fremlin measure theory book 2

8/12/2019 fremlin measure theory book 2

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MEASURE THEORY

Volume 2

D.H.Fremlin


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By the same author:Topological Riesz Spaces and Measure Theory, Cambridge University Press, 1974.Consequences of Martins Axiom, Cambridge University Press, 1982.

Companions to the present volume:Measure Theory, vol. 1, Torres Fremlin, 2000;Measure Theory, vol. 3, Torres Fremlin, 2002;Measure Theory, vol. 4, Torres Fremlin, 2003;Measure Theore, vol. 5, Torres Fremlin, 2008.

First edition May 2001

Second edition January 2010


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MEASURE THEORY

Volume 2

Broad Foundations

D.H.Fremlin

Research Professor in Mathematics, University of Essex


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Dedicated by the Author

to the Publisher

This book may be ordered from the printers, http://www.lulu.com/buy.

First published in 2001by Torres Fremlin, 25 Ireton Road, Colchester CO3 3AT, England

c D.H.Fremlin 2001The right of D.H.Fremlin to be identified as author of this work has been asserted in accordance with the Copyright,

Designs and Patents Act 1988. This work is issued under the terms of the Design Science License as published inhttp://www.gnu.org/licenses/dsl.html. For the source files see http://www.essex.ac.uk/maths/staff/fremlin/mt2.2010/index.htm.

Library of Congress classification QA312.F72

AMS 2010 classification 28-01

ISBN 978-0-9538129-7-4

Typeset byAMS-TEXPrinted by Lulu.com


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5

Contents

General Introduction 9

Introduction to Volume 2 10

*Chapter 21: Taxonomy of measure spaces

Introduction 12211 Definitions 12Complete, totally finite,-finite, strictly localizable, semi-finite, localizable, locally determined measure spaces; atoms;elementary relationships; countable-cocountable measures.

212 Complete spaces 17Measurable and integrable functions on complete spaces; completion of a measure.

213 Semi-finite, locally determined and localizable spaces 23Integration on semi-finite spaces; c.l.d. versions; measurable envelopes; characterizing localizability, strict localizabil-

ity, -finiteness.

214 Subspaces 33Subspace measures on arbitrary subsets; integration; direct sums of measure spaces; *extending measures to well-ordered families of sets.

215 -finite spaces and the principle of exhaustion 43The principle of exhaustion; characterizations of -finiteness; the intermediate value theorem for atomless measures.

*216 Examples 46A complete localizable non-locally-determined space; a complete locally determined non-localizable space; a completelocally determined localizable space which is not strictly localizable.

Chapter 22: The Fundamental Theorem of Calculus

Introduction 52221 Vitalis theorem in R 52

Vitalis theorem for intervals in R.

222 Differentiating an indefinite integral 55Monotonic functions are differentiable a.e., and their derivatives are integrable; d

dx

xa

f =fa.e.; *the Denjoy-Young-Saks theorem.

223 Lebesgues density theorems 63f(x) = limh0

12h

x+hxh f a.e. (x); density points; limh0

12h

x+hxh|f f(x)| = 0 a.e. (x); the Lebesgue set of a

function.

224 Functions of bounded variation 67Variation of a function; differences of monotonic functions; sums and products, limits, continuity and differentiability

for b.v. functions; an inequality for

fg.225 Absolutely continuous functions 75

Absolute continuity of indefinite integrals; absolutely continuous functions on R; integration by parts; lower semi-

continuous functions; *direct images of negligible sets; the Cantor function.

*226 The Lebesgue decomposition of a function of bounded variation 84Sums over arbitrary index sets; saltus functions; the Lebesgue decomposition.

Chapter 23: The Radon-Nikodym theorem

Introduction 92231 Countably additive functionals 92

Additive and countably additive functionals; Jordan and Hahn decompositions.

232 The Radon-Nikodym theorem 96Absolutely and truly continuous additive functionals; truly continuous functionals are indefinite integrals; *theLebesgue decomposition of a countably additive functional.

233 Conditional expectations 105-subalgebras; conditional expectations of integrable functions; convex functions; Jensens inequality.

234 Operations on measures 112Inverse-measure-preserving functions; image measures; sums of measures; indefinite-integral measures; ordering ofmeasures.

