less than ı. Henceˇ

ˇSR-S .g; f;P/ � f .b/ g .b/ � f .a/ g .a/ � R-S

Z b

a

f dg

!ˇˇ

D

ˇˇnXiD1

g .ti / f .xi / �

nXiD1

g .ti / f .xi�1/

�f .xn/ g .tnC1/C f .x0/ g .t0/CR-SZ b

a

f dg

ˇˇ

D

ˇˇ R-S

Z b

a

f dg �nC1XiD1

f .xi�1/ Œg .ti / � g .ti�1/�

ˇˇ

D

ˇˇ R-S

Z b

a

f dg � SR-S�f; g;P�

�ˇˇ < ".Hence g is Riemann-Stieltjes integrable with respect to f over Œa; b� with

R-SZ b

a

g df D f .b/ g .b/ � f .a/ g .a/ � R-SZ b

a

f dg:

Example 58. By theorem 87 (page 233),

R-SZ 3

0

x2 dx2 D RZ 3

0

x22x dx D81

2:

By theorem 89,

R-SZ 3

0

x2 dx2 D 9 � 9 � 0 � 0 � R-SZ 3

0

x2 dx2:

Solving, we again find that

R-SZ 3

0

x2 dx2 D81

2:

8.4 Convergence theoremsWe will prove one convergence theorem for the Riemann-Stieltjes integral.In preparation, we establish two technical lemmas. Notice the alignmentof the hypotheses for these theorems with the integrability conditions fromSection 2.

Lemma 90. Suppose that f is a continuous function on Œa; b� and that g 2BV .Œa; b�/.2 Then ˇ

ˇ R-SZ b

a

f dg

ˇˇ � max

Œa;b�f � V ba g.

Proof. Exercise 31.

Lemma 91. Suppose that f is a continuous function on Œa; b� and "; ı > 0

with jf .x/ � f .y/j < " for all x; y 2 Œa; b� satisfying jx � yj < ı. Furthersuppose that g 2 BV .Œa; b�/. If P Df.ti ; Œxi�1; xi �/g is a tagged partitionof Œa; b� with mesh less than ı, thenˇ

ˇSR-S .f; g;P/ � R-SZ b

a

f dg

ˇˇ < "V ba g.

Proof. First note that maxx2Œxi�1;xi � jf .ti / � f .x/j < " and, by example51 (page 228),

f .ti / Œg .xi / � g .xi�1/� DR-SZ xi

xi�1

f .ti / dg:

Thus, by lemma 90 (page 238) and property (2) of theorem 60 (page 181),ˇˇSR-S .f; g;P/ � R-S

Z b

a

f dg

ˇˇ D

ˇˇXi

R-SZ xi

xi�1

.f .ti / � f / dg

ˇˇ

<Xi

"V xixi�1g

D "V ba g.

Since the Riemann-Stieltjes integral involves two functions, it is naturalthat convergence theorems consider two sequences of functions. Understand-ing the consequences of the convergence of fgng is critical in the study ofprobability theory. In particular, the central limit theorem, arguably the mostimportant theorem in probability and statistics, concerns just the sort of con-vergence of fgng identified in the following theorem.3

2 Recall that V .f;P/ DPk jf .xk/� f .xk�1/j where P DfŒxk�1; xk�g is a partition

of Œa; b�, V ba f D supP V .f;P/, and BV .Œa; b�/ is the set of all functions on Œa; b� forwhich V ba f is finite.

3 The relevant type of convergence is convergence in distribution. The sequence fgng con-verges in distribution to g if fgng converges pointwise to g at all points where g is contin-uous. In a probability theory context, g is monotone increasing so g is continuous except ata countable number of points and all the functions fgng and g have a variation of 1.

Theorem 92. Suppose that ffng is a sequence of continuous functions thatconverges uniformly to f on Œa; b�. Suppose further that fgng is a sequenceof functions converging to g on a dense subset S of Œa; b� that includes a andb. If there is a uniform bound M on the variations of fgng and g then

limn!1

R-SZ b

a

fn dgn DR-SZ b

a

f dg.

Proof. As f is continuous on Œa; b�, it is bounded by some value jf j 0 and choose ı > 0 so that jf .x/ � f .y/j < "

4Mfor all

x; y 2 Œa; b� satisfying jx � yj < ı. Let P Df.ti ; Œxi�1; xi �/gmiD1 be atagged partition of Œa; b� with mesh less than ı whose division points areselected from S: Because fgng converges to g on S , we can find a naturalnumberNg so that jgn .xi / � g .xi /j < "

4mBfor all n � Ng and 1 � i � m.

Since ffng converges uniformly to f we can find a natural number Nf sothat jfn � f j < "

4Mfor all n � Nf . Set N D max

˚Nf ; Ng

�and suppose

that n � N .By the triangle inequality and lemma 90 (page 238),ˇ

ˇ R-SZ b

a

fn dgn �R-SZ b

a

f dg

ˇˇ

�

ˇˇ R-S

Z b

a

.fn � f / dgn

ˇˇC

ˇˇ R-S

Z b

a

f d.gn � g/

ˇˇ

�"

4MM C

ˇˇ R-S

Z b

a

f d.gn � g/

ˇˇ .

For the second term, we use the triangle inequality and lemma 91(page 238) to conclude thatˇˇ R-S

Z b

a

f d.gn � g/

ˇˇ �

ˇˇ R-S

Z b

a

f dgn � SR-S .f; gn;P/ˇˇ

C jSR-S .f; gn � g;P/j

C

ˇˇ R-S

Z b

a

f dg � SR-S .f; g;P/ˇˇ

<"

4MV ba gn C

mXiD1

B"

4mBC

"

4MV ba g <

3

4".

Hence

limn!1

R-SZ b

a

fn dgn DR-SZ b

a

f dg:

Example 59. Compute limn!1R-SR 10 arctannx dxn.

Note that fxng converges pointwise to

g .x/ D

�0; 0 � x < 1

1; x D 1

on Œ0; 1� and V 10 xn D V 10 g D 1. Unfortunately, farctannxgn does not con-

verge uniformly on Œ0; 1�. However, it does converge uniformly to �2

on�12; 1�

and is uniformly bounded by �2

on�0; 12

�. Thus

limn!1

R-SZ 1

12

arctannx dxn D R-SZ 1

12

�

2dg D

�

2

and

0 � limn!1

R-SZ 1

2

0

arctannx dxn � limn!1

R-SZ 1

2

0

�

2dxn D R-S

Z 12

0

�

2dg D 0:

Hence

limn!1

R-SZ 1

0

arctannx dxn D�

2.

Example 60. Set

fn .x/ D

8<:0; x < 0

12; 0 � x < 1C 1

n

1; 1C 1n� x

and evaluate limn!1R-SR 2�1 fn .x/ dx5.

In this case, the functions ffng are not continuous and ffng convergespointwise but not uniformly on Œ0; 1� to

f .x/ D

8<:0; x < 0

12; 0 � x � 1

1; 1 < x:

We can still use theorem 92 by applying integration by parts.

limn!1

R-SZ 2

�1

fn .x/ dx5 D limn!1

�1 � 25 � 0 � .�1/5 � R-S

Z 2

�1

x5 dfn

D 25 � R-S

Z 2

�1

x5 df D 25 �

�0 �1

2C 1 �

1

2

D63

2

by theorem 88 (page 234).

8.5. Connecting to measure theory 241

Observe that if one thinks of the f in the previous example as a weightingfunction, then it places a weight of 1

2at x D 0 and at x D 1. In this case, it

is more common to work with the function

f � .x/ D

8<:0; x < 0;

12; 0 � x < 1;

1; 1 � x:

While ffng does not converge to f � at x D 12

, ffng does converge to f � ona dense set that includes �1 and 2. Thus we still have

limn!1

R-SZ 2

�1

fn .x/ dx5 D 25 � R-SZ 2

1

x5 df � D 25 �1

2

even though ffng fails to converge to f � on a set of weight 12

. While thissituation initially seems rather odd, it makes a bit more sense if we think of asequence of random variables fXng where Xn takes on the values 0 or 1C 1

n

with probability 12

each. The random variables converge in a natural sense4

to a random variable that takes on the values 0 and 1with probability 12

each.

8.5 Connecting to measure theoryThe fact that f fails to be integrable with respect to g over Œa; b� when fand g have a common point of discontinuity is rather awkward. This runscounter to all our previous experience where a few discontinuities could beignored. One way to remedy this difficulty would be to develop a theory ofthe gauge-Stieltjes integral (see exercise 10). We will take a different path.

The fundamental idea underlying the Riemann-Stieltjes integral is that theinterval .s; t � is treated according to its weight, w ..s; t �/ D g .t/ � g .s/,rather than its length, l ..s; t �/ D t � s. This statement should remind you ofthe Lebesgue integral. In fact, we can develop a general theory of measurevery similar to that of Lebesgue measure using w instead of l . For this de-velopment, we assume that g is monotone increasing and right-continuous(g�xC�D g .x/ for x in the domain of g). For the rest of this section, we

will assume that g satisfies these conditions without reiteration in subsequenttheorems and lemmas.

If we are to mimic the development of Lebesgue measure, we need toknow that the weight function w ..s; t �/ D g .t/ � g .s/ shares some funda-mental properties with the length function l ..s; t �/ D t � s.

4 They converge in distribution.

Theorem 93 (Properties of w). Under the assumptions of this section,

1. w ..a; b�/ D w ..a; c�/C w ..c; b�/ whenever a < c < b.

2. If .a; b� [nkD1

.ak ; bk �, then w ..a; b�/ �Pk w ..ak ; bk �/.

3. If the intervals in f.ak ; bk�gnkD1 are disjoint and [k .ak ; bk� .a; b�,

thenPk w ..ak ; bk�/ � w ..a; b�/.

4. If .a; b� [1kD1

.ak ; bk �, then w ..a; b�/ �Pk w ..ak ; bk �/.

5. If the intervals in f.ak ; bk�g1kD1 are disjoint and [k .ak ; bk� .a; b�,

thenPk w ..ak ; bk�/ � w ..a; b�/.

Proof. We will prove only part (4). The other properties are more straight-forward and are left as exercises.

Fix " > 0. Since g is increasing and right-continuous, we can find ana� > a so that g .a�/ < g .a/ C " and for each k 2 N we can choosea b�

k> bk so that w

��ak ; b

�k

�� w ..ak ; bk�/ C

"

2k. Then

˚�ak ; b

�k

��,

being an open cover of the compact interval Œa�; b�, has a finite subcover˚�aj ; b

�j

��njD1

. Then from (2),

w ..a; b�/ � w��a; a�

��C

nXjD1

w��aj ; b

�j

�� "C

1XkD1

w ..ak ; bk�/C ".

Since " > 0 is arbitrary, w ..a; b�/ �Pk w ..ak ; bk �/.

Since w behaves sufficiently like l , we can use the weight function basedon g to define the outer measure of any subset of R.

Definition 32 (Outer measure). Let A � R. The Lebesgue-Stieltjes outermeasure of A is

��g .A/ D inf�Xk

w ..ak ; bk �/ W˚.ak ; bk �

�is a countable cover of A by finite intervals

.

Example 61 (.a; b�). For any outer measure ��g , we have ��g ..a; b�/ Dw ..a; b�/ D g .b/ � g .a/.

To see why, note that .a; b� covers itself so that ��g ..a; b�/ � w ..a; b�/.The reverse inequality follows from (4) of theorem 93.

Example 62 (Œa; b�). For any outer measure ��g , we have ��g .Œa; b�/ Dg .b/ � g .a�/.

To see why, note that for any ı > 0; Œa; b� � .a � ı; b� so that��g .Œa; b�/ � g .b/ � g .a � ı/. Since the inequality holds for all ı > 0,we can conclude that ��g .Œa; b�/ � g .b/ � g .a

�/. For the other inequality,observe that if f.ak ; bk�g is a countable cover of Œa; b�, then there is an in-dex k� so that a 2 .ak� ; bk� �. Then Œa; b� � .ak� ; b� [k .ak ; bk �. Henceg .b/ � g .a�/ � g .b/ � g .ak�/ � w ..ak� ; b�/ �

Pk w ..ak ; bk �/ by (4)

of theorem 93.

Example 63 (Point mass). Define

g .x/ D

�0; x < 0

1; 0 � x

and let A be a subset of R. Then

��g .A/ D

�1; 0 2 A

0; 0 62 A.

To see why, note that

1. w ..a; b�/ D g .b/ � g .a/ is 1 if a < 0 � b and is zero otherwise.

2. A [k2Z .k � 1; k� andPk w ..k � 1; k�/ D 1 so ��g .A/ � 1.

3. If 0 2 A and A [k .ak ; bk �, then 0 2 .ak� ; bk� � for some k�. ThusPk w ..ak ; bk �/ � 1 so ��g .A/ � 1.

4. If 0 62 A, then A [n˚��n;� 1

n

�[ .0; n�

�so ��g .A/ D 0.

Example 64 (Bernoulli distribution). Let 0 < p < 1 and define

g .x/ D

8<:0; x < 0

1 � p; 0 � x < 1

1; 1 � x.

Let A be a subset of R. An analysis similar to that of the previous exampleshows that

��g .A/ D

8<:

0; 0; 1 62 A

1 � p; 0 2 A; 1 62 A

p; 0 62 A; 1 2 A

1 0; 1 2 A.

Thus ��g describes the probabilities associated with a random variable thattakes on the value 1 with probability p and the value 0 with probability1 � p.5

Of course, the outer measure generated by g need not be translation in-variant. However, it will satisfy the remaining properties of Lebesgue outermeasure.

Theorem 94 (Properties of outer measure). Any Lebesgue-Stieltjes outermeasure ��g satisfies

1. ��g .;/ D 0.

2. If A B then ��g .A/ � ��g .B/.

3. ��g .[kAk/ �Pk ��g .Ak/ for any countable collection of sets fAkg.

Proof. The verification of these properties follows very closely the proof ofthe corresponding properties for Lebesgue outer measure and is left as anexercise.

As with Lebesgue measure, one of the critical problems is to identify thecollection of sets on which �g is countably additive. We can address thisquestion the same way—with the Caratheodory criterion.

Definition 33 (Caratheodory’s measurability condition). A set E R is(�g ) measurable if

��g .A/ D ��g .A \E/C �

�g .A \E

c/

for every subsetA R. WhenE is measurable, we write �g .E/ D ��g .E/.

Depending on the function g, the set of �g-measurable sets can be quitedifferent from the Lebesgue-measurable sets.

Example 65 (Point mass). Define

g .x/ D

�0; x < 0

1; 0 � x

as in example 63 (page 243). Then every subset E of R is measurable. Tosee why, let A be an arbitrary subset of R. If 0 62 A, then

��g .A/ D ��g .A \E/ D �

�g .A \E

c/ D 0:

5 In other words, ��g describes the probabilities associated with the binomial distributionB .1;p/ .

If 0 2 A, then 0 is in exactly one of A \E or A \Ec . In either case,

��g .A/ D ��g .A \E/C �

�g .A \E

c/

so that E is measurable.

Review the development of Lebesgue measure on pages 98 through 104.Except for the proof that .�1; a/ is a measurable set and the proof of theo-rem 22 (page 104), none of the proofs uses properties or lengths of intervals.The proofs depend only on the Caratheodory measurability condition, mono-tonicity, and subadditivity. We have already verified that��g is monotone andsubadditive so we can assert that

1. �g is countably additive on �g-measurable sets and

2. the collection of �g-measurable sets is a sigma algebra.

Once we verify that .�1; a/ is �g -measurable, we can conclude that, inaddition,

3. The collection of �g-measurable sets contains the Borel sets.

Theorem 95. Intervals of the form .�1; a/ are �g-measurable.

Proof. Since we know that the collection of �g-measurable sets is a sigmaalgebra, it will suffice to show that intervals of the form .�1; a� are �g-measurable. Let A be a subset of R and choose a countable cover fIkg of Aby intervals of the form Ik D .ak ; bk � such thatX

k

w .Ik/ � ��g .A/C ":

Since A [kIk , monotonicity and subadditivity imply that

��g .A \ .�1; a�/C ��g .A \ .a;C1//

� ��g ..[kIk/ \ .�1; a�/C ��g ..[kIk/ \ .a;C1//

�Xk

��g .Ik \ .�1; a�/CXk

��g .Ik \ .a;C1// :

Now

.ak ; bk� \ .�1; a� D

8<:.ak ; bk � ; bk � a

.ak ; a� ; ak � a < bk;; a < ak

and

.ak ; bk � \ .a;C1/ D

8<:;; bk � a

.a; bk � ; ak � a < bk.ak ; bk� ; a � ak :

Hence ��g and w agree for all the sets in the summation and, by part (1) oftheorem 93 (page 242), w .Ik/ D w .Ik \ .�1; a�/ C w .Ik \ .a;C1//.Moreover, all the terms in the summations are nonnegative so they can bereordered. Therefore we can continue the chain of inequalities with

DXk

w .Ik \ .�1; a�/CXk

w .Ik \ .a;C1//

DXk

Œw .Ik \ .�1; a�/C w .Ik \ .a;C1//�

DXk

w .Ik/ � �� .A/C ".

Since " > 0 was arbitrary, we conclude that

�� .A \ .�1; a�/C �� .A \ .a;C1// � �� .A/ .

The reverse inequality is a consequence of subadditivity. The interval.�1; a� is measurable.

Since the collection of �g-measurable sets is a sigma algebra containingintervals of the form .�1; a�, the collection of �g-measurable sets containsall the Borel sets. As we have seen in earlier examples, there may be addi-tional �g-measurable sets. It is relatively straightforward to prove that anyset of ��g -measure zero is �g-measurable. These sets fill out the collectionof �g -measurable sets as they do for Lebesgue measure.

Theorem 96. A subset E of R is �g-measurable if and only if E can bewritten as E D B [Z where B is a Borel set and Z is a set of �g-measurezero.

Proof. Let E be a �g-measurable set and set F D Ec . First suppose that�g .F / <1. Then for any natural number n there is a countable cover fIkgof F by intervals of the form Ik D .ak ; bk� such that

�g .F / � �g .[kIk/ �Xk

w .Ik/ < �g .F /C1

n.

8.6. Integration with measures 247

If we set Gn D [kIk , then Gn is a Borel set satisfying F Gn and�g .Gn/ < �g .F / C

1n

. When �g .F / D 1, apply the same argumentto Fk D F \.k; k C 1� for k 2 Z to find Borel sets

˚Gn;k

�1kD�1

containingFk such that ��

�Gn;knFk

�< 1

n2kC2. Then Gn D [kGn;k is a Borel set

containing F with �� .GnnF / < 1n

.The measurability of F now implies that

�g .F /C1

n> �g .Gn/ D �g .Gn \ F /C �g .Gn \ F

c/

D �g .F /C �g .GnnF / .

Subtracting, we find that �g .GnnF / < 1n

. Set G D \nGn and Z DGnF D G \ E. Since GnF GnnF , we have �g .Z/ D �g .GnF / D 0.Taking complements,Gc F c D E so thatGc[Z D Gc[.G \E/ D E.Thus E is the union of a Borel set and a set of �g-measure zero.

The converse is a consequence of the facts that sets of �g-measure zeroare measurable and that the collection of �g-measurable sets is a sigmaalgebra.

8.6 Integration with measures

The motivation for considering a more general theory of measure is the ob-servation that the properties of the length function l and the weight functionw are quite similar. This suggests the possibility of constructing a theoryof integration based on �g to extend the Riemann-Stieltjes integral like theLebesgue integral extends the Riemann integral. As with the Lebesgue in-tegral, the critical condition that must be satisfied before considering theintegral of a function f is that f be measurable.

Definition 34 (Measurable function). Let X be a �g-measurable subset ofreal numbers and let f W X ! R. Then f is �g-measurable if the preimageunder f of any interval is �g-measurable.

Review the section on Lebesgue measurable functions on page 105 andfollowing. You will see that all the proofs depend solely on

1. the properties of inverse images of functions,

2. the fact that all Borel sets are measurable, and

3. the fact that the collection of measurable sets is a sigma algebra.

Since these facts are also true for �g-measurable sets, all the proofs fromSection 6 of Chapter 6 transfer without modification into the context of �g-measurable functions. In particular,

1. All increasing functions are �g-measurable.

2. All continuous functions are �g-measurable.

3. Linear combinations of �g-measurable functions are �g-measurable.

4. If f1 is a �g-measurable function and f1 D f2 except on a set of �g-measure zero,6 then g is �g -measurable:

5. Supremums and infimums of sets of �g-measurable functions are again�g-measurable:

6. Limits, limsups, and liminfs of sequences of �g-measurable functionsare again �g-measurable:

Because of the close parallel between Lebesgue and �g-measurability,one commonly drops the modifier and speaks only of measurability. Theparticular sigma algebra of measurable sets is understood from the context.

At this point, we have essentially replicated the infrastructure for theLebesgue integral found in Sections 2 through 6 of Chapter 6 for the moregeneral measure �g . Given that we have the same supporting structure, wecan define the Lebesgue-Stieltjes integral by making slight modifications tothe definition of the Lebesgue integral.

Definition 35 (Lebesgue-Stieltjes integral). Let f be a nonnegative, �g-measurable function defined on Œa; b�. Let P be the partition of Œ0;C1/defined by the division points fykg

nkD1 and set Ek D f �1 .Œyk�1; yk//, k D

1; 2; : : : ; n, and EnC1 D f �1 .Œyn;1//. The (lower) Lebesgue-Stieltjessum of f with respect to P is SL-S

�f; �g ;P

�DPnC1kD1 yk�1 ��g .Ek/. The

(�g -) integral of f over Œa; b� is

MZ b

a

f d�g D supPSL-S

�f; �g ;P

�where the supremum is taken over all partitions P of Œ0;C1/ : The functionf is said to be (�g ) integrable if M

R ba f d�g < C1:7

6 Alternatively, f1 D f2 ��g-a.e. or f1 D f2 a.e. ��g .7 We use M

R ba f d�g since the integral is based on a measure. Serendipitously, M also corre-

sponds to the Greek letter �.

8.6. Integration with measures 249

When f W Œa; b� ! R and the integral of at least one of f C D12.jf j C f / or f � D 1

2.jf j � f / is finite, we set

MZ b

a

f d�g DMZ b

a

f C d�g �MZ b

a

f � d�g

and say that f is (�g-) integrable if MR ba jf j d�g < C1 or, equivalently,

if the integrals of both f C and f � are finite.

The simple and Young versions of the Lebesgue integral can be similarlytransformed and the resulting integrals are again equivalent to the Lebesgue-Stieltjes integral. The Lebesgue-Stieltjes integral is a generalization of theLebesgue integral since �g D � when g .x/ D x.

It would appear that when g is an increasing, right-continuous functionand f is Riemann-Stieltjes integrable with respect to g over Œa; b� thenMR ba f d�g D R-S

R ba f dg. This is not quite correct since the Riemann-

Stieltjes integral will not capture any weight the measure �g places at a.Instead,

MZ b

a

f d�g DR-SZ b

a

f dg C f .a/�g .fag/ :

Thus the Lebesgue-Stieltjes integral agrees with the Riemann-Stieltjes inte-gral over Œa; b� exactly when g is continuous at a. As shown by the followingexample, the Lebesgue-Stieltjes integral is a strict extension of the Riemann-Stieltjes integral.

Example 66 (Step functions). Let

f .x/ D

�1; 0 � x � 1

0; 1 < x � 2

and let

g .x/ D

�0; 0 � x < 1

1; 1 � x � 2:

In example 54 (page 229), we showed that f is not Riemann-Stieltjes inte-grable with respect to g over Œ0; 2�. However M

R 20f d�g does exist and is 1.

Let P be a partition of Œ0;C1/ defined by the division points fykgnkD1

with at least one division point in .0; 1/. Set Ek D f �1 .Œyk�1; yk//,k D 1; 2; : : : ; n, and EnC1 D f �1 .Œyn;1//. Now y0 D 0 < y1and there is exactly one index k1 satisfying yk1�1 � 1 < yk1 . Thus�g .E1/ D �g ..1; 2�/ D 0, �g

�Ek1

�D �g .Œ0; 1�/ D 1, and �g .Ek/ D

�g .;/ D 0 for all other values of k. Hence SL-S�f; �g ;P

�D yk1�1 so

MR baf d�g D supP SL-S

�f; �g ;P

�D 1.

Replacing � by �g and LR ba f by M

R ba f d�g in the proofs of theorems

33 and 34 (page 135), we see that, as with the Lebesgue integral, we canignore the behavior of f on a set of �g-measure zero. In other words, iff1 is �g-integrable and f1 D f2 a.e. �g , then f2 is �g -integrable andMR baf2 d�g D M

R baf1 d�g . Note, however, that a single point can have

positive measure. Changing the value of the function there will produce adifferent value for the integral.

What about the convergence properties of the Lebesgue-Stieltjes integral?Since the infrastructure is essentially the same, the convergence theorems forthe Lebesgue integral extend to the Lebesgue-Stieltjes integral. The state-ments and proofs require only minor changes. The most significant changeis that .b � a/ must be replaced by �g .Œa; b�/ in the proof of theorem 36(page 136).8

Theorem 97 (Bounded convergence). Let ffkg be a uniformly bounded se-quence of�g -measurable functions that converges pointwise to f �g-a.e. onŒa; b�. Then

limk

MZ b

a

fk d�g DMZ b

a

limkfk d�g D

MZ b

a

f d�g .

Theorem 98 (Monotone convergence). Let ffkg be a monotone increasingsequence of nonnegative �g-measurable functions that converges a.e. �g tof on Œa; b�. Then

MZ b

a

f d�g D limk

MZ b

a

fk d�g

where a value ofC1 is permissible.

Corollary 99. Let ffkg be a sequence of nonnegative, �g -measurable func-tions. Then

MZ b

a

limk

fk d�g � limk

MZ b

a

fk d�g .

Theorem 100 (Dominated convergence). Suppose that ffkg is a sequenceof �g-measurable functions that converges to f �g-a.e. on Œa; b�. If there is

a measurable function h with MR ba h d�g < 1 and for which jfkj � h a.e.

8 This change implies that in any subsequent generalization we need to guarantee that�.Œa; b�/ is finite. This is not a problem for measures built from increasing functions onŒa; b� but can be an issue for general measures.

8.7. Extensions to other types of measures 251

�g on Œa; b� for all k; then

MZ b

a

f d�g D limk

MZ b

a

fk d�g .

The ease with which one can move from integration with respect to �to integration with respect to �g goes a long way toward explaining thepreference for the Lebesgue integral over the gauge integral. But we are notdone.

8.7 Extensions to other types of measures

While we have restricted our attention to measures of the form �g whereg is an increasing, right continuous function, this is not necessary. We candispense with the function g and develop a theory of integration starting withany triple .R;A; �/ called a measure space where A is a sigma algebra ofsubsets of R and � is a measure on A. In other words, � is a function fromA to the extended real numbers satisfying

1. � .;/ D 0,

2. � .E/ � 0 for all E 2 A,9 and

3. � .[iEi / DPi � .Ei / for any countable, disjoint collection of sets

fEig from A.

As long as the measure of every bounded interval is finite, everything thatwe have developed so far readily extends in a natural way to this new context.We simply replace the sum SL-S

�f; �g ;P

�DPnC1kD1 yk�1 � �g .Ek/ with

the almost imperceptible modification SM .f; �;P/ DPnC1kD1 yk�1 �� .Ek/.

Example 67. Let A be the sigma algebra of all subsets of R and define � by

� .E/ D

8ˆ<ˆ:

0; 0; 1 62 E

13; 0 2 E; 1 62 E

23; 0 62 E; 1 2 E

1; 0; 1 2 E;

9 The condition that �.E/ � 0 for all E 2 A is equivalent to �.A/ � �.B/ for allA;B 2 A satisfying A � B . See Exercise 65.

for E 2 A. Then � is a measure on A. As long as a � 0 and b � 1;

MZ b

a

1 d� D 1 �1

3C 1 �

2

3D 1;

MZ b

a

x d� D 0 �1

3C 1 �

2

3D2

3; and

MZ b

a

�x �

2

3

�2d� D

4

9�1

3C1

9�2

3D2

9.

To completely justify the value of the second integral from the defini-tion, set f .x/ D x and f1 D f C. Let P be the partition of Œ0;C1/defined by the division points fykg

nkD1 and set Ek D f �11 .Œyk�1; yk//,

k D 1; 2; : : : ; n, and EnC1 D f �11 .Œyn;1//. Then there are integers k0and k1 such that yk0�1 � 0 < yk0 and yk1�1 � 1 < yk1 .10 Assumingk0 < k1,

��Ek0

�D �

��a; yk0

��D1

3;

��Ek1

�D �

��yk1�1; yk1

��D2

3;

and � .Ek/ D 0 for all other values of k. Thus

SM�f C; �;P

�D 0 �

1

3C yk1�1 �

2

3D yk1�1 �

2

3:

Hence MR ba f

C d� D supP SM�f C; �;P

�D 2

3. Taking f2 D f �

and Fk D f �12 .Œyk�1; yk// D .�yk ;�yk�1� for k > 1; � .Fk/ D

� ..�yk ;�yk�1�/ D 0 so that

SM .f �; �;P/ D 0 � � .F1/ D 0.

Hence MR ba f

� d� D 0 and thus MR ba x d� D 2

3.

In practice, for discrete measures such as the measure in this example itis easier to take an approach similar to that in theorem 88 (page 234) forthe Riemann-Stieltjes integral. Note those points in Œa; b� that have positivemeasure and compute the sum over those points of the value of the functiontimes the point’s measure.

If you are familiar with probability distributions, you will note that � isthe probability measure associated with the binomial distribution B

�1; 13

�.

10 In fact, k0 D 1.

The three integrals reflect the facts that the total probability is 1, the mean ofthe distribution is 2

3, and the variance is 2

9.

Example 68. Let A be the sigma algebra M of Lebesgue measurable setsand define � as the measure on M generated by the weight function

w ..a; b�/ D arctan .b/ � arctan .a/C ..a; b�/

where places weights of 12; 1, and 1

2at �1; 0, and 1 respectively. Splitting

� into �1and �2 defined respectively by the weight functions w1 ..a; b�/ Darctan .b/� arctan .a/ and w1 ..a; b�/ D ..a; b�/ we can sum the two con-tributions to find that

MZ p3�p3

1 d� D MZ p3�p3

1 d�1 CMZ p3�p3

1 d�2 D2

3� C 2;

MZ p3�p3

x d� D MZ p3�p3

x d�1 CMZ p3�p3

x d�2

D MZ p3�p3

x

1C x2dx C 0 D 0; and

MZ p3�p3

x2 d� D MZ p3�p3

x2

1C x2dx C M

Z p3�p3

x2 d�2

D 2p3 �

2

3� C 1.

On the one hand, any generalization gained by starting from .R;A; �/instead of

�R;A; �g

�is largely illusory since, as long as A contains the

Borel sets and bounded intervals have finite measure, � D �g where

g .x/ D

8<:� ..0; x�/ ; 0 < x

0; x D 0

�� ..x; 0�/ ; x < 0:

On the other hand, the pure measure-theoretic point of view invites othertypes of generalizations. The most simple generalization is to replace R byan alternate set such as Rn or C. In this case, A must be a sigma algebra ofsubsets of the underlying space and � must be a measure on A.

Example 69. Let A be the sigma algebra of all subsets of C and define� .A/to be the number of elements in A \ f1;�1; i;�ig.11 Then � is a measure

11 Alternatively, � places a weight of 1 on each of the points 1;�1; i , and �i .


on A and

MZC

1 d� D 4;

MZC

z d� D 0;

MZC

z2 d� D 0; and

MZC

jzj d� D 4.

Even greater generalizations can be obtained by allowing � to take onother types of values. We have seen hints of this idea when working withthe Riemann-Stieltjes integral when g is nonmonotone but of bounded vari-ation. This generalization produces a theory of signed measures. Similarly, itis relatively easy to construct a theory of complex-valued measures by con-sidering measures of the form � D �1 � �2 C i�3 � i�4 where the �i arenonnegative measures on a common sigma algebra A.

We will not pursue this path. Instead, we conclude with a brief introduc-tion of the idea of projection-valued measures. For simplicity, we will re-strict our attention to finite-dimensional vector spaces. For such spaces, theprojection-valued measure will always be discrete. The true power of theseideas starts to become evident when working with continuous linear map-pings between complete, infinite-dimensional vector spaces (Hilbert spaces),but the development of this theory is beyond the scope of this text.12

A projection P on a vector space is a linear mapping satisfying P 2 DP . The idea behind the algebraic expression is that, once a vector has beenprojected into a subspace, repeating the projection will have no additionaleffect.

Definition 36 (Projection-valued measure). Given a setX and a sigma al-gebra A of subsets of X , a projection-valued measure on A is a function �from A to the set of projections onto subspaces of a fixed vector space V thatsatisfies

1. � .;/ D 0,13

12 In infinite-dimensional (or finite dimensional) contexts, the inner product can be used toground the theory of projection-valued measures in the theory of complex-valued measures.For any pair of vectors v andw from the vector space V , the complex-valued measure �v;wis defined by �v;w .E/ D h�.E/v;wi. The operatorA D

Rf d� is defined by the way

that it acts on vectors: hAv;wi DRf d�v;w .

13 The zero in this case is the zero mapping, not the real number 0.

2. � .E/ � 0 for all E 2 A (i.e. the inner product h� .E/ v; viis nonneg-ative for all v 2 V ),14 and

3. � .[iEi / DPi � .Ei / for any countable, disjoint collection of sets

fEig from A.

Example 70. Let

P2 D1

2

�1 1

1 1

and

P4 D1

2

�1 �1

�1 1

:

Then P2 and P4 are projections onto the spaces spanned by˚� 11

��and˚� 1

�1

��respectively. For E R, define

� .E/ D

8<:0; 2; 4 62 E

P2; 2 2 E; 4 62 E

P4; 2 62 E; 4 2 E

P2 C P4; 2; 4 2 E:

Then � is a projection-valued measure on A, the sigma algebra of all subsetsof R. In this case,

MZ 5

0

1 d� D P2 C P4 D I;

MZ 5

0

x d� D 2P2 C 4P4 D

�3 �1

�1 3

, and

MZ 5

0

x2 d� D 4P2 C 16P4 D

�10 �6

�6 10

.

To justify the value of the second integral, set f .x/ D x. Let P be thepartition of Œ0;C1/ defined by the division points fykg

nkD1 and set Ek D

f �1 .Œyk�1; yk//, k D 1; 2; : : : ; n, and EnC1 D f �1 .Œyn;1//. Then, as-suming P has division points between 2 and 4, there are two values k2 andk4 such that yk2�1 � 2 < yk2 and yk4�1 � 4 < yk4 . Then �

�Ek2

�D P2,

P�Ek4

�D P4, and � .Ek/ D 0 for all other values of k. Thus the analog

of the Lebesgue-Stieltjes sum is SM .f; �;P/ D yk2�1 � P2 C yk4�1 � P4which will converge to 2P2 C 4P4 as the as the mesh of P gets finer. ThusMR 50 x d� D 2P2 C 4P4.

14 Alternatively, .� .E//� D �.E/ where A� is the adjoint of the operator A. Or (2) can beomitted with the stipulation that �.E/ is an orthogonal projection.


If in the preceding example we setA D MR 50 x d�, thenA2 D M

R 50 x

2 d�.

This observation suggests that we can compute f .A/ as MR 50 f d�. This

is indeed the case. We shall not prove this result as it would involve theintroduction of theorems from operator theory. Instead we give a verifiableexample.

Example 71. Under the same conditions as example 70,

MZ 5

0

px d� D

p2P2 C 2P4 D

"12

p2C 1 1

2

p2 � 1

12

p2 � 1 1

2

p2C 1

#.

A quick calculation shows that

"12

p2C 1 1

2

p2 � 1

12

p2 � 1 1

2

p2C 1

#2D

�3 �1

�1 3

D A.

While a more complete development of these ideas lies outside the scopeof this text, we note that the notion of a projection-valued measure togetherwith convergence theorems like those of the previous section provide a sig-nificant set of tools for the study of the structure of linear operators. In par-ticular, one way to search for invariant subspaces of a linear operator A (asubspace W such that AW W ) is to search for a projection P that com-mutes with A (AP D PA).

If A D MRX z d�, then A will commute with any operator of the

form p .A/ D MRXp d� where p is a polynomial. Given any charac-

teristic function 1E , the operator P D MRX1E d� is a projection since

P 2 D MRX.1E /

2 d� D P . Thus if one can find a sequence fpkg of polyno-mials that converges �-a.e. to 1E , then P will commute with A. In this casewe have found an invariant subspace for the linear operator A.

Beyond purely mathematical applications, projection-valued measures areused to model measurement in the study of quantum theory in physics.

8.8 Exercises

8.0 Stieltjes-Type integrals: filling the gaps1. Verify the standard integral properties for the Riemann-Stieltjes integral.

(a) Uniqueness. The value of the Riemann-Stieltjes integral is unique(if it exists).

8.8. Exercises 257

(b) Linearity I. If f and g are Riemann-Stieltjes integrable with respectto h over the interval Œa; b� and c 2 R, then so are f C g and cf .Moreover,

R-SZ b

a

.f C g/ dh D R-SZ b

a

f dhC R-SZ b

a

g dh

and

R-SZ b

a

cf dh D c R-SZ b

a

f dh:

(c) Linearity II. If f is Riemann-Stieltjes integrable with respect to gand h over the interval Œa; b� and c 2 R, then f is Riemann-Stieltjesintegrable with respect to g C h and cg. Moreover,

R-SZ b

a

f d .g C h/ D R-SZ b

a

f dg C R-SZ b

a

f dh

and

R-SZ b

a

f d .cg/ D c R-SZ b

a

f dg:

(d) Monotonicity. If f and g are Riemann-Stieltjes integrable with re-spect to h over the interval Œa; b� with f .x/ � g .x/ for x 2 Œa; b�,then it is not necessarily true that R-S

R baf dh � R-S

R bag dh: Find

and justify sufficient conditions for monotonicity.(e) Triangle inequality. Find and justify sufficient conditions such that

if f and jf j are Riemann-Stieltjes integrable with respect to h over

Œa; b�, thenˇ

R-SR ba f dh

ˇ� R-S

R ba jf j dh.

8.0 Stieltjes-Type integrals: deeper reflections2. Suppose that g .x/ represents the mass in the interval .�1; x/. What

would g .xi / � g .xi�1/ represent?

3. Find a pair of functions f and g on Œ0; 1� so that f is not Riemann-Stieltjes integrable with respect to g over Œ0; 1� but jf j is.

8.1 Examples: filling the gaps4. In example 51 (page 228), why is

PniD1�g D .g .b/ � g .a//?

5. Fill in the details of example 54 (page 229).

(a) Explain why SR-S .f; g;P1/ D 1 and SR-S .f; g;P2/ D 0.


(b) Use part (a) with partitions of arbitrarily small mesh to explain whyf cannot be Riemann-Stieltjes integrable with respect to g overŒ0; 2�.

6. Fill in the details of example 55 (page 230) to show that R-SR 20 f1 dg D

�1 and R-SR 20 f2 dg D �2.

8.1 Examples: deeper reflectionsWhen one encounters unexpected results such as those in example 54(page 229), it is good practice to try to place the phenomena in a largercontext. The next four exercises ask you to do this.

7. Example 54 (page 229) extends to the case where f and g are functionson Œa; b� with a common point of discontinuity. Give a proof of the factthat when f and g have a common point of discontinuity then f isnot Riemann-Stieltjes integrable over Œa; b� in the special case whereg is increasing. (Work with a partition of Œa; b� that straddles the dis-continuity.)

8. Explain example 54 (page 229) in light of theorem 20 (page 63). (Whatis the weight/measure of f1g?)

9. Example 54 (page 229) shows that Riemann-Stieltjes integrabilityover Œa; b� and Œb; c� is not sufficient to conclude Riemann-Stieltjesintegrability over Œa; c�. Prove that if f is Riemann-Stieltjes inte-grable with respect to g over all three intervals, then R-S

R caf dg D

R-SR ba f dg C R-S

R cb f dg.

10. Define f and g as in example 54 (page 229). Show that f is gauge-Stieltjes integrable with respect to g over Œ0; 1� where gauge-Stieltjesintegrable has the obvious interpretation. What is g-S

R ba f dg? (Use tag

forcing.)