235 Measurable transformations 122The formula

g(y)(dy) =

J(x)g((x))(dx); detailed conditions of applicability; inverse-measure-preserving func-

tions; the image measure catastrophe; using the Radon-Nikodym theorem.


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Chapter 24: Function spaces

Introduction 133241 L0 andL0 133

The linear, order and multiplicative structure ofL0; Dedekind completeness and localizability; action of Borel func-

tions.

242L1 141The normed lattice L1; integration as a linear functional; completeness and Dedekind completeness; the Radon-

Nikodym theorem and conditional expectations; convex functions; dense subspaces.

243L 151The normed latticeL; norm-completeness; the duality between L1 and L; localizability, Dedekind completenessand the identification L=(L1).

244Lp 158The normed lattices Lp, for 1 < p


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265 Surface measures 316Normalized Hausdorff measure; action of linear operators and differentiable functions; surface measure on a sphere.

*266 The Brunn-Minkowski inequality 324Arithmetic and geometric means; essential closures; the Brunn-Minkowski inequality.

Chapter 27: Probability theory

Introduction 328271 Distributions 329

Terminology; distributions as Radon measures; distribution functions; densities; transformations of random variables;*distribution functions and convergence in measure.

272 Independence 335Independent families of random variables; characterizations of independence; joint distributions of (finite) independent

families, and product measures; the zero-one law; E(XY), Var(X+ Y ); distribution of a sum as convolution ofdistributions; Etemadis inequality; *Hoeffdings inequality.

273 The strong law of large numbers 3481

n+1

ni=0Xi0 a.e. if the Xn are independent with zero expectation and either (i)

n=0

1(n+1)2

Var(Xn)


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2A4 Normed spaces 511Normed spaces; linear subspaces; Banach spaces; bounded linear operators; dual spaces; extending a linear operatorfrom a dense subspace; normed algebras.

2A5 Linear topological spaces 514Linear topological spaces; topologies defined by functionals; convex sets; completeness; weak topologies.

2A6 Factorization of matrices 517Determinants; orthonormal families; T =P DQ where D is diagonal and P, Q are orthogonal.

Concordance 519

References for Volume 2 521

Index to Volumes 1 and 2Principal topics and results 524General index 528


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General introduction 9

General introduction In this treatise I aim to give a comprehensive description of modern abstract measure theory,with some indication of its principal applications. The first two volumes are set at an introductory level; they areintended for students with a solid grounding in the concepts of real analysis, but possibly with rather limited detailedknowledge. As the book proceeds, the level of sophistication and expertise demanded will increase; thus for the volumeon topological measure spaces, familiarity with general topology will be assumed. The emphasis throughout is on themathematical ideas involved, which in this subject are mostly to be found in the details of the proofs.

My intention is that the book should be usable both as a first introduction to the subject and as a reference work.For the sake of the first aim, I try to limit the ideas of the early volumes to those which are really essential to the

development of the basic theorems. For the sake of the second aim, I try to express these ideas in their full naturalgenerality, and in particular I take care to avoid suggesting any unnecessary restrictions in their applicability. Of coursethese principles are to to some extent contradictory. Nevertheless, I find that most of the time they are very nearlyreconcilable,providedthat I indulge in a certain degree of repetition. For instance, right at the beginning, the puzzlearises: should one develop Lebesgue measure first on the real line, and then in spaces of higher dimension, or shouldone go straight to the multidimensional case? I believe that there is no single correct answer to this question. Moststudents will find the one-dimensional case easier, and it therefore seems more appropriate for a first introduction, sinceeven in that case the technical problems can be daunting. But certainly every student of measure theory must at afairly early stage come to terms with Lebesgue area and volume as well as length; and with the correct formulations,the multidimensional case differs from the one-dimensional case only in a definition and a (substantial) lemma. Sowhat I have done is to write them both out (114-115). In the same spirit, I have been uninhibited, when settingout exercises, by the fact that many of the results I invite students to look for will appear in later chapters; I believethat throughout mathematics one has a better chance of understanding a theorem if one has previously attempted

something similar alone.The plan of the work is as follows:

Volume 1: The Irreducible MinimumVolume 2: Broad FoundationsVolume 3: Measure AlgebrasVolume 4: Topological Measure SpacesVolume 5: Set-theoretic Measure Theory.