11. Find a pair of functions f and g on Œ0; 1� such that f is Riemann-Stieltjes integrable with respect to g over Œ0; 1� and F .x/ D R-S

R x0 f dg

is discontinuous at every point in Œ0; 1�.

12. Prove that if f and g are functions on Œa; b� where f is bounded, gis absolutely continuous, and f is Riemann-Stieltjes integrable with re-spect to g over Œa; b�, then F .x/ D R-S

R x0f dg is continuous on Œa; b�.

(Bound and telescope.)

8.8. Exercises 259

8.2 Integrability theorems: filling the gaps13. Prove theorem 84 (page 231):

14. Prove theorem 85 (page 231). (Extend partitions of Œc; d � to partitionsof Œa; b� using a common extension.)


(a) Why can’t we just define �P1 DP

P infŒxi�1;xi � f � 1Œxi�1;xi �?(b) Explain why SR-S

��P1 ; g;P1

�� SR-S .f; g;P1/ �

SR-S��P1; g;P1

�.

(c) Explain why SR-S��P1 ; g;P1

�D SR-S

��P1 ; g;P3

�.

(d) Explain why SR-S��P1 ; g;P3

�� SR-S .f; g;P3/.

8.2 Integrability theorems: deeper reflections16. Prove that if f is continuous and g is monotone increasing on Œa; b�,

then there is a c 2 Œa; b� that satisfies

R-SZ b

a

f dg D f .c/ R-SZ b

a

dg.

17. Modify the proof of theorem 20 (page 63) to prove that f is Riemann-Stieltjes integrable with respect to g over Œa; b� if and only if f is con-tinuous �g-a.e. on Œa; b�. (See Section 5.)

8.3 Evaluation theorems: filling the gaps18. In the proof of theorem 87 (page 233)

(a) Why can we choose ı > 0 so that jf .x/ � f .y/j < "B.b�a/

when-ever x; y 2 Œa; b� with jx � yj < ı?

(b) Why isˇˇXP

ZIi

Œf .ti / � f � g0

ˇˇ <X

P

ZIi

"

B .b � a/B D "?

19. Why are f�cC�

and f .c�/ always defined when f is a monotonefunction?


(a) Why can we assume that M > 0?(b) Why is f .x/ D f

�c�i�

for x 2 Œa; b� with 0 < ci � x < ı?

(c) Verify that if ck D x0, thenˇf .t1/ Œg .x1/ � g .x0/� � f .ck/

�g�cCk

�� g

�c�k

�� ˇ< "

2n.

(d) Explain whyˇSR-S

�f; g;P

��PniD1 f

�ci��g�cCi�� g

�c�i��ˇ< ".

21. In g1 of example 57 (page 235), why were 3x and x2 C 2 used insteadof 3x C 1 and x2?


(a) Why does P� have mesh less than ı?(b) Why is xk�1 a tag for Œtk�1; tk �? Pay particular attention to the cases

k D 1 and k D nC 1.(c) It is possible that some of the intervals in P� are degenerate. How?

Why is this not a problem?

8.3 Evaluation theorems: deeper reflections23. Compute the following Riemann-Stieltjes integrals. (dxe is the ceiling

function that returns the least integer greater than or equal to x.)

(a) R-SR 2�1 x

3 dx2:

(b) R-SR 1�1 jxj ddxe:

(c) R-SR 1�1 dxe djxj:

(d) R-SR 20 sin�x ddxe:

(e) R-SR 20 x ddx2e:

(f) R-SR �0

sin x d sin x:

(g) R-SR 10 c dc where c is the Cantor function.

(h) R-SR 10 x dc where c is the Cantor function.

24. Prove that theorem 87 (page 233) can be extended by allowing g tohave a finite number of points of nondifferentiability as long as g iscontinuous on all of Œa; b�. (Split the partition P into those subintervalsthat contain a point where g is not differentiable and those that do not.)

25. Prove that theorem 87 (page 233) can be extended to allow g to be differ-entiable a.e. as long as g is continuous on all of Œa; b� and g0 is Riemannintegrable when defined to be zero where g is not differentiable. (Makeuse of the Lebesgue integral.)

26. Extend exercise 25 by proving the conclusion still holds when f hasa finite number of discontinuities but remains bounded. (Divide andconquer.)

8.8. Exercises 261

27. The hypotheses of theorem 88 (page 234) can be relaxed. Keeping thesame hypothesis on g, prove that it is necessary and sufficient that f becontinuous at fcig

nkD1 :

28. Use theorem 92 (page 239) to extend theorem 88 (page 234) to thecase where g has a countable number of discontinuities fckg andPk

ˇg�cCk

�� g

�c�k

�ˇ<1. (For each point ck of discontinuity define

hk .x/ D

8<:

0; x < ck

g .ck/ � g�c�k

�; x D ck

g�cCk

�� g

�c�k

�; ck < x

and define gn DPk�n hk .)

29. Suppose that f is a bounded, continuous function on Œa; b�. Justify asimple expression for evaluating R-S

R baf df . Can you extend your result

to allow f to have a finite or countable set of discontinuities?

8.4 Convergence theorems: filling the gaps30. Why is R-S

R ba f dg of lemma 90 (page 238) defined? In other words,

why is f Riemann-Stieltjes integrable with respect to g over Œa; b�?

31. Prove lemma 90 (page 238).

32. Justify the inequalityˇˇSR-S .f; g;P/ � R-S

Z b

a

f dg

ˇˇ D

ˇˇXi

R-SZ xi

xi�1

.f .ti / � f / dg

ˇˇ

<Xi

"V xixi�1g

from the proof of lemma 91 (page 238).

33. Why do the integrals in theorem 92 (page 239) exist?

34. In the proof of theorem 92 (page 239), why is it possible to select atagged partition of Œa; b�with mesh less than ı and whose division pointsare from S?

8.4 Convergence theorems: deeper reflections35. Why didn’t we prove theorem 92 (page 239) by first proving the lem-

mas 90 and 91 (page 238) and theorem 92 (page 239) for monotoneincreasing functions fgng and g? The extension to functions of boundedvariation would follow immediately.

36. Let frkg be an enumeration of the rational numbers in Œ0; 1� and setg .x/ D

Prk�x

1

2k. Find an integral-free expression for R-S

R 10 x

2 dg.

(Work with gn .x/ DPrk�x; k�n

1

2k.)

37. Show how any infinite seriesPk ak can be represented by a Riemann-

Stieltjes integral. (The weighting function

g .x/ D

8<ˆ:

0; x D 0

1 � 1

2k; x 2

1 � 1

2k�1; 1 � 1

2k

i; k 2 N

1; x D 1

provides one possible approach.)

8.5 Measure theory: filling the gaps38. Prove parts (1)-(3) and (5) of theorem 93 (page 242).


(a) Why can we find an a� > a so that g .a�/ < g .a/C "?(b) Why can we choose a b�

k> bk so thatw

��ak ; b

�k

�� w ..ak ; bk �/C

"

2k?

(c) Why is˚�ak ; b

�k

��an open cover of Œa�; b�?

(d) Why is w ..a; b�/ � w ..a; a��/ CPnjD1w

aj ; b

�j

i��P1kD1w ..ak ; bk �/C 2"?

40. Prove theorem 94 (page 244).

41. Why is it sufficient to prove that intervals of the form .�1; a� are �g-measurable in the proof of theorem 95 (page 245)?

42. Prove that any set of ��g -outer measure zero is �g-measurable. In otherwords, prove that if ��g .E/ D 0 then E is �g-measurable.

43. If Fk D F \ .k; k C 1� and ��Gn;knFk

�< 1

n2kC2for all k 2 Z,

explain why �� .GnnF / < 1n

where Gn D [kGn;k .


(a) Why is F measurable?(b) Why is Gc a Borel set?

8.8. Exercises 263

8.5 Measure theory: deeper reflections45. Modify the proof of theorem 22 (page 104) so that it applies to �g-

measurability. (Use the right-continuity of g to move from half-open toopen intervals.)

46. Express the following in terms of g.

(a) ��g ..a; b// :(b) ��g .fag/ :(c) ��g .Œa; b// :

47. Define

g .x/ D

�2x; x < 0

x C 1; 0 � x.

Calculate

(a) ��g ..�1; 1// :(b) ��g .f0g/ :(c) ��g .f1g/ :(d) ��g .frational numbers in Œ0; 1�g/.

48. Define

g .x/ D

8<:0; x < 0

x C 1; 0 � x < 1

3; 1 � x.

Calculate

(a) ��g ..�1; 1// :(b) ��g .f0g/ :(c) ��g ..�1;1// :(d) ��g .C / where C is the Cantor set.

49. Let g .x/ D dxe. Calculate

(a) ��g .Œ�1; 1�/ :(b) ��g .f0g/ :(c) ��g .C / where C is the Cantor set.(d) Give a general description of the value of ��g .A/.

50. Let c be the Cantor function. Calculate

(a) ��c .Œ0; 1�/.


(b) ��c .C / where C is the Cantor set.(c) ��c .frational numbers in Œ0; 1�g/.

51. Let g .x/ D dxe. Prove that every subset of R is �g-measurable.

8.6 Integration with measures: filling the gaps52. Prove that any increasing function f is �g-measurable.

53. Prove that any continuous function f is �g-measurable.

54. Prove that if a function f is continuous �g-a.e., then f is �g-measurable.

55. Prove that if g is an increasing, right-continuous function and f isRiemann-Stieltjes integrable with respect to g over Œa; b�, then f is�g-measurable and M

R ba f d�g D R-S

R ba f dg C f .a/�g .fag/. (See

exercises 17 and 54.)

56. Prove that if f1 is a �g-integrable function on Œa; b� and f1 D f2 a.e.

�g , then f2 is �g-integrable and MR baf1 d�g D M

R baf2 d�g .

57. Prove that MRE[F f d�g D M

RE f d�gC M

RF f d�g wheneverE and

F are disjoint �g-measurable sets and f is �g-integrable over E [ F .

8.6 Integration with measures: deeper reflections58. Suppose that if f1 is �g-measurable and f2 D f1 except on a countable

set S . Prove that f2 is measurable even if �g .S/ > 0. (Why is fag ameasurable set?)

59. Modify the definition of the Lebesgue-Young integral to apply to themeasure �g and prove that it is equivalent to the Lebesgue-Stieltjes in-tegral for �g-measurable functions.

60. Modify the definition of the simple-Lebesgue integral to apply to themeasure �g and prove that it is equivalent to the Lebesgue-Stieltjes in-tegral for �g-measurable functions.

61. Without using exercise 55 for the evaluation, give an example of a func-tion f and an increasing, right-continuous function g on R such thatMR ba f d�g ¤ R-S

R ba f dg.

62. Prove that if f is nonnegative and �g-integrable over Œa; b�, then for anyv > 0

�g .fx 2 Œa; b� W f .x/ > vg/ �1

vMZ b

a

f d�g :

8.8. Exercises 265

63. Give an example of two functions f1 and f2 and an increasing, right-continuous function g for which f1 D f2 except at a single point, butfor which M

R ba f1 d�g ¤ M

R ba f2 d�g .

64. Example 54 (page 229) shows that Riemann-Stieltjes integrability overŒa; b� and Œb; c� is not sufficient to conclude Riemann-Stieltjes integra-bility over Œa; c�.

(a) Prove that if f is Lebesgue-Stieltjes integrable over Œa; b� and Œb; c�,then f is Lebesgue-Stieltjes integrable over Œa; c�.

(b) Give an example where f is Lebesgue-Stieltjes integrable over Œa; b�and Œb; c� ; but M

R ca f d�g ¤ M

R ba f d�g C M

R cb f d�g .

(c) Explain how (b) can be true in spite of exercises 9 and 57.(d) Suggest a way to resolve this seeming contradiction.

8.7 Extensions: filling the gaps65. Prove that � .E/ � 0 for all E 2 A is equivalent to � .A/ � � .B/ for

all A;B 2 A satisfying A B .

66. Prove that� in example 67 (page 251) is a measure on A. In other words,show that � satisfies the three properties on page 251.

67. How do we know that � in example 68 (page 253) is a measure on M?

68. Justify the integral calculations in example 68 (page 253).

69. Let .R;A; �/ be a measure space such that A contains the Borel setsand every bounded interval has finite measure. Let

g .x/ D

8<:� ..0; x�/ ; 0 < x

0; x D 0

�� ..x; 0�/ ; x < 0:

Prove that

(a) g is increasing and right-continuous. (Use monotonicity and that factthat

Pn �

��aC 1

nC1; aC 1

n

��D � ..a; aC 1�/ <1.)

(b) � D �g .

70. Prove that .C;A; �/ of example 69 (page 253) is a measure space.

71. Verify that P1 and P2 of example 70 (page 255) are projections.

72. Verify that � of example 70 (page 255) is a projection-valued measureon A.

8.7 Extensions: deeper reflections73. For the measure in example 69 (page 253)

(a) Find the smallest positive integer n so that MRCzn d� ¤ 0.

(b) Compute MRC1z2

d�:

74. Prove that if .R;A1; �1/ and .R;A2; �2/ are two measure spaces then

(a) .R;A1 \A2; �1 C �2/ is a measure space.(b) For any �-measurable function f , M

RE f d� D M

RE f d�1 C

MRE f d�2 where � D �1 C �2.

75. Suppose that � is a discrete measure that assigns weights to the pointsfcig

niD1 and for which �

�Rn fcig

niD1

�D 0. Prove that for any function

f defined on R, MRRf d� D

Pi f .ci / � .fcig/.

76. Suppose that

� .A/ D

8<:0; 1; 2 62 A

1; 1 2 A; 2 62 A

2; 1 62 A; 2 2 A

3; 1; 2 2 A:Compute

(a) MR 20 1 d�:

(b) MR 10 x d�:

(c) MR 20 x d�:

(d) MR 20x2 d�:

77. Let A be the sigma algebra of all subsets of C and define

� .E/ DX

nCim2En;m2Znf0g

;

1

nm.

(a) Prove that � is a measure on A.Let D2 D fz 2 C W jzj � 2g. Compute

(b) MRD21 d�.

(c) MRD2z d�.

(d) MRD2z2 d�.

78. Consider the measure space .R;A; �/ where A Df;;Rg and

� .A/ D

�1; A D R

0; A D ;:

8.8. Exercises 267

(a) Characterize the �-measurable functions.(b) Give a simple formula for evaluating M

RRf d� when f is �-

measurable.

79. Consider the measure space .R;A; �/ where A is the collection of allsubsets of R and � .A/ is the number of elements inA when A is a finiteset and � .A/ D1 otherwise.

(a) Characterize the �-measurable functions.(b) Provide an alternate expression for M

RRf d� when f is �-

integrable.

80. Give an example of a measure space on which the bounded convergencetheorem (page 250) fails.

81. Let .R;A; �/ be a measure space where A contains all intervals and� .Œ0; 1�/ < 1. Prove that if � is translation invariant, then � is a mul-tiple of the Lebesgue measure.

82. Let X D Œ0; 1� Œ0; 1� and let w be the weight function that assigns aweight of x2y to the rectangle Œ0; x� Œ0; y�. (Note that the degeneraterectangles Œ0; 0� Œ0; y� and Œ0; x� Œ0; 0� have weights of zero.) Whatare the measures of Œ0; b� .c; d �, .a; b� Œ0; d �, and .a; b� .c; d �?

83. Let

P�2 D1

4

26641 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

3775 ;

P2 D1

4

26641 �1 1 �1

�1 1 �1 1

1 �1 1 �1

�1 1 �1 1

3775 , and

P4 D1

2

2664�1 0 1 0

0 �1 0 1

1 0 �1 0

0 1 0 �1

3775 .

Define�by taking�.fzg/DPz ,zD�2; 2; 4, setting�.Rnf�2; 2; 4g/D0, and extending � in the natural way to all subsets of R.


(a) Computei. M

R 5�5 1 d�;

ii. A D MR 5�5 x d�; and

iii. B D MR 5�5

px d�:

(b) Show that B2 D A.

84. Let

P3 D1

3

264 2 1 �1

1 2 1

�1 1 2

375 , and

P6 D1

3

264 1 �1 1

�1 1 �1

1 �1 1

375 .

Define � by setting � .fzg/ D Pz , z D 3; 9, � .Rn f3; 9g/ D 0. Extend� in the natural way to all subsets of R.

(a) Computei. M

R 100 1 d�;

ii. A D MR 100 x d�, and

iii. B D MR 100

px d�:

(b) Show that B2 D A. (You may want to use a computer algebrasystem.)

(c) Find log3A using an integral.(d) Another approach to finding log2Awould be to approximate it using

log2A D1

ln2 lnA and the power series

ln x D1XkD1

.�1/kC1

k.x � 1/k :

Why does this not work?

85. Prove that if .X;A, �/ is a measure space where� is a projection-valuedmeasure, then � .A/ and � .B/ are orthogonal projections when A andB are disjoint elements of A. (Two projectionsP1 andP2 are orthogonalif P1P2 D 0.)

8.9. References 269

8.8 Other directions: deeper reflections86. Show that the Dirichlet function is gauge-Stieltjes15 integrable with

respect to g over Œa; b� if and only if the rational numbers in Œa; b�have �g-measure zero. What must be true about g in order to have�g .Q\ Œa; b�/ D 0?

87. Modify the definition of the Darboux integral to a definition of theDarboux-Stieltjes integral.

(a) Prove that the Darboux-Stieltjes and Riemann-Stieltjes integrals arethe same.

(b) State and prove a Cauchy criterion for the Darboux-Stieltjes integral.(c) Prove that a function f is Darboux-Stieltjes integrable with respect

to an increasing, right-continuous function g over Œa; b� if and onlyif f is continuous �g a.e.

88. The weight functions in this chapter have a specific form. Namely, theyare constructed using w ..a; b�/ D g .b/ � g .a/ where g is an increas-ing, right-continuous function. When one moves away from this form ofweight function, odd things can happen. Let w ..a; b�/ D

pb � a and

define

��w .A/ D inf

(Xk

w ..ak ; bk �/ Wfw ..ak ; bk �/g is a countablecover of A by finite intervals

)for any A R.16

(a) Prove that ��w ..a; aC 1�/ D 1 for any a 2 R. (px � x for 0 �

x � 1.)(b) Show that .0; 1� is not measurable. (Apply the Caratheodory crite-

rion to A D .0; 2�.)

8.9 ReferencesBurk, F.E. (2007). A Garden of Integrals. Mathematical Association of

America.Brandt, H.E. (1999). Positive operator valued measure in quantum informa-

tion processing. Am. J. Phys. 67, 434–440.

15 Use the most obvious modification of the Riemann-Stieltjes integral.16 This exercise is based on an example from Measure, Topology, and Fractal Geometry by

Gerald Edgar (page 139).


DePree, J. and C. Swartz (1988). Introduction to Real Analysis. John Wiley& Sons.

Halmos, P.R. (1976). Measure Theory. Graduate Texts in Mathematics 18,Springer.

Hildebrandt, T.H. (1938). Definitions of Stieltjes integrals of the Rie-mann type. The American Mathematical Monthly 45 (5): 265–278.JSTOR 2302540.

Protter, Jr., M.H. and C.B. Morrey (1991). A First Course in Real Analysis.Springer.

Taylor, A.E. (1965). General Theory of Functions and Integration. Dover.Teschl, G. (2009). Mathematical Methods in Quantum Mechanics with

Applications to Schrodinger Operators. Graduate Studies in Mathe-matics 157, American Mathematical Society. www.mat.univie.ac.at/�gerald/ftp/book-schroe/.

CHAPTER 9

A Look Back

Historically, the Riemann and Darboux integrals (and the equivalent Cauchyintegral that preceded them) were introduced to solve a different set of prob-lems than were the Lebesgue and gauge integrals. The Riemann and Darbouxintegrals were developed in response to foundational questions such as thoseidentified by Bishop Berkeley (page 7). The Lebesgue and gauge integralsaddress issues of convergence arising from Fourier series. It should then beno great surprise that the integrals have somewhat different properties. Thisconcluding chapter presents a comparative overview of the integrals coveredin this text.

9.1 Basic approaches

Given a function f on Œa; b�, how do we defineR ba f ?

Riemann integralThe Riemann integral partitions Œa; b� into subintervals fIkg. Then, usingtags ftkg with tk 2 Ik , the sum

Pk f .tk/�xk is computed. If all such

sums approximate some fixed real number A when the subintervals in thepartition are suitably controlled, the function f is integrable over Œa; b� andthe value of the integral is A. Specifically, given an " > 0 there must be aı > 0 such that, when all the subintervals in a partition have width less thanı, the sum must be within " of A independent of the the choice of the tags.When proving results about the Riemann integral it is common to spenda significant amount of energy identifying a possible value for A. Cauchysequences and cluster points figure prominently in this process.

Darboux IntegralLike the Riemann integral, the Darboux integral partitions Œa; b� intosubintervals fIkg. Instead of selecting tags, we compute the sums

271

272 CHAPTER 9. A Look Back

Pk infIk f �xk and

Pk supIk f �xk . A function is Darboux integrable

on Œa; b� ifsup

Xk

infIkf �xk D inf

Xk

supIk

f �xk

where the outer supremum and infimum are taken over all possible partitionsof Œa; b� by subintervals. In this case, the Darboux integral is the commonvalue.

The Riemann and Darboux integrals are equivalent. However, proofs forthe two integrals can follow quite different paths. For example, when prov-ing theoretical results about the Darboux integral, there is no need to con-struct a value using Cauchy sequences or some related technique since thesupremum and infimum already identify a value. Even apparent analogs havesignificant differences. For the Darboux integral, a partition can play therole played by the ı in the Riemann integral. A function is Darboux inte-grable on Œa; b� if and only if given any " > 0, there is a partition fIkg suchthat

ˇPk infIk f �xk �

Pk supIk f �xk

ˇ< " (theorem 12, page 55). Un-

like the case for the Riemann integral, however, the inequality involves onlythe single partition fIkg and not other associated partitions.

Lebesgue integralThe Lebesgue integral takes what one might call an orthogonal approach.Given a measurable function satisfying ˛ < f < ˇ, the interval .˛; ˇ� ispartitioned using disjoint intervals of the form .yk ; ykC1�. From such a par-tition, the sum

Pk yk�1� .Ek/ is computed whereEk is the preimage of the

kth interval and � is Lebesgue measure. The sums will always converge inthe sense that there is a real value A that the sums approach as the maximumlength of the subintervals of the partition decreases to zero. By allowing Ato be an extended real number, this approach can be generalized to handlefunctions that are not bounded above. (See page 132.)

Alternatively, one can partition Œa; b� with measurable sets fEkg and com-pute sup

Pk

�infEk f

�� .Ek/ where the supremum is taken over all mea-

surable partitions of Œa; b�. For non-negative (possibly unbounded) functions,the supremum always exists in the extended real numbers. An arbitrary mea-surable function f is Lebesgue integrable if both f C and f � are (finitely)Lebesgue integrable. (See page 133.)

The major catch with the Lebesgue integral is that the significant frame-work of measurability and Lebesgue measure must be constructed in order tomake sense of � .Ei /. The work done to develop measure theory is roughly

9.2. Integrable functions 273

equivalent to the effort required to develop the Darboux integral. Once themeasure-theoretic foundation has been laid, one finds that the defining sumsonly make sense for measurable functions. Then one can begin the task ofinvestigating the properties of the Lebesgue integral applied to measurablefunctions.

Gauge integralThe gauge integral returns to a Riemann-style approach where Œa; b� is parti-tioned with tagged intervals. Unlike the Riemann integral, the gauge integraldoes not permit a free choice of tags once the intervals have been selected.Instead, the intervals and tags are selected together subject to a gauge re-striction. A gauge is a function that associates a positive value ı .t/ witheach point t 2 Œa; b�. A tagged partition f.tk ; Ik/g is ı-fine if for all k thelength of Ik is less than ı .tk/. A function is gauge integrable if there is areal value A such that given any " > 0, it is possible to construct a gaugeı such that j

Pk f .tk/�xk � Aj < " for any ı-fine partition. Once again,

when proving general results, one must invoke Cauchy sequences (or someanalog) to construct the value A. However, there is no need to develop thesomewhat complicated (and for many students initially mystifying) theoryof measure. Instead, one needs to develop facility in constructing gauges.

9.2 Integrable functions

In 1823, before Riemann introduced his integral, Cauchy gave a related defi-nition of an integral that essentially uses right endpoints as tags. (See page 8.)While the Cauchy integral is equivalent to the Riemann and Darboux inte-grals (see exercise 38 on page 49 and exercise 32 on page 70), Cauchy as-sumed that all functions to be integrated were continuous. Only later, whenthe need to deal with functions arising from sources like Fourier series cameinto focus, was an effort made to determine the extent of the class of func-tions for which an integral makes sense. The Dirichlet function gave an earlyand relatively simple example of a function that is not Riemann integrable.

Riemann-Darboux integralIt is fairly easy to see that monotone and continuous functions are Riemann-Darboux integrable, but a complete characterization of the Riemann-Darboux integrable functions had to wait over 30 years. In 1902, Lebesgueproved that a function is Riemann-Darboux integrable over Œa; b� if and only


if it is bounded on Œa; b� and continuous except on a set of measure zero1

(theorem 20, page 63).

Lebesgue integralThe Lebesgue integral greatly extends the class of integrable functions. Anybounded, measurable function is Lebesgue integrable. No non-measurablefunction is Lebesgue integrable since the sum

Pi yi� .Ei / is not defined if

one of the Ei is not measurable. For non-negative functions, the restrictionthat the integrand be bounded can be removed if integrals are permitted totake on values in the extended real numbers. In general, a function f isLebesgue integrable if and only if f is measurable and both L

R ba f

C andLR ba f

� are finite.

Gauge IntegralThe gauge integral modestly extends the class of integrable functions. Ev-ery Lebesgue integrable function is gauge integrable (theorem 77, page 200)and all gauge integrable functions are measurable (theorem 78, page 200),but it is possible for f to be gauge integrable while both g

R ba f

C and gR ba f

�

are infinite. In particular, if f is a differentiable function, then f 0 is gaugeintegrable. Example 47 (page 192) illustrates that not all derivatives areLebesgue integrable.

9.3 Convergence theoremsGiven a sequence ffng of integrable functions that converges to a functionf , when is f integrable and when is

limn

Z b

a

fn D

Z b

a

limnfn?

Riemann-Darboux integralFor the Riemann-Darboux integral, a sequence of integrable functions mustconverge uniformly to guarantee that the limit function is integrable and thatthe order of the limit and integral may be interchanged. The biggest obsta-cle to convergence results for the Riemann-Darboux integral is that the limitfunction may not be integrable. In the absence of uniform convergence, intro-ducing additional restrictions such as requiring the sequence to be uniformly

1 As the Dirichlet function shows, being continuous except on a set of measure zero is not thesame as being equal almost everywhere to a continuous function.


bounded and increasing will not ensure that the limit function is integrable.(See example 8 on page 41.) Even when the limit function is integrable,example 9 (page 42) shows that the limit and integral cannot always be in-terchanged. This second mode of failure becomes predominant as the classof integrable functions grows.

Lebesgue integralThe issue of integrability is substantially reduced for the Lebesgue integral.If a sequence of measurable functions converges almost everywhere, thenthe limit function is again measurable (theorem 25, page 107). The only wayfor the limiting function f to fail to be Lebesgue integrable is for L

R ba f

CorLR ba f

� to be infinite. But even when the limit function is integrable, un-bounded values of the functions ffng are potential barriers to the interchangeof limits and integrals. (See example 9, page 42.) Therefore, the convergenceresults for the Lebesgue integral tend to focus on controlling the size of thefunctions ffng. If the functions in fjfnjg are uniformly bounded by a con-stant, then the limit function f is Lebesgue integrable and the order of theintegral and limit can be swapped (theorem 36, page 136). The bound neednot be uniform. An integrable function bounding all the functions in fjfnjgwill suffice (theorem 39, page 140). Boundedness can be dropped altogetherif we restrict our attention to an increasing sequence of non-negative func-tions and allow integrals to assume values in the extended real numbers (the-orem 37, page 138). Even if the sequence ffkg of non-negative functionsis not increasing and does not converge, we can still say (theorem 38, page139) that

LZ b

a

limk

fk � limk

LZ b

a

fk .

These are far more powerful results than those that are available for theRiemann-Darboux integral.

Gauge integralSimilar results hold for the gauge integral. However, the proof techniquesare quite different. The gauge integral has no analog to the fact that the limitof measurable functions is always measurable. Nevertheless, the existenceof gauge integrable functions g1 and g2 on Œa; b� such that g1 � fk � g2for all k is sufficient to ensure that the limit function is gauge integrableand that the order of the integral and the limit can be reversed (theorem 69,page 189). This result is stronger than the corresponding theorem for the


Lebesgue integral since g2 need not be non-negative as must be the caseif jfkj is to be bounded. In the case of a monotone sequence of functionsffkg, the boundedness of

˚ gR ba fk

�is both necessary and sufficient for the

limit function to be integrable and for the order of the integral and limitto be reversible (theorem 67, page 186). Note that, unlike the case for theLebesgue integral, these results do not require

˚ gR baf Ck

�or˚ gR b

af �k

�to be

bounded.

9.4 The fundamental theoremsThe first fundamental theorem of calculus states that, under appropriate con-ditions,

R xaF 0 D F .x/�F .a/ for all x 2 Œa; b�. At a minimum, F must be

continuous everywhere as illustrated by example 7 (page 40). Beyond thisbasic requirement, the conditions vary by the type of integral.

Riemann-Darboux integralFor the Riemann-Darboux integral, the required condition is that F be dif-ferentiable on Œa; b� with F 0 bounded on Œa; b� and continuous a.e. These arethe conditions that ensure that F 0 is integrable. As illustrated by Volterra’sfunction (page 81), F 0 need not be Riemann-Darboux integrable even if Fhas a bounded derivative at every point in Œa; b�.

Lebesgue integralFor the Lebesgue integral, the appropriate condition is that F be abso-lutely continuous on Œa; b�. Then F will be differentiable a.e. on Œa; b�

and F 0 will be Lebesgue integrable with LR xa F

0 D F .x/ � F .a/ (theo-rem 47, page 148). Conversely, the second fundamental theorem of calculusfor the Lebesgue integral implies that F must be absolutely continuous ifLR xaF 0 D F .x/ � F .a/.

Gauge integralThe gauge integral has the simplest version of the first fundamental theoremof calculus. If F is differentiable on Œa; b�, then F 0 is gauge integrable onŒa; b� and g

R xa F

0 D F .x/ � F .a/ (theorem 71, page 191). The discus-sion following the proof of theorem 75 (page 197) shows the existence ofdifferentiable functions that are not absolutely continuous and so for whichthe fundamental theorem will fail when using the Lebesgue integral. As longas F remains continuous, the first fundamental theorem of calculus for thegauge integral can be generalized to admit a countable number of points at

9.5. Conclusion 277

which F is not differentiable (theorem 72, page 193). Since Lebesgue inte-grable functions are also gauge integrable, absolutely continuous functionsmust also satisfy g

R xa F

0 D F .x/ � F .a/.

The second fundamental theoremGiven an integrable function f on Œa; b�, the second fundamental theorem ofcalculus says that, under appropriate conditions, the function F .x/ D

R xaf

will be differentiable with F 0 D f . In the cases of both the Riemann-Darboux and the Lebesgue integrals, F is absolutely continuous and so mustbe differentiable a.e. (theorem 42, page 143). Moreover, F 0 .x0/ D f .x0/

whenever f is continuous at x0. Since Riemann-Darboux integrable func-tions are continuous a.e., F 0 D f a.e. Even though Lebesgue integrablefunctions need not be continuous anywhere (consider the Dirichlet function,for example), any Lebesgue integrable function f satisfies F 0 D f a.e.(theorem 42, page 146). For gauge integrable functions f , the functionsF .x/ D g

R xa f may not be absolutely continuous. Nevertheless, except

for a set of measure zero, F is differentiable with F 0 D f (theorem 75,page 197).

9.5 ConclusionIn this text, we have investigated four integrals: Riemann, Darboux,Lebesgue, and gauge. In each case, the succeeding integral definition of-fers a distinct improvement over the previous integral. The Darboux inte-gral supports more efficient proof techniques than the Riemann integral. TheLebesgue integral allows for a much larger class of integrable functions andhas much stronger convergence properties than the Riemann-Darboux in-tegral. The gauge integral modestly enlarges the class of integrable func-tions and has somewhat more flexible convergence properties while, at thesame time, avoiding the need to develop measure theory. While the option ofavoiding the early introduction of measure theory is appealing, the desirabil-ity of doing so is somewhat diminished by the fact that, as seen at the end ofChapter 8, measure theory itself has powerful and useful generalizations.

CHAPTER 10

Afterword: L2 Spaces andFourier Series

As is evident in the previous chapters, the task of resolving issues relatedto the interactions between Fourier series and integration was a significantmotivator in the development of definitions and theories of integration. Insome sense, we have completed the task. We have two integrals, Lebesgueand gauge, that interact well with sequences and series in general and so withFourier series in particular.

But these integrals do more than resolve the issues related to the inter-change of integrals and limits. They also provide a context,L2 spaces, withinwhich we can develop an alternative notion of convergence that is partic-ularly suited to the study of Fourier series. In this new setting, functionshave Fourier series whose coefficients can be computed using integrationand which converge to the original function. The properties of the Lebesgueand gauge integrals are critical in developing L2 spaces.

The goal of this epilogue is to introduce L2 spaces and to examine howthey resolve fundamental issues raised by Fourier series that, initially, seemto be unrelated to questions of integration.

Given a Lebesgue-integrable function f on the interval Œ0; 1�, the integralsak D

R 10f .x/ sin .2�kx/ dx and bk D

R 10f .x/ cos .2�kx/ dx are de-

fined. Moreover, as long as a0 C 2P1kD1 .ak sin .2�kx/C bk cos .2�kx//

converges appropriately to a function g, then the convergence theorems de-veloped in prior chapters along with a few basic facts about the integrals of

279

280 CHAPTER 10. Afterword: L2 Spaces and Fourier Series

trigonometric functions1 show that, for any n 2 N,Z 1

0

g .x/ sin .2�nx/ dx

D

Z 1

0

a0 C 2

1XkD1

.ak sin .2�kx/C bk cos .2�kx//

!sin .2�nx/ dx

D a0

Z 1

0

sin .2�nx/ dx

C 2

1XkD1

ak

Z 1

0

sin .2�kx/ sin .2�nx/ dx

C 2

1XkD1

bk

Z 1

0

cos .2�kx/ sin .2�nx/ dx

D an.

A similar derivation shows thatR 10 g .x/ cos .2�nx/ dx D bn. Thus f and

g produce the same coefficients. By translating and scaling, we can drawanalogous conclusions for functions defined on any compact interval.

While the issues related to integration have been resolved, other questionsremain. The relevant integrals now make sense and, with a bit of care, weare justified in interchanging the integrals and the summations. However, wehave not verified what we really want to know: namely, does f equal g? Infact we cannot know that f D g. If the function f is modified on a set ofmeasure zero, then all the associated integrals, and thus the coefficients akand bk , remain unchanged. Consequently, it is clear that the best we couldhope for is that f D g a.e.

The words “as long as” that preceded “converges appropriately” at thebottom of page 279 are even more problematic. How do we know that thesum converges at all?

The question of the convergence of trigonometric series received a greatdeal of attention in the early 1800s. For example, Neils Henrik Abel pub-lished a result that, when combined with trigonometric identities like

nXkD1

.�1/k�1 cos

�.2k � 1/ �x

2

�D1 � .�1/ n cos .�nx/

2 cos .�x=2/

1 For integers k and n,R 10 cos .2�kx/ sin .2�nx/ dx D 0,

R 10 sin2 .2�nx/ dx D 1

2, and,

when k ¤ n,R 10 sin .2�kx/ sin .2�nx/ dx D

R 10 cos .2�kx/ cos .2�nx/ dx D 0.

10.1. L2 spaces 281

provided a useful convergence test (Dirichlet’s test) for the convergenceof trigonometric series.2 But knowing that a0 C 2

P1kD1.ak sin.2�kx/ C

bk cos.2�kx// converges is not the same as knowing the series convergesto f .

We will not develop results like Dirichlet’s test here as the Lebesgue andgauge integrals provide another path to resolving the questions raised byFourier series. Moreover, since this chapter is not concerned with compar-ing the properties of various integrals, we will work exclusively with theLebesgue integral and use the notation

R baf d�.

While we worked exclusively with real-valued functions in the main bodyof this text, L2 spaces, the context within which we will be working in thischapter, are typically defined for complex-valued functions. Since this gen-eralization requires little extra effort, going forward we will assume thatwe are working with complex-valued functions unless otherwise noted. Acomplex-valued function f D u C iv is measurable if the real-valuedfunctions u and v are both measurable. When u and v are integrable,R ba f d� D

R ba u d� C i

R ba v d�. A review of some of the basic proper-

ties of complex numbers is provided in exercises 1 through 3. Also, consultAppendix A.6, which contains a quick overview of the basic facts aboutcomplex numbers used in this chapter.

10.1 L2 spacesFix a compact interval Œa; b� and let � denote the Lebesgue measure.3

Definition 37. A measurable function f W Œa; b� ! C is called square-integrable if

R ba jf j

2 d� <1. The set of all square-integrable functions onŒa; b� is denoted by L2 .Œa; b�/.4

More generally, for any measurable set X , Lp .X/ is defined as the set ofmeasurable functions f satisfying

RX jf j

p d� < 1. We will not developthe theory of Lp spaces in the main text, but some of the central ideas areintroduced in the exercises.

2 David Bressoud provides a well-written account of this work in pages 165–173 of A RadicalApproach to Real Analysis, The Mathematical Association of America, Washington, D.C.,1994.

3 While we will develop the theory of L2 spaces using the Lebesgue integral, one could usethe gauge integral. The Lebesgue integral is selected here since, with little additional effort,what follows could be extended to any finite measure � on a compact setX .

4 It is easy to see that if f D u C iv where u and v are real-valued functions, then f issquare-integrable if and only if u and v are both square integrable. (See Exercise 40.)

The set L2 .Œa; b�/ has much more structure than its set description in-dicates. First, L2 .Œa; b�/ is a vector space. In other words, using the usualoperations of scalar multiplication and addition of functions, L2 .Œa; b�/ sat-isfies the algebraic properties of closure, commutativity, associativity, anddistributivity and has an additive identity and inverses.

In what follows, we assume the reader is familiar with the basic propertiesof vector spaces and with (orthogonal) projections in inner product spaces.Exercises 4 through 19 provide a review of the salient properties. Also, con-sult Appendix A.7:

Theorem 101. The set of square-integrable functions forms a vector spaceunder the usual operations of scalar multiplication and addition of functions.

Proof. Suppose that f and g are square-integrable functions over Œa; b� andthat c is a scalar. Note that cf and f C g are measurable functions. Theproof that the set of square-integrable functions is closed under scalar multi-plication is left as a straightforward exercise (exercise 20).

To prove that L2.Œa; b�/ is closed under addition, note thatZ b

a

jf C gj2 d� �Z b

a

.jf j C jgj/2 d�

�

Z b

a

.2max fjf j ; jgjg/2 d�

D

Z b

a

.4maxfjf j2; jgj2g/ d�

�

Z b

a

.4.jf j2 C jgj2// d�

� 4

Z b

a

jf j2 d�CZ b

a

jgj2 d�

!<1.

The set of square-integrable functions inherits the remainder of the vectorspace properties from the vector space of all functions on Œa; b�.

In fact, L2 .Œa; b�/ is more than a vector space. We can use the Lebesgueintegral to create an inner product on L2 .Œa; b�/.

Definition 38. Let f; g 2 L2 .Œa; b�/. Define hf; gi DR ba f g d� where

f .x/ D f .x/.

Theorem 102. The function h�; �i is an inner product on L2 .Œa; b�/. In otherwords, for any functions f; g, and h 2 L2 .Œa; b�/ and any scalar c 2 C

10.1. L2 spaces 283

1. hf; gi is finite,

2. hg; f i D hf; gi,

3. hcf; gi D c hf; gi,

4. hf C g; hi D hf; hi C hg; hi, and

5. hf; f i � 0 with hf; f i D 0 if and only if f D 0 a.e.

Proof. To verify (1), note that, since .jf j � jgj/2 � 0, jfgj � 2 jfgj �

jf j2 C jgj2. Also, jf gj D jfgj. Hence

jhf; gij D

ˇˇZ b

a

f g d�

ˇˇ �

Z b

a

jfgj d� �Z b

a

.jf j2 C jgj2/ d� <1.