Volume 1 is intended for those with no prior knowledge of measure theory, but competent in the elementary techniquesof real analysis. I hope that it will be found useful by undergraduates meeting Lebesgue measure for the first time.Volume 2 aims to lay out some of the fundamental results of pure measure theory (the Radon-Nikod ym theorem,Fubinis theorem), but also gives short introductions to some of the most important applications of measure theory(probability theory, Fourier analysis). While I should like to believe that most of it is written at a level accessible

to anyone who has mastered the contents of Volume 1, I should not myself have the courage to try to cover it in anundergraduate course, though I would certainly attempt to include some parts of it. Volumes 3 and 4 are set at arather higher level, suitable to postgraduate courses; while Volume 5 assumes a wide-ranging competence over largeparts of analysis and set theory.

There is a disclaimer which I ought to make in a place where you might see it in time to avoid paying for this book.I make no attempt to describe the history of the subject. This is not because I think the history uninteresting orunimportant; rather, it is because I have no confidence of saying anything which would not be seriously misleading.Indeed I have very little confidence in anything I have ever read concerning the history of ideas. So while I am happy tohonour the names of Lebesgue and Kolmogorov and Maharam in more or less appropriate places, and I try to includein the bibliographies the works which I have myself consulted, I leave any consideration of the details to those bolderand better qualified than myself.

For the time being, at least, printing will be in short runs. I hope that readers will be energetic in commenting onerrors and omissions, since it should be possible to correct these relatively promptly. An inevitable consequence of thisis that paragraph references may go out of date rather quickly. I shall be most flattered if anyone chooses to rely onthis book as a source for basic material; and I am willing to attempt to maintain a concordance to such references,indicating where migratory results have come to rest for the moment, if authors will supply me with copies of paperswhich use them. In the concordance to the present volume you will find notes on the items which have been referredto in other published volumes of this work.

I mention some minor points concerning the layout of the material. Most sections conclude with lists of basicexercises and further exercises, which I hope will be generally instructive and occasionally entertaining. How manyof these you should attempt must be for you and your teacher, if any, to decide, as no two students will have quitethe same needs. I mark with a >>> those which seem to me to be particularly important. But while you may not needto write out solutions to all the basic exercises, if you are in any doubt as to your capacity to do so you should take


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10 General introduction

this as a warning to slow down a bit. The further exercises are unbounded in difficulty, and are unified only by apresumption that each has at least one solution based on ideas already introduced. Occasionally I add a final problem,a question to which I do not know the answer and which seems to arise naturally in the course of the work.

The impulse to write this book is in large part a desire to present a unified account of the subject. Cross-referencesare correspondingly abundant and wide-ranging. In order to be able to refer freely across the whole text, I have chosena reference system which gives the same code name to a paragraph wherever it is being called from. Thus 132E is thefifth paragraph in the second section of the third chapter of Volume 1, and is referred to by that name throughout. Letme emphasize that cross-references are supposed to help the reader, not distract her. Do not take the interpolation

(121A) as an instruction, or even a recommendation, to lift Volume 1 off the shelf and hunt for121. If you are happywith an argument as it stands, independently of the reference, then carry on. If, however, I seem to have made rathera large jump, or the notation has suddenly become opaque, local cross-references may help you to fill in the gaps.

Each volume will have an appendix of useful facts, in which I set out material which is called on somewhere in thatvolume, and which I do not feel I can take for granted. Typically the arrangement of material in these appendices isdirected very narrowly at the particular applications I have in mind, and is unlikely to be a satisfactory substitute forconventional treatments of the topics touched on. Moreover, the ideas may well be needed only on rare and isolatedoccasions. So as a rule I recommend you to ignore the appendices until you have some direct reason to suppose that afragment may be useful to you.

During the extended gestation of this project I have been helped by many people, and I hope that my friends andcolleagues will be pleased when they recognise their ideas scattered through the pages below. But I am especiallygrateful to those who have taken the trouble to read through earlier drafts and comment on obscurities and errors.

Introduction to Volume 2

For this second volume I have chosen seven topics through which to explore the insights and challenges offered bymeasure theory. Some, like the Radon-Nikodym theorem (Chapter 23) are necessary for any understanding of thestructure of the subject; others, like Fourier analysis (Chapter 28) and the discussion of function spaces (Chapter 24)demonstrate the power of measure theory to attack problems in general real and functional analysis. But all haveapplications outside measure theory, and all have influenced its development. These are the parts of measure theorywhich any analyst may find himself using.

Every topic is one which ideally one would wish undergraduates to have seen, but the length of this volume makesit plain that no ordinary undergraduate course could include very much of it. It is directed rather at graduate level,where I hope it will be found adequate to support all but the most ambitious courses in measure theory, though it isperhaps a bit too solid to be suitable for direct use as a course text, except with careful selection of the parts to becovered. If you are using it to teach yourself measure theory, I strongly recommend an eclectic approach, looking for

particular subjects and theorems that seem startling or useful, and working backwards from them. My other objective,of course, is to provide an account of the central ideas at this level in measure theory, rather fuller than can easily befound in one volume elsewhere. I cannot claim that it is definitive, but I do think I cover a good deal of ground in waysthat provide a firm foundation for further study. As in Volume 1, I usually do not shrink from giving best results, likeLindebergs conditions for the Central Limit Theorem (274), or the theory of products of arbitrary measure spaces(251). If I were teaching this material to students in a PhD programme I would rather accept a limitation in thebreadth of the course than leave them unaware of what could be done in the areas discussed.

The topics interact in complex ways one of the purposes of this book is to exhibit their relationships. There is nocanonical linear ordering in which they should be taken. Nor do I think organization charts are very helpful, not leastbecause it may be only two or three paragraphs in a section which are needed for a given chapter later on. I do at leasttry to lay the material of each section out in an order which makes initial segments useful by themselves. But the orderin which to take the chapters is to a considerable extent for you to choose, perhaps after a glance at their individualintroductions. I have done my best to pitch the exposition at much the same level throughout the volume, sometimesallowing gradients to steepen in the course of a chapter or a section, but always trying to return to something whichanyone who has mastered Volume 1 ought to be able to cope with. (Though perhaps the main theorems of Chapter 26are harder work than the principal results elsewhere, and286 is only for the most determined.)

I said there were seven topics, and you will see eight chapters ahead of you. This is because Chapter 21 is ratherdifferent from the rest. It is the purest of pure measure theory, and is here only because there are places later inthe volume where (in my view) the theorems make sense only in the light of some abstract concepts which are notparticularly difficult, but are also not obvious. However it is fair to say that the most important ideas of this volumedo not really depend on the work of Chapter 21.

As always, it is a puzzle to know how much prior knowledge to assume in this volume. I do of course call on theresults of Volume 1 of this treatise whenever they seem to be relevant. I do not doubt, however, that there will be


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Introduction to Volume 2 11

readers who have learnt the elementary theory from other sources. Provided you can, from first principles, constructLebesgue measure and prove the basic convergence theorems for integrals on arbitrary measure spaces, you ought to beable to embark on the present volume. Perhaps it would be helpful to have in hand the results-only version of Volume1, since that includes the most important definitions as well as statements of the theorems.

There is also the question of how much material from outside measure theory is needed. Chapter 21 calls for somenon-trivial set theory (given in2A1), but the more advanced ideas are needed only for the counter-examples in 216,and can be passed over to begin with. The problems become acute in Chapter 24. Here we need a variety of resultsfrom functional analysis, some of them depending on non-trivial ideas in general topology. For a full understanding of

this material there is no substitute for a course in normed spaces up to and including a study of weak compactness.But I do not like to insist on such a preparation, because it is likely to be simultaneously too much and too little.Too much, because I hardly mention linear operators at this stage; too little, because I do ask for some of the theoryof non-locally-convex spaces, which is often omitted in first courses on functional analysis. At the risk, therefore, ofwasting paper, I have written out condensed accounts of the essential facts (2A3-2A5).Note on second printing, April 2003

For the second printing of this volume, I have made two substantial corrections to inadequate proofs and a largenumber of minor amendments; I am most grateful to T.D.Austin for his careful reading of the first printing. In addition,I have added a dozen exercises and a handful of straightforward results which turn out to be relevant to the work oflater volumes and fit naturally here.