To prove the second half of (5), suppose that hf; f i D 0. Set Z Dfx 2 Œa; b� W f .x/ ¤ 0g and note that Z is the disjoint union of Zn D˚x 2 Œa; b� W 1

n� jf .x/j < 1

n�1

�, n 2 N. (Here we interpret 1

n�1as 1

when n D 1.) Since

0 D hf; f i D

Z b

a

jf j2 d� �ZZn

1

n2d� D

1

n2� .Zn/ � 0,

we conclude that � .Zn/ D 0 for n 2 N. Hence � .Z/ DPn � .Zn/ D 0

and f D 0 a.e.The proofs of the remaining properties are left as an exercise (exercise 21)

as they are straightforward consequences of properties of the Lebesgueintegral.

Since L2 .Œa; b�/ has an inner product, it also has the standard norminduced by the inner product.

Definition 39. Let f 2 L2 .Œa; b�/. Define kf k2 Dphf; f i D

.R ba jf j

2 d�/1=2.

Theorem 103. kf k2 is a norm on L2 .Œa; b�/. That is, for f; g 2 L2 .Œa; b�/and any scalar c 2 C,

1. kcf k2 D jcj kf k2 and

2. kf C gk2 � kf k2 C kgk2.

Proof. Exercises 11 and 16.


We refer to kf k2 as the L2 norm of f .

While we have referred to the elements ofL2 .Œa; b�/ as functions (and wewill continue to do so), the elements of L2 .Œa; b�/ are more accurately char-acterized as equivalence classes of functions that are equal a.e. In a vectorspace with a norm k�k, one wants kvk to be zero if and only if the vector vis zero. By (5) of theorem 102, kf � gk2 D 0 is equivalent to f D g a.e:Hence, for example, the zero function and the Dirichlet function

d .x/ D

�1; x 2 Q

0; x 62 Q

are considered to be the same function in L2 .Œa; b�/.At this point it is worthwhile pausing to consider how all this is related to

Fourier series. The functions involved in creating a Fourier series (or, moregenerally, a trigonometric series) on the interval Œ0; 1� are 1; sin .2�kx/, andcos .2�kx/ where k D 1; 2; 3;. . . . As elements of the inner product spaceL2 .Œ0; 1�/, these functions are mutually orthogonal as

h1; sin .2�kx/i DZ 1

0

sin .2�kx/ d� D 0,

h1; cos .2�kx/i DZ 1

0

cos .2�kx/ d� D 0, and

hsin.2�jx/; cos .2�kx/i DZ 1

0

sin.2�jx/ cos .2�kx/ d� D 0

for all j; k 2 N. Additionally, for j ¤ k 2 N,

hsin.2�jx/; sin .2�kx/i DZ 1

0

sin.2�jx/ sin .2�kx/ d� D 0 and

hcos.2�jx/; cos .2�kx/i DZ 1

0

cos.2�jx/ cos .2�kx/ d� D 0.

Similar calculations show that h1; 1i D 1 and that

hsin .2�kx/ ; sin .2�kx/i D hcos .2�kx/ ; cos .2�kx/i D1

2

for k 2 N.Returning to our original definition of Fourier series, we see that

hf; 1i D

Z 1

0

f .x/ d� D a0

10.1. L2 spaces 285

and, for k 2 N,

hf; sin .2�kx/i DZ 1

0

f .x/ sin .2�kx/ d� D ak , and

hf; cos .2�kx/i DZ 1

0

f .x/ cos .2�kx/ d� D bk .

Applying the theory of (orthogonal) projections5 from linear algebra (seeexercises 17 to 19), we see that fn.x/ D a0 C 2

PnkD1.ak sin.2�kx/ C

bk cos.2�kx// is the projection of f into the finite-dimensional subspaceVn spanned by

Bn D fcos.2�nx/; : : : ; cos.2�2x/; cos.2�/x; 1; sin.2�x/;

sin.2�2x/; : : : ; sin.2�nx/g:

(See exercise 19.) In this reframed context, the principle question becomes:When does the sequence ffng of projections into the nested sequence ofsubspaces fVng converge to f ?

We begin by computing the norm of fn. Using the orthogonality of theelements of Bn, we find that

kfnk22 D hfn; fni

D

*a0 C 2

nXkD1

.ak sin .2�kx/C bk cos .2�kx// ;

a0 C 2

nXkD1


+

Dˇa20ˇC

nXkD1

.jakj2 C jbkj

2/.

Thus the L2 norm of fn is kfnk2 Dqˇa20ˇCPnkD1.jakj

2 C jbkj2/, the

usual Euclidean norm of the vector bf n D .bn; : : : ; b1; a0; a1; : : : ; an/.The dimension of the vector bf n changes with n, but it is fairly nat-ural to add zeros on both ends of bf n so that we can view bf n D.: : : ; 0; 0; 0; bn; : : : ; b1; a0; a1; : : : ; an; 0; 0; 0; : : :/ as belonging to a singleinfinite dimensional space that does not depend on n. In this context, the

5 Given a subspace W of an inner product space V and v 2 V the (orthogonal) projection ofv intoW is the unique vector Ov 2W such that v� Ov is orthogonal to every element inW .

original function f naturally corresponds to the doubly infinite sequencebf D .: : : ; b2; b1; a0; a1; a2; : : :/. When we consider the norm of bf we areled to the definition of l2 spaces.

Definition 40 (l2 spaces). Let I be a countable index set. A sequencefxigi2I of scalars is square-summable if

Pi2I jxi j

2 is finite. The set ofall square-summable sequences is denoted by l2 .I/.

The most common choices for the index set I are N and Z. Using thenatural addition and scalar multiplication of sequences, l2 .I/ becomes avector space. The proof that l2 .I/ is closed under addition follows the samecontours as the proof of theorem 101. The standard dot product and norm forvectors in Rn and Cn extend to l2 .I/.

Definition 41. Let x Dfxigi2I and y Dfyigi2I be elements of l2 .I/.Define hx; yi D

Pi2I xiyi and kxk2 D

qPi2I jxi j

2.

Theorem 104. hx; yi is an inner product on l2 .I/ with associated normkxk2 D

phx; xi. That is, for x; y; z 2 l2 .I/ and any scalar c 2 C,

1. hx; yi is finite,

2. hy; xi D hx; yi,

3. hcx; yi D c hx; yi,

4. hxC y; zi D hx; zi C hy; zi,

5. hx; xi � 0 with hx; xi D 0 if and only if x D 0,

6. kcxk2 D jcj kxk2, and

7. kxC yk2 � kxk2 C kyk2 (triangle inequality).

Proof. The only properties whose proofs require significant effort are (1) and(7). The proofs of the remaining properties involve straightforward algebraicmanipulation of series and are left as exercises. (exercise 24.)

To prove (1), note that .jxi j � jyi j/2 � 0 so that 2 jxi j jyi j � jxi j

2Cjyi j2.

Hence

jhx; yij D

ˇXi2I

xiyi

ˇ�Xi2Ijxiyi j�

Xi2I

2 jxi j jyi j�Xi2I.jxi j

2Cjyi j2/<1.

10.1. L2 spaces 287

For (7), we may assume that y ¤ 0 and set z D x� hx;yihy;yiy. Observe that

hz; yi D�x�hx; yihy; yi

y; y�D hx; yi �

hx; yihy; yi

hy; yi D 0.

Hence z is orthogonal to y so that we can apply the Pythagorean theorem tox D hx;yi

hy;yiyC z to conclude that

kxk22 D

ˇhx; yihy; yi

ˇ2kyk22Ckzk

22 �

ˇhx; yihy; yi

ˇ2kyk22 D

jhx; yij2

kyk22.

Multiplying by kyk22 and taking square roots, we derive the Cauchy-Schwarzinequality

jhx; yij � kxk2 kyk2 .

Now use the Cauchy-Schwarz inequality to determine that

kxC yk22 D hxC y; xC yi

D hx; xi C hx; yi C hy; xi C hy; yi

� kxk22 C 2 kxk2 kyk2 C kyk22

D .kxk2 C kyk2/2 .

Property (7) follows by taking square roots.

In this context, there is a natural analog to the question of the convergenceof the projections ffng of f . Namely, if x D .: : : ; b2; b1; a0; a1; a2; : : :/ 2

l2 .Z/ and xn D .: : : ; 0; 0; bn; : : : ; b2; b1; a0; a1; a2; : : : ; an; 0; 0; : : :/, doesthe sequence fxng

1nD1 converge to x? In this case, it is clear that the answer

is yes since, by the definition of convergence for infinite sums,

limnkx � xnk

22 D lim

n

Xk>n

.jakj2 C jbkj

2/

D limn

1XkD1

.jakj2 C jbkj

2/ �

nXkD1

.jakj2 C jbkj

2/

!D 0.

This fact leads naturally to the question of whether or not the vector of co-efficients bf D .: : : ; b2; b1; a0; a1; a2; : : :/ used in the Fourier series for thefunction f belongs to l2 .Z/. Shortly, we will answer this question in theaffirmative and so conclude that f Ofng converges to bf . However, the conver-gence of f Ofng to bf does not imply that ffng converges to f .

To see why not, consider the problem of expressing f .x/ D sin 2�x asan infinite linear combination of the functions in

B� D f1; cos .2�x/ ; cos .2�2x/ ; cos .2�3x/ ; : : :g .

Since hsin .2�x/ ; 1i D hsin .2�x/ ; cos .2�nx/i D 0 for all k 2 N,we see that the (one-sided) sequence of coefficients relative to B� isbf D.0; 0; 0; 0; : : :/. Consequently bf n D .0; 0; 0; 0; : : :/ and f Ofng not only con-verges tobf but is uniformly identical tobf . However, fn is always the zerofunction so that ffng most certainly does not converge to f . At this point,we have no reason to believe that a similar, but perhaps more subtle, phe-nomenon does not occur when using the set

B Df: : : ; cos .2�2x/ ; cos .2�x/ ; 1; sin .2�x/ ; sin .2�2x/ ; : : :g :

10.2 CompletenessContinuing our analysis of

fn.x/ D a0 C 2

nXkD1

.ak sin.2�nx/C bk cos.2�kx/q/;

we note that

fkC1 .x/ � fk .x/ D akC1 sin .2� .k C 1/ x/C bkC1 cos .2� .k C 1/ x/ :

Thus fkC1 � fk and fjC1 � fj are mutually orthogonal when j ¤ k.Additionally, fn and f � fn are orthogonal since fn is a projection of fonto the subspace Vn. By the Pythagorean theorem for inner product spaces(see exercise 13), we can conclude that

kf k22 D kfnk22 C kf � fnk

22

� kfnk22

D

n�1XkD0

kfkC1 � fkk22 C kf0k

22 .

HenceP1kD0 kfkC1 � fkk

22 converges. Since kfkC1 � fkk

22 D jakC1j

2 C

jbkC1j2, we see that bf 2 l2 .Z/. As noted previously, this fact implies that

f Ofng converges to bf .Since

P1kD0 kfkC1 � fkk

22 converges, given " > 0 we can find an N so

that kfm � fnk22 D

Pm�1kDn kfkC1 � fkk

22 < "2 or kfm � fnk2 < " for all

m > n � N . In other words, ffng is a Cauchy sequence.This observation raises the question of whether or not all Cauchy se-

quences in L2 .Œa; b�/ converge to a function in L2 .Œa; b�/. They do, but

10.2. Completeness 289

the convergence need not be pointwise. Exercise 51 gives an example of aCauchy sequence in L2 .Œ0; 1�/ that does not converge at any point in Œ0; 1/.However, Cauchy sequences will converge in norm to a function inL2 .Œ0; 1�/.

Definition 42. A sequence of functions fgng from L2 .Œa; b�/ is said to con-verge in norm to the function g 2 L2 .Œa; b�/ provided that, for each " > 0,it is possible to find a natural number N so that kgn � gk2 < " whenevern > N . In other words, fgng converges in norm to g if kgn � gk2 convergesto 0.

Theorem 105 (L2.Œa; b�/ is complete). Any Cauchy sequence of functionsfrom L2 .Œa; b�/ converges in norm to a function in L2 .Œa; b�/.

Proof. Let ffng be a Cauchy sequence of functions from L2 .Œa; b�/. Webegin by creating a related, positive function to be used in the dominatedconvergence theorem (theorem 39, page 140).

Since ffng is a Cauchy sequence, we can find a strictly increasing se-quence of indices fnkg such that kfnkC1 � fnkk2 < 2�k for k � 1. Setfn0 D 0 and define g .x/ D

P1kD1

ˇfnk .x/ � fnk�1 .x/

ˇ. It may be that

g .x/ is unbounded for some values of x. However, g is defined in the ex-tended real numbers and, being the limit of measurable functions, is measur-able. Moreover, by Fatou’s lemma (theorem 38, page 139),

kgk2 D

0@Z b

a

1XkD1

ˇfnk � fnk�1

ˇ!2d�

1A1=2

�

0@limn

Z b

a

nXkD1

ˇfnk � fnk�1

ˇ!2d�

1A1=2

D limn

0@Z b

a

nXkD1

ˇfnk � fnk�1

ˇ!2d�

1A1=2

D limn

��nXkD1

ˇfnk � fnk�1

ˇ��2

� limn

nXkD1

��fnk � fnk�1��2��fn1��2 C 1.

Thus g2 is integrable and so the set of values for which g .x/ is not finite hasmeasure zero. In other words, g is finite a.e.

Define f .x/ DP1kD1.fnk .x/ � fnk�1 .x// if g .x/ is finite and

f .x/ D 0 otherwise. Then˚fnk

�converges a.e. to f . Moreover, jfnk j �Pk

jD1 jfnj � fnj�1 j � g for all k 2 N, so jf j � g. Thus f 2 L2 .Œa; b�/with

kf k2 D

Z b

a

jf j2 d�

!1=2�

Z b

a

jgj2 d�

!1=2� kgk2 .

Similarly, kfnkk2 � kgk2 for all k 2 N. Since jfnk � f j2 � .g C g/2 D

4g2 for all k 2 N, the dominated convergence theorem implies that

limkkfnk � f k2 D

limk

Z b

a

jfnk � f j2 d�

!1=2D 0.

Hence the subsequence ffnk g converges in norm to f .To show that the original Cauchy sequence ffng converges in norm to

f , suppose that " > 0 and, noting that nn � n, select an N such thatkfnn � fnk2 < "=2 and kfnn � f k < "=2 whenever n > N . Then

kfn � f k2 � kfnn � fnk2 C kfnn � f k2 < "

for all n > N . In other words, ffng converges in norm to f .

A set S with a norm is called complete if every Cauchy sequence in Sconverges to an element of S . Since every Cauchy sequence in L2 .Œa; b�/converges to a function in L2 .Œa; b�/, we see that L2 .Œa; b�/ is complete.

It is worth noting that the proof of theorem 105 provides the tools to con-clude that while a sequence of functions ffng that converges in norm to afunction f in L2 .Œ.a; b/�/ need not converge at any point in Œa; b�, there isa subsequence of ffng that will converge to f a.e.

Theorem 106. Suppose that ffng is a sequence of functions fromL2 .Œ.a; b/�/ that converges in norm to f 2 L2 .Œ.a; b/�/. Then there is astrictly increasing sequence of indices fnkg such that ffnk g converges to fa.e.

Proof. Since ffng converges in norm in L2 .Œa; b�/, ffng is a Cauchy se-quence. Thus we can apply the construction used in the proof of theo-rem 105 to find a strictly increasing sequence of indices fnkg such that

10.3. Density 291

P1kD1.fnk .x/�fnk�1 .x// converges both almost everywhere and in norm

to a function f �. But

mXkD1

�fnk .x/ � fnk�1 .x/

�D fnm ,

so that ffnk g converges almost everywhere and in norm to f �. The proofwill be complete when we show that f D f � a.e.

Since kf � f �k2 � kf � fnkk2 C kf� � fnkk2 and both kf � fnkk2

and kf � � fnkk2 can be made arbitrarily small by selecting a suffi-ciently large value for k, we conclude that kf � f �k2 D 0 and hencef D f � a.e.

Note that the proof of theorem 106 also shows that if ffng converges innorm to f then all a.e.-convergent subsequences of ffng converge a.e. to f .

10.3 DensityWe now know that, when f 2 L2 .Œa; b�/, the sequence ffng defined byfn .x/ D a0C2

PnkD1 .ak sin .2�kx/C bk cos .2�kx// converges in norm

to some function g 2 L2 .Œa; b�/.6 The remaining question is whether or notg D f . Since fn is the projectionof f onto the space spanned by

Bn D fcos.2�nx/; : : : ; cos.2�2x/; cos.2�x/; 1;

sin.2�x/; sin.2�2x/; : : : ; sin.2�nx/g,

the condition that f D g for all f 2 L2 .Œa; b�/ is equivalent to thecondition that linear combinations of the functions in f: : : ; cos.2�2x/;cos.2�x/; 1; sin.2�x/; sin.2�2x/; : : :g are dense in L2 .Œa; b�/.

The first step in verifying this density is to note that the three trigonometricidentities

2 cos˛ cosˇ D cos .˛ � ˇ/C cos .˛ C ˇ/ ;

2 sin˛ sinˇ D cos .˛ � ˇ/ � cos .˛ C ˇ/ ; and

2 sin˛ cosˇ D sin .˛ C ˇ/C sin .˛ � ˇ/

can be used to prove that the vector space V spanned by

B Df: : : ; cos .2�2x/ ; cos .2�x/ ; 1; sin .2�x/ ; sin .2�2x/ ; : : :g

6 Recall that the coefficients ak and bk are computed from the function f . See page 280.

is also an algebra. In other words, V is closed not only under scalar multi-plication and addition of functions but also under the multiplication of func-tions. (See exercise 39.)

Instead of beginning with the trigonometric polynomials and asking whichfunctions can be approximated by them, we will approach the problem fromthe other end by asking what sets of functions will generate algebras thatare dense in L2 .Œa; b�/. The algebra generated by a set S of functions isthe smallest set of functions that contains S and is also closed under scalarmultiplication, function addition, and function multiplication.

We will identify a finite sequence of sets each of which generates an alge-bra whose members can approximate elements of the previous set arbitrarilyclosely. Expressed alternatively, we will create a finite sequence of sets eachof which generates an algebra whose closure in theL2 norm contains the pre-vious set. We will choose the initial set so that it generatesL2 .Œa; b�/. Hencethe final set also generates an algebra whose closure contains L2 .Œa; b�/.

Begin by noting that if f D u C iv 2 L2 .Œa; b�/, we can express f DuC � u� C ivC � iv� with each of the component functions belongingto L2 .Œa; b�/. (See exercise 40.) Hence the algebra generated by the non-negative, measurable functions in L2 .Œa; b�/ is L2 .Œa; b�/. Next, note that,given any non-negative function f , the sequence fjf � fnj

2g where fn isdefined by fn D min ff; ng converges pointwise to 0. Since jf � fnj

2 <

jf j2 and jf j2 is integrable, we can use the dominated convergence theorem(theorem 39, page 140) to conclude that

limnkf � fnk2 D

limn

Z b

a

jf � fnj2 d�

!1=2D 0:

Thus the closure of the algebra generated by the set of bounded, non-negative, measurable functions on Œa; b� contains the set of non-negative,measurable functions in L2 .Œa; b�/.

Now suppose f is a measurable function with 0 � f � B .The measurable simple function 'n D

Pn�1kD0 k

Bn� 1Ek where Ek D

f �1��k Bn; .k C 1/ B

n

��satisfies

kf �'nk2D

Z b

a

jf �'nj2 d�

!1=2�

Z b

a

�B

n

�2d�

!1=2DBpb � a

n:

Since we can make Bpb�an

arbitrarily small by taking n sufficiently large, theclosure of the algebra generated by the measurable characteristic functionscontains the set of bounded, non-negative, measurable functions.

10.3. Density 293

Theorem 107. The closure of the algebra of continuous functions on Œa; b�contains the measurable characteristic functions.