The regular process of revision of this work has led me to make a couple of notational innovations not de-scribed explicitly in the early printings of Volume 1. I trust that most readers will find these immediately com-

prehensible. If, however, you find that there is a puzzling cross-reference which you are unable to match withanything in the version of Volume 1 which you are using, it may be worth while checking the errata pages inhttp://www.essex.ac.uk/maths/staff/fremlin/mterr.htm.

Note on second edition, January 2010

For the new (Lulu) edition of this volume, I have eliminated a number of further errors; no doubt many remain.There are many new exercises, several new theorems and some corresponding rearrangements of material. The newresults are mostly additions with little effect on the structure of the work, but there is a short new section (266) onthe Brunn-Minkowski inequality.


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12 Taxonomy of measure spaces

*Chapter 21

Taxonomy of measure spaces

I begin this volume with a starred chapter. The point is that I do not really recommend this chapter for beginners.It deals with a variety of technical questions which are of great importance for the later development of the subject,but are likely to be both abstract and obscure for anyone who has not encountered the problems these techniques aredesigned to solve. On the other hand, if (as is customary) this work is omitted, and the ideas are introduced only whenurgently needed, the student is likely to finish with very vague ideas on which theorems can be expected to apply to

which types of measure space, and with no vocabulary in which to express those ideas. I therefore take a few pages tointroduce the terminology and concepts which can be used to distinguish good measure spaces from others, with afew of the basic relationships. The only paragraphs which are immediately relevant to the theory set out in Volume 1are those on complete, -finite and semi-finite measure spaces (211A, 211D, 211F, 211Lc,212, 213A-213B, 215B),and on Lebesgue measure (211M). For the rest, I think that a newcomer to the subject can very well pass over thischapter for the time being, and return to it for particular items when the text of later chapters refers to it. On theother hand, it can also be used as an introduction to the flavour of the purest kind of measure theory, the study ofmeasure spaces for their own sake, with a systematic discussion of a few of the elementary constructions.

211 Definitions

I start with a list of definitions, corresponding to the concepts which I have found to be of value in distinguishing

different types of measure space. Necessarily, the significance of many of these ideas is likely to be obscure until youhave encountered some of the obstacles which arise later on. Nevertheless, you will I hope be able to deal with thesedefinitions on a formal, abstract basis, and to follow the elementary arguments involved in establishing the relationshipsbetween them (211L).

In 216C-216E below I will give three substantial examples to demonstrate the rich variety of objects which thedefinition of measure space encompasses. In the present section, therefore, I content myself with very brief descriptionsof sufficient cases to show at least that each of the definitions here discriminates between different spaces (211M-211R).

211A DefinitionLet (X, , ) be a measure space. Then, or (X, , ), is (Caratheodory)complete if wheneverAE and E= 0 then A; that is, if every negligible subset ofX is measurable.

211B DefinitionLet (X, , ) be a measure space. Then (X, , ) is a probability spaceifX= 1. In this case is called a probabilityor probability measure.

211C DefinitionLet (X, , ) be a measure space. Then , or (X, , ), is totally finite ifX


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211L Definitions 13

211F Definition Let (X, , ) be a measure space. Then , or (X, , ), is semi-finite if whenever E andE=there is an F Esuch that F and 0< F


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14 Taxonomy of measure spaces 211L

Set

F =iIFi XiX

and H=X\ F.We see thatF Xi= Fi Xi for each iI (becauseXiiI is disjoint), so F and H. For any E E,

(E\ H) = (E F) = iI(E F Xi) = iI(E Fi Xi) = 0because every Fi belongs toF. Thus F F. If G and (E\G) = 0 for every E E, then X\ G andF = F

(X

\G) belong to

F. So(F

Xi)

i for each i

I. But also (F

Xi)

sup

nN (Fin

Xi) =i, so

(F Xi) = (F Xi) for each i. BecauseXi is finite, it follows that ((F\ F) Xi) = 0, for each i. Summingover i, (F\ F) = 0, that is, (H\ G) = 0.

Thus H is an essential supremum forE in . AsEis arbitrary, (X, , ) is localizable.(ii) IfE andE=, then there is some iIsuch that

0< (E Xi)Xi


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211Rb Definitions 15

(c) is strictly localizable. PPPSet Xx={x}for everyxX. ThenXxxX is a partition ofX, and for anyEX(E Xx) = 1 ifxE, 0 otherwise.