Proof. Given a non-empty subset S of Œa; b� and t 2 Œa; b�, define dS .t/ Dinf fjx � t j W x 2 Sg. We claim that dS is a continuous function. To see why,fix t0 2 Œa; b� and " > 0. Suppose that t 2 Œa; b� with jt0 � t j < "=2. Selectx0 2 S such that dS .t0/ � jx0 � t0j < dS .t0/ C "=2. Then the triangleinequality implies that

dS .t/ � jx0 � t j � jx0 � t0j C jt0 � t j < dS .t0/C ".

A symmetric argument shows that dS .t0/ < dS .t/C ". Hence dS is contin-uous.

If dF .x/ D 0, then there is a sequence fxng of points in F such thatlimn jx � xnj D 0. Thus if F is a closed set, dF .x/ D 0 implies that x 2 F .

Now let E be an arbitrary non-empty, measurable subset of Œa; b� and let" > 0. By theorem 22 (page 104), we can find an open setG and a closed setF satisfying F E G, � .GnE/ < "2=2, and � .EnF / < "2=2. Then� .GnF / < "2. Set H D Gc and note that, since F and H are disjoint,closed sets, dF .x/C dH .x/ > 0. Consequently,

f .x/ DdF .x/

dF .x/C dH .x/

is a continuous function on Œa; b� satisfying 0 � f � 1 and f D 1E onF [H . Thus

k1E �f k2 D

Z b

a

j1E �f j2 d�

!1=2�

�ZGnF

1 d�

�1=2Dp� .GnF /< ".

Hence the closure of the continuous functions on Œa; b� contains the measur-able characteristic functions on Œa; b�.

At this point, we know that the closure of the algebra of continuous func-tions contains L2 .Œa; b�/. To finish the task of verifying that the algebra offunctions generated by

B D f: : : ; cos .2�2x/ ; cos .2�x/ ; 1; sin .2�x/ ; sin .2�2x/ ; : : :g

is dense in L2 .Œ0; 1�/, we will prove three versions of the Stone-Weierstrasstheorem. The first two versions of the theorem apply to real-valued functionsand the third to complex-valued functions. The Stone-Weierstrass theoremsdraw the conclusion that an algebra of functions is uniformly dense in theset of continuous functions. In other words, any continuous function can beuniformly approximated by a function from the algebra. Since a sequence of

functions converging uniformly on Œa; b� to a function f will also convergeto f in the L2 norm, the algebra of functions is also dense in the set ofcontinuous functions when using the norm of L2 .Œa; b�/.

The proof of the first version of the Stone-Weierstrass theorem proceedsby dividing the range of f into thirds, finding a function from the algebrathat roughly approximates f on the three corresponding preimages, and thenrepeating the process on the error.

Theorem 108 (Stone-Weierstrass 1). Let X be a nonempty, compact setand let C be an algebra of continuous real-valued functions that satisfies

1. 1 2 C,

2. if f; g 2 C, then max ff; gg 2 C, and

3. the functions of C separate the points of X in the sense that, if x0 andy0 are distinct elements of X , then there is a function f 2 C such thatf .x0/ ¤ f .y0/.

Then given any continuous function f on X and any " > 0; we can finda g 2 C such that jf � gj < ". In other words, C is uniformly dense in thecontinuous real-valued functions on X .7

Proof. The algebra C contains all constant functions. So suppose that f isa non-constant, real-valued, continuous function on X . Because f is a con-tinuous function on the compact set X , f is bounded and takes on its supre-mum and infimum. Since the algebra C is closed under vertical scaling andtranslation, we may assume that f ’s maximum and minimum are 1 and �1respectively.

Define E D f �1.Œ13; 1�/ and F D f �1.Œ�1;�1

3�/. Then E and F; being

closed (non-empty) subsets of X; are compact. Now for every x 2 E andy 2 F there is a function gx;y 2 C such that gx;y .x/ ¤ gx;y .y/. Scale andtranslate gx;y to define

hx;y D2

3�4

3

gx;y � gx;y .x/

gx;y .y/ � gx;y .x/.

By construction, hx;y .x/ D 23

and hx;y .y/ D �23 and, since C is an algebra,hx;y 2 C

Temporarily fix y 2 F . For each x 2 E, the continuity of hx;y impliesthe existence of a neighborhood Ux of x on which hx;y > 1

3. Since fUxgx2E

7 Note that in this theorem and the next we are taking the set of scalars to be R rather than C.

10.3. Density 295

is an open cover of the compact set E, there are points x1; x2; : : : ; xn suchthat E � [niD1Uxi . Let 'y D max

˚hx1;y ; hx2;y ; : : : ; hxn;y

�. Then 'y 2 C,

'y .y/ D �23

, and 'y .x/ > 13

for all x 2 E.Now for each y 2 F , the continuity of 'y implies that we can find a

neighborhood Uy of y on which 'y < �13 . Again, the compactness of Fmeans that we can find a set of points y1; y2; : : : ; ym 2 F such that F �[miD1Uyi . Then

� D min˚'y1 ; 'y2 ; : : : ; 'ym

�D �max

˚�'y1 ;�'y2 ; : : : ;�'ym

�satisfies � 2 C, � .x/ > 1

3for all x 2 E, and � .y/ < �1

3for all y 2 F .

Finally, define f1 D min˚max

˚�;�1

3

�; 13

�. Then f1 satisfies f1 2 C,

�13� f1 �

13

, f1 .x/ D 13

for all x 2 E, and f1 .y/ D �13 for all y 2 F . Inaddition , the maximum and minimum of f � f1 are 2

3and �2

3respectively.

Now note that 32.f � f1/ is a continuous function with a maximum and

minimum of 1 and �1 respectively. Apply the previous procedure to cre-ate a new function f2 2 C such that 3

2.f � f1/ � f2 has a maximum and

minimum of 23

and �23

respectively. Thus

�

�2

3

�2� f �

�f1 C

2

3f2

��

�2

3

�2and the upper and lower bounds are attained. Repeating the process on�

3

2

�2 �f �

�f1 C

2

3f2

��will produce another function f3 2 C such that

�

�2

3

�3� f �

f1 C

2

3f2 C

�2

3

�2f3

!�

�2

3

�3where, again, f � .f1 C 2

3f2 C

�23

�2f3/ takes on the values of the bounds.

The construction can be iterated to produce a sequence of functions ffng inC with

�

�2

3

�n� f �

nXkD1

�2

3

�k�1fk �

�2

3

�n.

Since limn!1

�23

�nD 0, we see that any continuous function on X can be

uniformly approximated by an element of C.

An alternate way of expressing the conclusion of theorem 108 is to statethat the uniform closure of C contains all the continuous real-valued func-tions on X . In the next version of the Stone-Weierstrass theorem, we use thefact that the closure of an algebra is again an algebra (see exercise 45) todrop the hypothesis that if f; g 2 C, then max ff; gg 2 C.

Theorem 109 (Stone-Weierstrass 2). Let X be a nonempty, compact setand let C be an algebra of continuous real-valued functions that satisfies

1. 1 2 C, and


Then the uniform closure of C contains all the continuous real-valuedfunctions on X . In other words, given any continuous function f on X andany " > 0 we can find a g 2 C such that jf � gj < ".

Proof. We begin by noting that, since max ff; gg D 12.jf � gj C .f C g//

and jf j D max ff; 0g C max f�f; 0g, the two statements f; g 2 C impliesmax ff; gg 2 C and f 2 C implies jf j 2 C are equivalent. We will show thatif f is a continuous function on X , then jf j can be uniformly approximatedby a polynomial in f . In other words, given " > 0, we can find a polynomialp" such that jp" .f / � jf jj < ". Since p" .f / 2 C, jf j is in C, the uniformclosure of C. We can then conclude that C contains all the continuous func-tions on X because C is an algebra satisfying the conditions of theorem 108and the closure of C is C (see exercises 45 and 46).

Let f be an arbitrary continuous function on X . Since f is bounded andjf j D Bjf

Bj for any positive constantB , we may assume that jf j is bounded

by 1. We will construct a sequence of polynomials fpng that converges uni-formly to

pt on Œ0; 1�. Then

˚pn ı f

2�

is a sequence of functions in C that

converges uniformly to jf j Dpf 2.

Define p0 D 0 and pnC1 .t/ D pn .t/ C12.t � p2n .t// for n � 0. Then

pn .t/ �pt � 1 for 0 � t � 1. To see why, observe that p0 .t/ D 0 �

pt

and, by induction,

pt � pnC1 .t/ D

pt � pn .t/ �

1

2

�t � p2n .t/

�D .pt � pn .t//

�1 �

1

2

pt C pn .t/

��

10.3. Density 297

is nonnegative for 0 � t � 1 and n � 0. Hence fpng, being an increasingand bounded sequence of functions on Œ0; 1�, must converge on Œ0; 1� to somefunction p. Since p � 0 and must satisfy p .t/ D p .t/ � 1

2.t � p2 .t//, we

see that p .t/ Dpt on Œ0; 1�.

To verify that the convergence is uniform, suppose for the purpose of con-tradiction that the convergence is not uniform. In other words, assume thereis an " > 0, a strictly increasing sequence of indices fnkg, and a sequenceof points fxkg such that

pxk � pnk .xk/ > " for all k 2 N. Since X is

compact, fxkg must have a cluster point x�.Fix n 2 N. Then, for any k > n, the facts that fpkg is increasing and

nk � k > n imply that

pxk � pn .xk/ �

pxk � pnk .xk/ > ".

Becausept �pn .t/ is continuous and x� is a cluster point of fxkg, we con-

clude thatpx��pn .x

�/ � ". But n was arbitrary. So we have contradictedthe fact that fpng converges pointwise to

pt on Œ0; 1�. This contradiction

implies that fpng must converge uniformly topt .

The uniform closure of C contains all the continuous functions on X .

The final version of the Stone-Weierstrass theorem extends the result tocomplex-valued functions.

Theorem 110 (Stone-Weierstrass 3). Let X be a nonempty, compact setand let C be an algebra of complex-valued continuous functions that satisfies

1. 1 2 C,

2. if f 2 C, then Nf 2 C, and


Then the uniform closure of C contains all the complex-valued continuousfunctions on X . In other words, given any continuous function f on X andany " > 0 we can find a g 2 C such that jf � gj < ".

Proof. Let CR be the algebra of all real-valued functions in C (using R forthe scalars). By hypothesis, given distinct points x0 and y0 from X thereis a function f D u C iv 2 C such that f .x0/ ¤ f .y0/. This impliesthat either u .x0/ ¤ u .y0/ or v .x0/ ¤ v .y0/. Since u D 1

2.f C Nf / and

v D 12i.f � Nf / belong to C, u; v 2 CR. Thus CR fulfills the hypotheses

of theorem 109 and so the uniform closure of CR contains all real-valued,continuous functions on X . As C contains all functions of the form u C iv

with u; v 2 CR, the closure of C contains all complex-valued, continuousfunctions on X .

We now return to the original question of this sectionand prove that Span .B/ is dense in L2 .Œ0; 1�/ where B D

f: : : ; cos 2�2x; cos 2�x; 1; sin 2�x; sin 2�2x; : : :g. We cannot directlyapply the Stone-Weierstrass theorem because the functions in B do notseparate the points of Œ0; 1�. Every function f 2 B satisfies f .0/ D f .1/.Since there is only one problematic point and our interest is in approximat-ing a function in the L2 norm rather than uniformly, we can work aroundthis difficulty.

One way to deal with this issue is to identify the points 0 and 1 turning theinterval Œ0; 1� into a circle. Since the circle is compact, we can apply theorem110 to the algebra generated by B to conclude that the uniform closure ofthe algebra contains all continuous functions on the circle or, equivalently,all continuous functions g on Œ0; 1� with g .0/ D g .1/.

10

1/n 1 – 1/n

1

Figure 10.1. 'n

Now define a sequence of continuous functions f'ng that approximate1.0;1/ by taking 'n to be 1 on Œ1=n; 1 � 1=n�, 0 on f0; 1g, and linear on theintervals Œ0; 1=n� and Œ1 � 1=n; 1�. The sequence f'ng converges pointwiseto 1.0;1/. Given a continuous function f on Œ0; 1�, define fn D f � 'n. Thenff � fng D ff � .1 � 'n/g converges a.e. to zero. By the dominated con-vergence theorem,

limnkf � fnk2 D

�limn

Z 1

0

.jf j � .1 � 'n//2 d�

�1=2D 0.

Hence given any " > 0, we can find an n so that kf � fnk2 < "=2. More-over, fn is continuous with fn .0/ D 0 D fn .1/ so fn is in the uniform

10.4. Conclusion 299

closure of the algebra generated by B. Thus there is a function g in the alge-bra generated by B such that jfn � gj < "=2. But then

kfn � gk2 D

�Z 1

0

jfn � gj2 d�

�1=2<

�Z 1

0

"2

4d�

�1=2D "=2

so thatkf � gk2 � kf � fnk2 C kfn � gk2 < ":

Linear combinations of functions from B are dense in L2 .Œ0; 1�/.Reflecting back on the example just before Section 10.2, we

can see why linear combinations of the functions in B� D

f1; cos 2�x; cos 2�2x; cos 2�3x; : : :g are not dense in L2 .Œ0; 1�/. Thefunctions in B� do not separate points. Any function f 2 B� satisfiesf .x/ D f .1 � x/ for x 2 Œ0; 1� : If an identification similar to that usedabove is performed, we see that functions in Span .B�/ can be used touniformly approximate functions in L2.Œ0; 12 �/. This approximation can beextended to approximate functions f satisfying f .x/ D f .1 � x/ forx 2 Œ0; 1� but does not extend to all functions in L2 .Œ0; 1�/. (See exercises52 and 53.)

Returning to our original function f 2 L2 .Œ0; 1�/ and its correspondingFourier series, we know that the sequence ffng defined by

fn .x/ D a0 C 2

nXkD1


converges in norm to some function g 2 L2 .Œ0; 1�/. Moreover, since fn isthe projection into the subspace Vn spanned by

Bn D fcos.2�nx/; : : : ; cos 2�2x; cos 2�x; 1; sin 2�x;

sin 2�2x; : : : ; sin.2�nx/g;

we know that kf � fnk2 � kf � hk2 for all functions h 2 Vn.Let " > 0. Then there is a function h in the span of B such thatkf � hk2 < ". Now h 2 VN for some N and VN � Vn for n > N . Thusfor n > N we have kf � fnk2 � kf � hk2 < ". Hence ffng convergesin norm to f as desired. In L2 .Œ0; 1�/, Fourier series behave exactly as onewould hope.

10.4 ConclusionThe problem of making sense of Fourier series motivated mathematical de-velopments for over a century. In particular, a great deal of work in the area


of theories of integration was motivated by the desire to place Fourier serieson a firm theoretical foundation. The Lebesgue integral addresses the prob-lems related to the interaction between limits of functions and integration.In addition, the Lebesgue integral points toward a more satisfying resolu-tion of the problems related to the convergence of Fourier series. The con-vergence properties of the Lebesgue integral are instrumental in proving thatL2 spaces are complete. The completeness ofL2 .Œ0; 1�/, together with somebasic vector space theory and the Stone-Weierstrass theorem, create a robustcontext in which Fourier series always converge. L2 spaces provide a cleanfoundation for the theory of Fourier series. Specifically, any function be-longing to L2 .Œ0; 1�/ has a well-defined Fourier series that converges in theL2 norm to the original function. Moreover, the coefficients in the Fourierseries provide a norm-preserving linear correspondence between functionsin L2 .Œ0; 1�/ and sequences in l2 .Z/. In the end, we have a very satisfy-ing resolution of issues raised by Fourier’s approach to solving differentialequations.

10.5 Exercises10.01 Complex numbers: filling the gaps

The set C of complex numbers is the set of all numbers of the formx D aC ib where a; b 2 R and i2 D �1. Exercises 1 through 3 providea review of the basic properties of C needed for this chapter.

The complex conjugate of a complex number x D aCib is x D a�ib.

1. Let x and y be complex numbers. Prove that

(a) x C y D x C y,(b) xy D xy, and(c) xx � 0 with xx D 0 only if x D 0:

We define the modulus of a complex number x D a C ib to be jxj Dpxx D

pa2 C b2.

2. Let x and y be complex numbers. Prove that

(a) jxj D jxj,(b) jxyj D jxj jyj, and(c) jx C yj � jxj C jyj.

(Work with the squares of the expressions.)

10.5. Exercises 301

3. Let x D a C ib be a complex number. The real and imaginary partsof x, denoted by Rex and Im x, are Re x D a and Im x D b. Prove thatRe x D 1

2.x C x/ and Im x D 1

2i.x � x/.

10.02 Vector and inner product spaces: filling the gaps

A vector space is a set V of objects called vectors together with twooperations, called addition and scalar multiplication, subject to the fol-lowing ten axioms. In the axioms, u; v; and w are any elements of V anda and b are any scalars. (The set of scalars may be taken to be either R

or C.)

(a) uC v 2 V:(b) uC v D vC u:(c) .uC v/C w D uC .vC w/:(d) There is a vector 0 2 V such that uC 0 D u for all u 2 V .(e) For each u 2 V , there is a vector �u 2 V such that uC .�u/ D 0:(f) au 2 V:(g) a.uC v/ D auC av:(h) .aC b/v D avC bv:(i) a.bv/ D .ab/v:(j) 1v D v:

Exercises 4 through 19 provide a review of the properties of vectorspaces required in this chapter.

4. Prove that the set of scalar-valued functions on a fixed set X is a vectorspace.

A subset W of a vector space V is a subspace of V if W is a vectorspace using the operations of V .

5. Let V be a vector space. Prove that if W is a subset of V that containsthe vector 0 and is closed under addition and scalar multiplication, thenW is subspace of V .

Let S be a subset of a vector space V . Any vector v that can be expressedas v D c1v1 C c2v2 C � � � C cnvn where c1; c2; : : : ; cn are scalars andv1; v2; : : : ; vn 2 S is a linear combination of the elements of S . Thespan of S , written Span .S/, is the set of all possible linear combinationsof vectors from S .


6. Use exercise 5 to prove that Span .S/ is a subspace of V .

Let V be a vector space. An inner product on V is a mapping h�; �i WV V ! C (or R/ that for vectors u; v;w and scalar c satisfies

(a) hu; vi D hv;ui;(b) huC v;wi D hu;wi C hv;wi ;(c) hcu; vi D c hu; vi, and(d) hu;ui � 0, with hu;ui D 0 if and only if u D 0:

An inner product space is a vector space with an inner product.

7. Use the properties of an inner product space to prove that for vectors uand v and scalar c

(a) hu; vC wi D hu; vi C hu;wi, and(b) hu; cvi D c hu; vi.

The elements of a set S of vectors in an inner product space are mutuallyorthogonal if hu; vi D 0 for all u; v 2 S with u ¤ v.

8. Suppose that u and v are orthogonal vectors. Prove that au and bv arealso orthogonal for any choice of scalars a and b.

9. Suppose that v is orthogonal to all the vectors in a set S . Prove that v isalso orthogonal to any vector in Span .S/.

10. Let S be a set of vectors from an inner product space V . Prove that theset of vectors in V that are orthogonal to all the vectors in S is a subspaceof V .

Given a vector v in an inner product space V , define the norm of v tobe kvk D

phv; vi.

11. Prove that if v is a vector in an inner product space and c is a scalar, thenkcvk D jcj kvk. (Use properties (a) and (b) of an inner product space.)

12. Pythagorean theorem for inner product spaces. Let u and v be mem-bers of an inner product space V . Prove that if u and v are orthogonal,then kuC vk2 D kuk2 C kvk2.

13. Let fv1; v2; : : : ; vng be a set of orthogonal vectors in an inner prod-uct space V . Prove that k

PnkD1 ckvkk2 D

PnkD1 jckj

2 kvkk2 for any

choice of scalars c1; c2; : : : ; cn. (Use induction and exercise 9.)

10.5. Exercises 303

14. Prove the converse of exercise 12 for inner product spaces over the realnumbers.

15. Prove the Cauchy-Schwarz inequality. Given vectors u and v from aninner product space V , jhu; vij � kuk kvk. (Assuming v ¤ 0, set z Du� hu;vihv;viv. Show that z and v are orthogonal. Solve for u and use exercises

8 and 12 or exercise 13 to show that kuk2 kvk2 � jhu; vij2.)

16. Prove the triangle inequality for inner product spaces. If u and v areelements of an inner product space, then kuC vk � kuk C kvk. (Beginby expanding kuC vk2. Then use the Cauchy-Schwarz inequality fromexercise 15.)