By the definition of,

E=

xX(E Xx)for everyEX, andXxxX is a decomposition ofX. QQQ

Consequently is localizable, locally determined and semi-finite.

(d) is purely atomic. PPP{x} is an atom for every xX, and ifE > 0 then surely Eincludes{x} for some x.QQQ Obviously, is not atomless.

211O A non-semi-finite space Set X ={0}, ={, X}, = 0 and X =. Then is not semi-finite, asX=but Xhas no subset of non-zero finite measure. It follows that cannot be localizable, locally determined,-finite, totally finite nor a probability measure. Because =PX, is complete. X is an atom for , so is purelyatomic (indeed, it is point-supported).

211P A non-complete space WriteB for the -algebra of Borel subsets ofR (111G), and for the restrictionof Lebesgue measure toB (recall that by 114G every Borel subset ofR is Lebesgue measurable). Then (R, B, ) isatomless,-finite and not complete.

proof (a)To see that is not complete, recall that there is a continuous, strictly increasing bijection g : [0, 1]

[0, 1]such that g[C] > 0, where C is the Cantor set, so that there is a set A g[C] which is not Lebesgue measurable(134Ib). Now g1[A]Ccannot be a Borel set, since A = (g1[A])g1 is not Lebesgue measurable, therefore notBorel measurable, and the composition of two Borel measurable functions is Borel measurable (121Eg); so g1[A] is anon-measurable subset of the negligible set C.

(b) The rest of the arguments of 211M apply to just as well as to true Lebesgue measure, so is -finite andatomless.

*RemarkThe argument offered in (a) could give rise to a seriously false impression. The set A referred to there canbe constructed only with the use of a strong form of the axiom of choice. No such device is necessary for the resulthere. There are many methods of constructing non-Borel subsets of the Cantor set, all illuminating in different ways.In the absence of any form of the axiom of choice, there are difficulties with the concept of Borel set, and others withthe concept of Lebesgue measure, which I will come to in Chapter 56; but countable choice is quite sufficient for the

existence of a non-Borel subset ofR. For details of a possible approach see 423L in Volume 4.

211Q Some probability spacesTwo obvious constructions of probability spaces, restricting myself to the methodsdescribed in Volume 1, are

(a) the subspace measure induced by Lebesgue measure on [0, 1] (131B);(b) the point-supported measure induced on a setXby a functionh : X[0, 1] such that xXh(x) = 1, writing

E=

xEh(x) for every EX; for instance, ifXis a singleton{x} and h(x) = 1, or ifX= N and h(n) = 2n1.Of these two, (a) gives an atomless probability measure and (b) gives a purely atomic probability measure.

211R Countable-cocountable measure The following is one of the basic constructions to keep in mind whenconsidering abstract measure spaces.

(a)Let Xbe any set. Let be the family of those sets E

Xsuch that either EorX

\Eis countable. Then is

a-algebra of subsets ofX. PPP(i)is countable, so belongs to . (ii) The condition forEto belong to is symmetricbetween EandX\ E, soX\ E for everyE. (iii) LetEnnN be any sequence in , and set E=

nN En. If

every En is countable, then Eis countable, so belongs to . Otherwise, there is some n such thatX\ En is countable,in which case X\ E X\ En is countable, so again E . QQQ is called the countable-cocountable -algebraofX.

(b)Now consider the function : {0, 1} defined by writing E= 0 ifEis countable, E= 1 ifE and Eis not countable. Then is a measure. PPP (i) is countable so = 0. (ii) LetEnnN be a disjoint sequence in ,and E its union. () If every Em is countable, then so is E, so

E= 0 =

n=0 En.


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16 Taxonomy of measure spaces 211Rb

() If some Em is uncountable, then E Em also is uncountable, and E = Em = 1. But in this case, becauseEm, X\ Em is countable, so En, being a subset ofX\ Em, is countable for every n=m; thus En= 0 for everyn=m, and

E= 1 =

n=0 En.