Let W be a subspace of an inner product space V and let v 2 V . Thenthe projection of v intoW is the vector Ov 2 W such that v � Ov is orthog-onal to every element in W .8 Note that while the notations are similar,the projection Ov should not be confused with the vectorbf of Fourier co-efficients.

17. Prove that the word “the” in the definition of projection is justified. Inother words, prove that if both u and w are in the subspace W and bothv � u and v � w are orthogonal to every element of W , then u D w.(Use exercise 10 to show that u � w is orthogonal to itself and hencemust be 0.)

18. Let W be a subspace of an inner product space V and let v 2 V . Provethat the projection Ov of v into W satisfies kv � Ovk � kv � wk for allw 2 W . (Use exercise 12 on v � w D .v � Ov/C .Ov � w/.)

19. Suppose that fv1; v2; : : : ; vng is a set of non-zero orthogonal vectors inan inner product space V and that v 2 V . Prove that

Ov Dhv; v1ihv1; v1i

v1 Chv; v2ihv2; v2i

v2 C � � � Chv; vnihvn; vni

vn

is the projection of v onto Span fv1; v2; : : : ; vng. (Use exercise 9 to provethat v � Ov is orthogonal to all the vectors in Span fv1; v2; : : : ; vng.)

10.1 L2 -spaces: filling the gaps20. Prove that if f 2 L2 .Œa; b�/ and c is a scalar, then cf 2 L2 .Œa; b�/.

8 Such projections always exist when W is finite dimensional (spanned by a finite set of vec-tors). In general,W must be topologically closed to guarantee the existence of projections.

21. Use the properties of the Lebesgue integral to verify that h�; �i satis-fies the algebraic properties of an inner product on L2 .Œa; b�/. In otherwords, prove the remaining parts of theorem 102. (For (2) and (5), writef D u1 C iv1 and g D u2 C iv2.)

22. Prove that a measurable, complex-valued function f D uC iv is squareintegrable if and only if both u and v are square integrable.

23. Prove part (a) of theorem 103:

24. Prove that l2 .I/ is closed under scalar multiplication and vector addi-tion. In other words, prove that if x; y 2 l2 .I/ and c is a scalar, thenxC y 2 l2 .I/ and cx 2 l2 .I/. (Mirror the proof of theorem 101.)

25. In proving part (7) of theorem 104, why can we assume that y ¤ 0?

10.1 L2 -spaces: deeper reflections

Let X be a compact subset of R and let � be a finite measure on X:(See Sections 8.5 and 8.6 of Chapter 8.) Define L2 .X;�/ to be theset of all �-measurable, complex-valued functions f on X such thatRX jf j

2 d� <1.

26. Prove that L2 .X;�/ is a subspace of the vector space of all complex-valued functions onX . (This only requires that you verify thatL2 .X;�/contains the zero function and is closed under scalar multiplication andthe addition of functions.)

27. Prove that L2 .X;�/ is an inner product space under the inner producthf; gi D

RXf g d�. (See definition after exercise 5.)

28. Prove that Lp .Œa; b�/ � Lq .Œa; b�/ for 1 � q � p. In particular, anysquare integrable function is integrable.

29. Jensen’s inequality. Let f W Œ0;C1/ ! Œ0;C1/ be a continuousincreasing function that is also onto and let g D f �1.

(a) Suppose that f .a/ D b. Explain why ab DR a0 f d� C

R b0 g d�.

(Geometrically, placeR a0 f d� on the x-axis and

R b0 g d� on the y-

axis.)(b) Suppose that b > f .a/. Prove that ab <

R a0 f d�C

R b0 g d�. (First

explain why g .y/ > a for any y in the interval .f .a/ ; b/.)(c) Use symmetry to prove that ab <

R a0 f d�C

R b0 g d� for 0 � b <

f .a/.

10.5. Exercises 305

30. Prove that for any p; q > 1 with 1pC 1

qD 1 and a; b � 0,

ab �ap

pCbq

q.

(Apply Jensen’s inequality from exercise 29 to the function f .x/ Dxp�1.)

If f 2 Lp .Œa; b�/ for some 1 � p < 1, then kf kp is defined to

be kf kp D .R a0 jf j

p d�/1=p . L1 .Œa; b�/ is defined to be the set of allmeasurable functions f on Œa; b� for which jf j is bounded except on aset of measure zero. kf k1 is defined to be the smallest such bound.

31. Prove that if f 2 L1 .Œa; b�/ and g 2 L1 .Œa; b�/ then fg 2 L1 .Œa; b�/with

R ba jfgj d� � kf k1 kgk1.

32. Prove Holder’s inequality, a generalization of both exercise 31 and theCauchy-Schwarz inequality. Suppose that p; q � 1 with 1

pC 1

qD 1,

f 2 Lp .Œa; b�/, and g 2 Lq .Œa; b�/. Then fg 2 L1 .Œa; b�/ withR ba jfgj d� � kf kp kgkq . (Apply exercise 30 with a D jf j

kf kpand

b D jgjkgkq

and integrate the resulting inequality.)

33. Even though Lp .Œa; b�/ is not an inner product space, prove that

kf kp D .R ba jf j

p d�/1=p is a norm on Lp .Œa; b�/ for 1 � p < 1.In other words, prove that, for f; g 2 Lp .Œa; b�/ and any scalar ˛,

(a) k f kp D j˛j kf kp and(b) kf C gkp � kf kp C kgkp (Minkowski’s inequality).

(For part (b), modify the proof of theorem 101 to show that f C g 2Lp .Œa; b�/ and so jf C gjp�1 2 L p

p�1.Œa; b�/. Then apply Holder’s

inequality to the right side of jf C gjp � .jf j C jgj/ jf C gjp�1 D

jf j jf C gjp�1 C jgj jf C gjp�1 and relate k jf C gjp�1 k pp�1

to

kf C gkp .)

10.2 Completeness: filling the gaps34. In theorem 105

(a) Explain why g is finite a.e. (Let E D fx 2 Œa; b� W g .x/ D C1g.Show that for all n 2 N, � .E/ � 1

n2kgk22 :/

(b) Explain why˚fnk

�converges to f a.e.

35. Prove that if ffng converges to f in the norm of L2 .Œa; b�/ and ffnk gconverges a.e. to some function g, then f D g a.e.

10.2 Completeness: deeper reflections36. Modify the proof of theorem 105 to show that L2 .X;�/ is complete.

(See exercises 25 to 27.)

37. Modify the proof of theorem 105 to show that Lp .Œa; b�/ is complete.

10.3 Density: filling the gaps

Take B Df: : : ; cos 2�3x; cos 2�2x; cos 2�x; 1; sin 2�x; sin 2�2x;sin 2�3x; : : :g in exercises 38 through 49.

38. Prove that Span .B/ is dense in L2 .Œa; b�/ if and only if for any f 2L2 .Œa; b�/, the sequence ffng defined by

fn .x/ D a0 C 2

nXkD1


converges in norm to f . (Use exercises 18 and 19.)

39. Let V D Span .B/. Prove that V is an algebra. In other words, provethat V is closed under scalar multiplication, addition, and multiplica-tion. (Use the identities 2 cos˛ cosˇ D cos .˛ � ˇ/ C cos .˛ C ˇ/,2 sin˛ sinˇ D cos .˛ � ˇ/ � cos .˛ C ˇ/, and 2 sin˛ cosˇ D

sin .˛ C ˇ/C sin .˛ � ˇ/.)

40. Prove that if f D u C iv 2 L2 .Œa; b�/, then uC; u�; vC, and u� alsobelong to L2 .Œa; b�/. (Begin by explaining why .uC�u�/2 D .uC/2C.u�/2.)

41. Suppose that F , E, and U are measurable sets satisfying F E U ,� .U nE/ < "=2, and � .EnF / < "=2. Explain why � .U nF / < ".

42. Prove that if ffng is a sequence of functions on Œa; b� that convergesuniformly to the function f , then ffng also converges to f in the normof L2 .Œa; b�/.

43. Prove that if an algebra is uniformly dense in a set F of functions onŒa; b� then the algebra is also dense in F using the L2 .Œa; b�/ norm.(Show that any sequence that converges uniformly to f also convergesto f in the L2 norm.)

10.5. Exercises 307

44. In the proof of the Stone-Weierstrass theorem (theorem 108)

(a) Why does C contain all constant functions?(b) Why is it sufficient to prove that any continuous function f with

a maximum of 1 and a minimum of �1 can be uniformly approxi-mated by an element of C?

(c) Why are E and F closed and non-empty?

45. Prove that if A is an algebra of functions and A is the closure of A,then A is also an algebra as long as the operations of scalar multi-plication, addition, and multiplication are continuous in the sense thatlim cfn D c limfn, lim .fn C gn/ D lim fnClimgn, and lim .fngn/ D

limfn limgn. (The limit need not be a uniform limit. It may be point-wise, L2, or some other type of limit.)

46. Let S be a set of functions and let NS be the uniform closure of S . Provethat the uniform closure of NS is NS . In other words, prove that any func-tion that can be uniformly approximated by functions in NS can be uni-formly approximated by functions from S . (Use the triangle inequality.)

47. In theorem 109

(a) Supply the details explaining why it is sufficient to assume thatjf j � 1:

(b) Supply the details to prove that if fpng converges topx uniformly

on Œ0; 1� and jf j � 1, then fpn ıf 2g converges to jf j uniformly onthe domain of f .

48. Prove that if f is a continuous function, f .xk/ > " for k 2 N, andx� is a cluster point of fxkg, then f .x�/ � ". (If f is continuous andf .x�/ < ", then f < " on some neighborhood of x�.)

49. Let C be an algebra of complex-valued functions and let CR be the set ofreal-valued functions in C. Prove that CR is an algebra over R.

10.3 Density: deeper reflections50. The recursion used to approximate

pt in the proof of theorem 109 is

related to Newton’s method for finding a zero of a function.

(a) Derive the recursion of Newton’s method.(b) Explain why one would choose to employ the alternate recursion in

the proof of theorem 109 instead of using Newton’s recursion.(c) Rework the proof of theorem 109 using the recursion you found in

part (a) but starting with p0 D 1.

51. Define En D Œn�2k

2k; nC1�2

k

2k� where 2k � n < 2kC1.

(a) Prove that the sequence f1Eng converges in norm to 0 but does notconverge to 0 at any x 2 Œ0; 1�.

(b) Find a subsequence of f1Eng that converges to 0 everywhere.

In exercises 52 and 53, take B� D f1; cos 2�x; cos 2�2x; cos 2�3x;: : :g.

52. Suppose that f W Œ0; 1� ! C satisfies f .x/ D f .1 � x/. Further sup-pose that f can be uniformly approximated on

�0; 12

�by functions in

Span .B�/. Prove that f is also uniformly approximated on Œ0; 1� byfunctions in Span .B�/. (If x 2 Œ1

2; 1�, then 1 � x 2 Œ0; 1

2�.)

53. Suppose that f W Œ0; 1� ! C fails to satisfy f .x/ D f .1 � x/.Show that f cannot be uniformly approximated on Œ0; 1� by func-tions in Span .B�/. (Find an a 2 Œ0; 1� and an " > 0 so thatjf .a/ � f .1 � a/j > 2". Use these to show that if g 2 Span .B�/ andjf � gj < " on Œ0; 1

2�, then jf .x/ � g .x/j > " for some x 2 Œ1

2; 1�.)

54. Why does the following approach fail when trying to prove that Span .B/is dense in L2 .Œ0; 1�/?

(a) Given f 2 L2 .Œ0; 1�/, find a continuous function g that uniformlyapproximates f on Œ0; 1�.

(b) Use the fact that the functions in B separate points in�1n; 1�

to finda function 'n 2 Span .B/ that uniformly approximates g on

�1n; 1�.

(c) Show that limn kf � 'nk2 D 0.


Mathematical Association of America.Hewitt, E. and K. Stromberg (1975). Real and Abstract Analysis. Springer.Lay, D. (2012). Linear Algebra and Its Applications (4th ed.). Pearson.Stoock, D.W. (1999). A Concise Introduction to the Theory of Integration

(3rd ed.). Birkhauser.Young, N. (1988). An Introduction to Hilbert Space. Cambridge University

Press.

Appendices: A Compendiumof Definitions and Results

A.1 Sets of real numbers

While nearly all of the definitions that follow apply in a much broader con-text, you may assume here that all the sets consist of real numbers.

Boundary point: A point x is a boundary point of a setE if every open setcontaining x includes at least one point inE and one point in the complementof E.

Bounded set: A setE is bounded if there is a real numberB so that jxj < Bfor all x 2 E.

Closed set: A set F is closed if it is the complement of an open set. Alter-natively, a set F is closed if it contains all of its boundary points.

The complement of any open set is closed and the complement of anyclosed set is open.

The intersection of an arbitrary collection of closed sets is closed.The union of a finite number of closed sets is closed.

Closure: The closure of a set S , denoted by NS , is the smallest closed setcontaining S . Alternatively, NS consists of S together with all of its boundarypoints. In a space with a norm, NS consists of all points that are a limit of asequence of points from S .

Compact set: A set C is compact if, given any cover of C by open setsfG˛g, there is a finite subset

˚G˛i

�niD1

of fG˛g that is also a cover of C .A set of real numbers is compact if and only if it is both closed and

bounded.

Complement of a set: The complement Ec of a set E is the set of realnumbers that are not elements of E. Ec D fx 2 R W x 62 Eg. (See relativecomplement.)

309

310 Appendices: A Compendium of Definitions and Results

Contained: A set A is contained in a set B , denoted by A B , if everyelement of A is an element of B . If, in addition, there is at least one elementof B that is not an element of A; then we say that A is strictly or properlycontained in B and write A � B .

Cover: A collection of sets fA˛g is a cover of a set E if E is contained inthe union of the fA˛g. (E [˛A˛ .)

De Morgan’s Laws: For any collection of sets fA˛g, [˛Ac˛ D .\˛A˛/c

and \˛Ac˛ D .[˛A˛/c . In the special case of two sets A and B , Ac [Bc D

.A \ B/c and Ac \ Bc D .A [ B/c .

Dense: Given two sets A � B in a space X with norm k�k, the set A isdense in B if for any x 2 B there is a sequence fxig from A such thatlim kx � xnk D 0. In a more general context, A is dense in B if given anyx 2 B and any neighborhood U of x, U contains points of A.

Extended real numbers: The extended real numbers consist of the realnumbers together with�1 andC1which are understood to be respectivelyless than and greater than any real number. We denote the extended realnumbers by R.

Integers: The set of integers, denoted by Z, is the setZ Df: : : ;�3;�2;�1; 0; 1; 2; 3; : : :g.

Intersection: The intersection of a collection of sets fA˛g, written as\˛A˛ ,is the set of elements individually belonging to all the sets A˛ . The intersec-tion of a pair of sets A and B is written as A \ B .

Natural numbers: The set of natural numbers, denoted by N, is the setN Df1; 2; 3; 4; : : :g.

Neighborhood: Given a point x, a neighborhood of x is a open set contain-ing x.Nondegenerate: An interval I is nondegenerate if it contains an open in-terval. Alternatively, I contains more than a single point.

Nonoverlapping: Two intervals are nonoverlapping if their intersectionconsists of at most a single point. The common point will of necessity bean end point of both intervals.

Open set: A set E is open if, given any element x of E, there is an " > 0 sothat any point y satisfying jy � xj < " is also an element ofE. Alternatively,a set E is open if it contains none of its boundary points. (See relativelyopen.)

A.2. Infimums and supremums 311

The complement of any open set is closed and the complement of anyclosed set is open.

The union of an arbitrary collection of open sets is open.The intersection of a finite number of open sets is open.

Positive rational numbers: QC D fx 2 Q W x > 0g.

Rational numbers: A real number x is called rational if x can be expressedas the ratio of integers, x D p

qwhere p; q 2 Z, q ¤ 0. The set of rational

numbers is denoted by Q.

Real numbers: The set of real numbers is denoted by R.

Relative complement of a set: Given a pair of sets A and B , the relativecomplement of A with respect to B is BnA D fx 2 B W x 62 Ag D B \Ac ,the set of points belonging to B but not A. (See complement.)

Relatively open: A set E is relatively open in X if E D X \ U for someopen set U . Alternatively, given any x 2 U , there is " > 0 so that any pointy 2 X satisfying jx � yj < " is also an element of E. (See open.)

Union: The union of a collection of sets fA˛g, written as [˛A˛ , is the setof elements belonging to one or more of the sets A˛ . The union of a pair ofsets A and B is written as A [ B .

A.2 Infimums and supremumsA nonempty, bounded set of real numbers will have an infimum and a supre-meum.

Infimum: The infimum of a set S of real numbers, denoted by infS , is thelargest real number ˛ satisfying ˛ � x for all x 2 S .

The existence of infS as a real number for all nonempty and bounded Sis equivalent to the completeness of the real numbers.

In the extended real numbers, R, we write infS D �1 when S is notbounded below and infS D C1 when S is empty.

For any nonempty, bounded set S ,

infSD� sup .�S/D � sup f�x W x 2 Sg :

If A and B are sets with A B , then infA � infB .Sometimes, a set that determines the range of values to be considered will

appear under the inf. For example, infA f D inf ff .x/ W x 2 Ag where f isa real-valued function whose domain includes A.

If˚fj�1jD1

is a sequence of functions with a common domain, then gn D

infj�n fj is the function defined by gn .x/ D inf˚fj .x/ W n � j <1

�:

Supremum: The supremum of a set S of real numbers, denoted by supS ,is the smallest real number ˛ satisfying ˛ � x for all x 2 S .

The existence of supS as a real number for all nonempty and bounded Sis equivalent to the completeness of the real numbers.

In the extended real numbers, R, we write supS D C1 when S is notbounded above and supS D �1 when S is empty.

For any nonempty, bounded set S ,

supS D � inf .�S/ D � inf f�x W x 2 Sg :

If A and B are sets with A B , then supA � supB .Sometimes, a set that determines the range of values to be considered will

appear under the sup. For example, supA f D sup ff .x/ W x 2 Ag where fis a real-valued function whose domain includes A.

If˚fj�1jD1

is a sequence of functions with a common domain, then gn D

supj�n fj is the function defined by gn .x/ D sup˚fj .x/ W n � j <1

�:

The sequence fgng will be a decreasing sequence of functions.

A.3 Sequences of real numbersA sequence of real numbers is a function from N to R. The sequence istypically expressed using the form fang

1nD1 or fang where an is the value of

the function when the value n is input.

Bounded: A sequence fang is bounded if there is a real number B suchthat janj � B for all n 2 N. The sequence is said to be bounded above ifan � B for all n 2 N and bounded below if an � B for all n 2 N.

Every bounded sequence has at least one cluster point. If the cluster pointis unique, then the cluster point is the limit of the sequence.

Cauchy: A sequence fang is called a Cauchy sequence if, given any " > 0;there is a natural number n such that

ˇaj � ai

ˇ< " whenever i; j � n.

Every Cauchy sequence converges.

Cluster point: A point x is a cluster point (also known as an accumulationpoint) of a sequence fang if, given any " > 0 and any natural number n, thereis another natural number j with j � n for which

ˇaj � x

ˇ< ".

Every bounded sequence has at least one cluster point.

A.3. Sequences of real numbers 313

If a bounded sequence has a unique cluster point, then the sequence con-verges to that cluster point.

Complete: A set with a norm (like the absolute value) is complete if everyCauchy sequence converges. The set R of real numbers is complete as is anyclosed subset of R. The axiom that R is complete is equivalent to the axiomthat every bounded nonempty subset of R has a supremum.

Converge: A sequence fang is said to converge if there is a real value x forwhich, given any " > 0; there is a natural number n such that

ˇaj � x

ˇ< "

whenever j � n. In this case, fang is called a convergent sequence and x isits limit.

Any convergent sequence is bounded.A monotone sequence converges if and only if it is bounded.A sequence converges if and only if it is a Cauchy sequence.

Decreasing: A sequence fang is decreasing if an � anC1 for all n 2 N. Ifan > anC1 for all n 2 N, the sequence is said to be strictly decreasing.

Any bounded, decreasing sequence converges.

Increasing: A sequence fang is increasing if an � anC1 for all n 2 N. Ifan < anC1 for all n 2 N, the sequence is said to be strictly increasing.

Any bounded, increasing sequence converges.

Liminf: Let fang be a sequence that is bounded below. The limit infimumof fang is limnan D lim infn an D limn infj�n aj D limn inf

˚aj W j � n

�.

If fang is bounded below,˚infj�n aj

�is an increasing sequence so the

liminf always exists. If fang is not bounded below, we say that limnan D

�1.

Limit: A point x is the limit of a sequence fang if, given any " > 0, there isa natural number n for which

ˇaj � x

ˇ< " whenever j � n.

The limit of a sequence, when it exists, is unique.If fang is monotone and bounded, it has a limit. (See converges.)A sequence converges to a limit if and only if it is a Cauchy sequence.

Limsup: Let fang be a sequence that is bounded above. The limitsupremum of fang is limnan D lim supn an D limn supj�n aj Dlimn sup

˚aj W j � n

�.

If fang is bounded above,˚supj�n aj

�is a decreasing sequence so

the limsup always exists. If fang is not bounded above, we say thatlimnan D C1.

Monotone: A sequence is monotone if it is either an increasing sequence ora decreasing sequence.

A.4 Real-valued functionsGiven a subset X of R, a real-valued function f on X is a rule that assignsto each element of X a single value from R. The set X is called the domainof f . In the following, the set X is assumed to be the domain of f unlessotherwise specified.

Absolutely continuous: A function f is absolutely continuous on a setA if given any " > 0 there is a ı > 0 so that

Pk jf .xk/ � f .yk/j <

" for any finite choice of disjoint intervals f.xk ; yk/g from A satisfyingPk jxk � ykj < ı.

Bounded: A function f is bounded if there is a real number B such thatjf j � B . By jf j � B we mean that jf .x/j � B for all x 2 X . Alter-natively, f is bounded if the range of f , f .X/ D ff .x/ W x 2 Xg ; is abounded set.

If f is continuous on a compact domain D, then f .D/ is compact.If f is continuous on a compact domain, then f is bounded.

Continuous at x: A function f is continuous at a point x 2 X if any ofthe following equivalent statements is true. (See continuous.)

1. Given any " > 0 there is a ı > 0 so that jf .x/ � f .y/j < " for ally 2 X satisfying jx � yj < ı.

2. Given any open set V containing f .x/, there is a open setU containingx such that f .X \ U/ D ff .x/ W x 2 X \ U g V .

3. Given any sequence fxng from X that converges to x, limn f .xn/ D

f .x/.

Continuous: A function f is continuous if any of the following equivalentstatements is true.

1. The function f is continuous at each point x 2 X . (See continuous ata point and uniformly continuous.)

2. Given any x 2 X and " > 0 there is a ı > 0 so that jf .x/ � f .y/j < "for all y 2 X satisfying jx � yj < ı.

3. Given any open set V in R, its preimage, f �1 .V / D

fx 2 X W f .x/ 2 V g ; is an open set relative to X .

A.5. Sequences of functions 315

4. Given any x 2 X and any sequence fxng from X that converges to x,limn f .xn/ D f .x/.

If f is continuous on a compact domain, then f .D/ is compact.If f is continuous on a compact domain, then f is bounded.If f is continuous on a compact domain, then f is uniformly continuous.(See also intermediate value theorem.)

Decreasing: A function f is decreasing if f .x/ � f .y/ whenever x; y 2X with x < y. If f .x/ > f .y/, then we say that f is strictly decreasing.

Differentiable at a point: A function f is differentiable at the point a 2X if limx!a

f .x/�f .a/x�a

exists. In this case, f 0 .a/ D limx!af .x/�f .a/

x�ais

the derivative of f at a.

Differentiable: A function f is differentiable on a set if it is differentiableat every point in the set.

Increasing: A function f is increasing if f .x/ � f .y/ whenever x; y 2X with x < y. If f .x/ < f .y/, then we say that f is strictly increasing.

Intermediate value theorem: If f is continuous on Œa; b� and v is a valuebetween f .a/ and f .b/ then there is a c 2 .a; b/ such that f .c/ D v.

Monotone: A function is monotone if it is either increasing or decreasing.

Mean value theorem: If f is continuous on Œa; b� and differentiable on.a; b/, then there is a point c 2 .a; b/ satisfying f 0 .c/ D f .b/�f .a/

b�a.

Preimage: The preimage of a set E under a function f is f �1 .E/ Dfx 2 X W f .x/ 2 Eg.

Uniformly continuous: A function f is uniformly continuous if for any" > 0 there is a ı > 0 such that jf .x/ � f .y/j < " for all x; y 2 Xsatisfying jx � yj < ı.

In contrast to the definition of continuous, the ı in the definition of uni-formly continuous cannot depend on the value of x. Given " > 0, one valueof ı must serve for all choices of x. (See continuous.)

If f is continuous on a compact set X , then f is uniformly continuous.

A.5 Sequences of functions

In the following, we assume that ffng is a sequence of functions with acommon domain X .

Convergence: A sequence of functions ffng converges (pointwise) to afunction f if the sequence ffn .x/g of real numbers converges to f .x/ forall x 2 X . Alternatively, given any x 2 X and any " > 0, there is a nat-ural number n so that

ˇfj .x/ � f .x/

ˇ< " for all j � n. (See uniform

convergence.)

Decreasing: A sequence of functions ffng is decreasing if fn � fnC1 forall n 2 N. By fn � fnC1 we mean that fn .x/ � fnC1 .x/ for all x 2 X .

Increasing: A sequence of functions ffng is increasing if fn � fnC1 for alln 2 N. By fn � fnC1 we mean that fn .x/ � fnC1 .x/ for all x 2 X .

Infimum: The infimum of a sequence of functions ffng is the function g Dinfn fn defined by g .x/ D infn fn .x/ D inf ffn .x/ W n 2 Ng. The functioninfn fn is only defined as a real-valued function if ffn .x/g is bounded belowfor all x 2 X . We can always define infn fn as a function taking on extendedreal values.

Limit: A sequence of functions ffng has a limit if there is a real-valuedfunction f such that, given any x 2 X and any " > 0, there is a naturalnumber n so that

ˇfj .x/ � f .x/

ˇ< " for all j � n.

Monotone: A sequence of functions ffng is monotone if it is either increas-ing or decreasing.

Supremum: The supremum of a sequence of functions ffng is the func-tion g D supn fn defined by g .x/ D supn fn .x/ D sup ffn .x/ W n 2 Ng.The function supn fn is only defined as a real-valued function if ffn .x/g isbounded above for all x 2 X . We can always define supn fn as a functiontaking on extended real values.

Uniform convergence: A sequence of functions ffng converges uniformlyto a function f given any " > 0, there is a natural number n so thatˇfj .x/ � f .x/

ˇ< " for all j � n and for all x 2 X . (See convergence.)

Uniformly bounded: A sequence of functions ffng is uniformly boundedif there is a real number B such that jfnj � B for all n 2 N. Equivalently,jfn .x/j � B for all n 2 N and all x 2 X .

A.6 Complex numbers

The set C of complex numbers is the set of all numbers of the form x D

aC ib where a; b 2 R and i2 D �1.

A.7. Inner product spaces and projections 317

Complex conjugate: The complex conjugate of a complex number x DaC ib is x D a� ib. The complex conjugate of a complex-valued functionf is the function f where f .x/ D f .x/. For complex numbers x and y

x C y D x C y;

xy D xy, andxx � 0 with xx D 0 only if x D 0:

Imaginary part: Let x D a C ib be a complex number. The imaginarypart of x, denoted Imx, is Im x D b D 1

2i.x � x/.

Modulus: We define the modulus of a complex number x D a C ib tobe jxj D

pxx D

pa2 C b2. The modulus of x is the same as the Eucliean

distance from the point .a; b/ to the origin in the plane. For complex numbersx and y,

jxj D jxj,jxyj D jxj jyj, andjx C yj � jxj C jyj.

Real part: Let x D aCib be a complex number. The real part of x, denotedby Re x; is Re x D a D 1

2.x C x/.

A.7 Inner product spaces and projectionsThe following definitions and results are stated for complex vector spaces.By dropping any complex conjugates, the statements apply to real vectorspaces.

Inner product space: An inner product space is a vector space V togetherwith an inner product h�; �i W V V ! C that for any x; y; z 2 V and anyscalar ˛, satisfies

1. hy; xi D hx; yi,

2. h˛x; yi D ˛ hx; yi,

3. hxC y; zi D hx; zi C hy; zi, and

4. hx; xi � 0 with hx; xi D 0 if and only if x D 0.

Linear combination: Given a finite set of vectors fv1; v2; v3; : : : ; vngany vector v that can be expressed as v D ˛1v1 C ˛2v2 C � � � C ˛nvnwhere ˛1; ˛2; : : : ; ˛n are scalars is a linear combination of the vectors infv1; v2; v3; : : : ; vng.


Norm: A norm on a vector space V is a mapping k�k W V ! R that for anyvectors u and v from V and any scalar ˛ satisfies

1. kuk � 0 with kuk D 0 if and only if u D 0.

2. k˛uk D j˛j kuk, and

3. kuC vk � kuk C kvk (triangle inequality).

Any inner product space has an associated norm defined by kuk Dphu;ui.

Orthogonal set: A set of vectors fv˛g˛2I in an inner product space is or-thogonal if

˝v˛; vˇ

˛D 0 for ˛ ¤ ˇ. If vectors u and v are orthogonal then

kuC vk2 D kuk2 C kvk2.

Projection: Given a subspace W of an inner product space V and a vectorv 2 V , the projection of v into W is the unique vector Ov satisfying Ov 2 Wand v � Ov 2 W ? where W ? is the vector space of all vectors in V that areorthogonal to every vector in W . Note that, while the notation is similar, Ovshould not be confused with sequence of Fourier coefficients.

The projection Ov satisfies kv � Ovk � kv � wk for all w 2 W .If fv1; v2; : : : ; vng is a set of non-zero orthogonal vectors in an inner prod-

uct space V and v 2 V , then the projection of v into Span fv1; v2; : : : ; vngis

Ov Dhv; v1ihv1; v1i

v1 Chv; v2ihv2; v2i

v2 C � � � Chv; vnihvn; vni

vn.

Span: The span, Span.S/ of a set S of vectors is the set of all vectors thatcan be expressed as linear combinations of vectors from S .

Subspace: Let V be a vector space. A subset W of V is a subspace of V ifW , together with the operations of V , satisfies the vector space axioms.

To prove that a subsetW is a subspace of V , it is sufficient to verify thatWcontains the vector 0 and is closed under addition and scalar multiplication.

Index

NotationDR ba f , 52

DR ba f , 52

DR ba f , 52

gR ba f , 163

LR ba f , 119

MR ba f d�g , 248

MR ba f d�g , 248

L-YR ba f , 121

L-YR ba f , 120

L-YR ba f , 120

NR baf .x/ dx, 4

RR ba f , 27

R-SR ba f dg, 227

s-LR ba f , 123

��g -a.e., 248SL .f;P/, 91, 119SL-S

�f; �g ;P

�, 248

SL.f;P/, 91

SL-Y .f;P/, 120SL-Y .f;P/, 120SR-S .f; g;P/, 227SD .f;P/, 52SD.f;P/, 52

SR.f;P/, 26PP infIk f �xk , 52PP supIk f �xk , 52PP f �x, 26

1A, 121Ec , 309f .c�/, 234f�cC�, 234

f �1 .A/, 315

f �, 123f C, 123ı .t/, 162� .t/, 163� .E/, 98�� .A/, 95�g .A/, 244��g .A/, 242ha;b .x/, 82infS , 311infA f , 311limnan, 313lim infn an, 313limnan, 313lim supn an, 313N, 310Q, 311QC, 311R, 311NR, 310Z, 310kPk, 26PL, 29PR, 29AnB , 311supS , 312supA f , 312\˛A˛ , 310[˛A˛ , 311[, 29V ba f , 179V .f;P/, 179

Abel, Neils Henrik, 280absolute continuity, 143, 145, 201

319

320 Index

absolutelycontinuous, 314integrable, 178, 183–185ntegrable, 201

algebra, 292–294, 296–299, 306, 307generated by a set, 292, 293, 298

almost everywhere, 63, 108, 135,174, 175

Archimedes, 1

Berkeley, Bishop George, 7binomial formula, 6Borel sets, 103, 105, 113, 246bound-and-telescope, 37, 59, 61,

130, 167boundary point, 309bounded

function, 314sequence, 312set, 309uniformly, 316variation, 179–183, 201

Cantor, Georg, 77function, 79, 86–88, 148, 158set, 77, 78, 86

Caratheodory, Constantin, 97measurability condition, 98, 244

Cauchy, Augustin-Louis, 8criterion, 55, 61, 172, 215, 231sequence, 312

Cauchy-Schwarz inequality, 287,303

characteristic function, 121closed, 309closure, 292, 293, 296, 297, 307,

309cluster point, 312compact, 309comparison test, 184, 189complement, 309

relative, 311

complete, 289, 290, 300, 306, 313complex conjugate, 317conditionally integrable, 178conjugate, complex , 317contained, 310continuous, 314

absolutely, 143, 145, 201, 314at a point, 314distributions, 233function, 32, 57, 106, 231uniformly, 315

convergencein norm, 289–291, 294, 299, 300,

306, 308of a sequence of functions, 316of real sequence, 313pointwise, 316uniform, 42, 316

convergence theoremsgauge dominated, 189gauge monotone, 186, 189Lebesgue bounded, 136Lebesgue dominated, 140Lebesgue monotone, 138Lebesgue summation, 141measure theory bounded, 250measure theory dominated, 251measure theory monotone, 250Riemann uniform, 42Riemann-Stieltjes, 239

countable additivity, of measure, 93Cousin’s theorem, 167cover, 310covering

Vitali, 195, 205, 206

Darboux, Gaston, 10, 51sum, 52, 56, 58, 66, 67

Darboux-Stieltjes integral, 270De Morgan’s Laws, 310decreasing

function, 315

Index 321

sequence of functions, 316sequence of real numbers, 313

ı-fine, 163, 209, 212dense, 291–294, 306, 310differentiable, 315

at a point, 315Dini derivatives, 204–206Dirichlet function, 28, 49, 53, 71, 83,

92, 164, 229, 284discrete distribution, 234divide-and-conquer, 130, 134, 136,

144, 194, 197, 203division point, 25

Egoroff’s theorem, 135extended real numbers, 310

forced tags, 168Fourier, Joseph, 9

series, 9, 12, 73, 84, 279fundamental theorem of calculus

gaugederivative form, 194, 197evaluation form, 191, 193

Lebesguederivative form, 143, 146evaluation form, 141, 148

Newton-Leibniz, 4Riemann

derivative form, 40, 47evaluation form, 39, 47

� -fine, 163, 209, 212gauge, 162

integral, 163, 172, 199, 200,208

gauge-Stieltjes integral, 259, 270

Holder’s inequality, 305Henstock’s lemma, 175, 176, 184,

187, 188, 195, 197, 216Henstock, Ralph, 13, 162

Hippocrates, 1lunes of, 17

imaginary part, 317increasing

function, 32, 57, 106, 179, 182,231, 315

sequence of functions, 316sequence of real numbers, 313

infimum, 108, 311of a sequence of functions, 316

inner measure, 116, 117inner product space, 302, 317integers, 310integrability

absolute gauge, 178, 183–185Darboux

continuous a.e., 63height-width bounds, 62

gaugeCauchy criterion, 172equal a.e., 175zero a.e., 174

Lebesgue, 119, 133equal a.e., 135zero a.e., 135

over subintervals, 37, 46, 67, 173,174, 231

Riemanncontinuous, 32increasing, 32

integralCauchy, 8, 49, 70Darboux, 10, 52

upper/lower, 52Darboux-Stieltjes, 270gauge, 14, 163, 172, 199, 200, 208gauge-Stieltjes, 259, 270Lebesgue, 12, 124, 126, 131, 132,

199–201Lebesgue-Stieltjes, 248Lebesgue-Young, 120, 126, 133

322 Index

integral (cont.)Newton, 4Riemann, 10, 27, 124, 128, 172Riemann-Stieltjes, 14, 227simple-Lebesgue, 123, 128, 131

integration by parts, 236intermediate value theorem, 315intersection, 310

Jensen’s inequality, 304

Kurzweil, Jaroslav, 13, 162

L2 spaces, 279, 281l2 spaces, 286Lebesgue, Henri, 12, 91

integral, 124, 131, 132, 199–201partition, 91, 109sum, 91, 93, 109, 119

Lebesgue-Stieltjesintegral, 248sum, 248

Lebesgue-Youngintegral, 120, 126, 133sum, 120

Leibniz, Gottfried, 4integral, 126

limitinfimum, 313of a sequence of functions, 316of real sequence, 313pointwise, 107supremeum, 313

linear combination, 288, 301, 317lunes of Hippocrates, 1, 17

mean value theorem, 315measurability conditions, 98, 104,

105, 111measurable

function, 105–108, 114, 247, 265partition, 120

setsexamples, 98, 99, 101, 244–246properties of, 98, 99, 101–103,

199, 200, 244–246measure

inner, 116, 117Lebesgue, 98outer, 95, 110, 242–244projection-valued, 255, 256space, 251, 253, 254, 266–269zero, 63, 68, 69, 105, 108, 169,

246mesh, 26, 31, 58, 67modulus, 317monotone

function, 315sequence of functions, 316sequence of real numbers, 314

monotonicityof integral, 44, 66, 149, 214, 258of measure, 93

natural numbers, 310neighborhood, 294, 295, 307, 310Newton, Isaac, 4, 11non-measurable set, 94nondegenerate, 310nonoverlapping, 310norm, 284, 318

inner product, 302L2, 283–285, 289–291, 294,

298–300, 306, 308l2, 286Lp , 305

open, 310relatively, 311

orthogonal, 284, 302, 303, 318outer measure, 95, 96, 98, 101–103,

110, 242–244

partial tagged partition, 175

Index 323

partition, 25Lebesgue, 91, 109measurable, 120partial tagged, 175tagged, 25

point mass, 227, 243, 244pointwise convergence, 316preimage, 91, 315projection, 285, 287, 291, 303,

318projection-valued measure, 255,

256

quadrature of the parabola, 2, 17

rational numbers, 311real numbers, 311real part, 317refinement, 30, 56, 58, 66, 125, 126,

149relative complement, 311relatively open, 311Riemann, Bernhard, 10, 51

integral, 172sum, 26, 30, 31, 44

Riemann-Stieltjesintegral, 14, 227integration by parts, 236sum, 227

sigma algebra, 100, 103, 111–113,117

simple function, 121simple-Lebesgue integral, 123, 128,

131span, 318splitting tags, 168standard properties, 44, 66, 149, 214,

257step function, 180, 229, 234, 249Stieltjes, Thomas Joannes, 14

Stone-Weierstrass theorem, 294,296, 297

straddle lemma, 191subadditivity, 96subintervals, integrability over, 37,

38, 46, 67, 173, 174, 231subspace, 285, 302, 304, 318sum

Darboux, 52, 56, 58, 66, 67Lebesgue, 91, 93, 109, 119Lebesgue-Stieltjes, 248Lebesgue-Young, 120Riemann, 26, 30, 31, 44Riemann-Stieltjes, 227

supremum, 108, 312dominated, 189of a sequence of functions, 316

tag, 25forcing, 168splitting, 168

tagged partition, 25translation invariance, 93truncation, 130, 144

uniform convergence, 316uniformly

bounded, 316continuous, 315

union, 311

variation, 179, 181Vitali

covering, 195, 205, 206covering theorem, 195–197, 201,

203, 205, 207Volterra, Vito, 81

function, 81, 88

Weierstrass, Karl, 75weight function, 241, 242, 270

About the Author

C. Ray Rosentrater is a Professor of Mathematics at Westmont Collegewhere he has also served as department chair and Associate Dean for Cur-riculum. He has been recognized as Westmont’s Teacher of the Year inthe Natural and Behavioral Sciences and has received the Faculty ResearchAward. He earned a PhD in mathematics from Indiana University and anMSc in computer science from the University of Toronto. Awarded a Ful-bright Fellowship in 1995, he now serves as Westmont’s Fulbright ProgramAdvisor. He served multiple terms on the ACMS board including terms asVice President and President. His other publications include papers in oper-ator theory and articles connecting analysis to computer science and linearalgebra to statistics. He co-wrote two chapters in Mathematics through theEyes of Faith.

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