AsEnnN is arbitrary, is a measure. QQQ This is the countable-cocountable measureon X.(c) If X is any uncountable set and is the countable-cocountable measure on X, then is a complete, purely

atomic probability measure, but is not point-supported. PPP (i) IfA

E and E= 0, then E is countable, so A also

is countable and belongs to . Thus is complete. (ii) BecauseX is uncountable, X = 1 and is a probabilitymeasure. (iii) IfE >0, then F =E= 1 whenever F is a non-negligible measurable subset ofE, so Eis itself anatom; thus is purely atomic. (iv)X= 1> 0 =

xX{x}, so is not point-supported. QQQ

211X Basic exercises >>>(a) Let (X, , ) be a measure space. Show that is -finite iff there is a totally finitemeasureonXwith the same measurable sets and the same negligible sets as .

>>>(b)Let g : R R be a non-decreasing function and g the associated Lebesgue-Stieltjes measure (114Xa). Showthatg is complete and -finite. Show that

(i)g is totally finite iffg is bounded;(ii) g is a probability measure iff limx g(x) limx g(x) = 1;(iii)g is atomless iffg is continuous;

(iv) ifEis any atom for g, there is a point xEsuch that gE= g{x};(v)g is purely atomic iff it is point-supported.>>>(c)Let be counting measure on a set X. Show that is always complete, strictly localizable and purely atomic,

and that it is -finite iffX is countable, totally finite iffX is finite, a probability measure iffX is a singleton, andatomless iffX is empty.

(d)Show that a point-supported measure is always complete, and is strictly localizable iff it is semi-finite.

(e)Let Xbe a set. Show that for any -idealIof subsets ofX(definition: 112Db), the set =I {X\ A: A I}

is a -algebra of subsets ofX, and that there is a measure : {0, 1}given by settingE= 0 ifE

I, E= 1 ifE

\ I.

Show thatIis precisely the null ideal of, that is complete, totally finite and purely atomic, and is a probabilitymeasure iffX / I.

(f) Show that a point-supported measure is strictly localizable iff it is semi-finite.

(g)Let (X, , ) be a measure space such that X >0. Show that the set of conegligible subsets ofXis a filter onX.

211Y Further exercises (a) Let (X, , ) be a measure space, and for E, F write E F if(EF) = 0.Show that is an equivalence relation on . Let A be the set of equivalence classes in for; for E , writeE A for its equivalence class. Show that there is a partial ordering on A defined by saying that, for E, F ,

E

F

(E

\F) = 0.

Show that is localizable iff for every AAthere is an hAsuch that (i)ah for everyaA (ii) whenever g Ais such thatag for every a A, then h g.

(b)Let (X, , ) be a measure space, and construct Aas in 211Ya. Show that there are operations , , \ on Adefined by saying that

E F= (E F), E F = (E F), E\ F = (E\ F)for all E, F . Show that ifAA is any countableset, then there is an hA such that (i) ah for every aA(ii) whenever g A is such that ag for every a A, then h g. Show that there is a functional : A [0, ]defined by saying that (E) = E for everyE. ((A,) is called the measure algebra of (X, , ).)


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212B Complete spaces 17

(c)Let (X, , ) be a semi-finite measure space. Show that it is atomless iff whenever >0, E and E 0 there is adecomposition ofXconsisting of sets of measure at most .

(e)Let be the countable-cocountable -algebra ofR. Show that [0, [ /. Let be the restriction of countingmeasure to . Show that (R, , ) is complete, semi-finite and purely atomic, but not localizable nor locally determined.

211 Notes and comments The list of definitions in 211A-211K probably strikes you as quite long enough, even thoughI have omitted many occasionally useful ideas. The concepts here vary widely in importance, and the importance ofeach varies widely with context. My own view is that it is absolutely necessary, when studying any measure space, toknow its classification under the eleven discriminating features listed here, and to be able to describe any atoms whichare present. Fortunately, for most ordinary measure spaces, the classification is fairly quick, because if (for instance)the space is -finite, and you know the measure of the whole space, the only remaining questions concern completenessand atoms. The distinctions between spaces which are, or are not, strictly localizable, semi-finite, localizable andlocally determined are relevant only for spaces which are not-finite, and do not arise in elementary applications.

I think it is also fair to say that the notions of complete and locally determined measure space are technical; I mean,that they do not correspond to significant features of the essential structure of a space, though there are some interestingproblems concerning incomplete measures. One manifestation of this is the existence of canonical constructions forrendering spaces complete or complete and locally determined (212C, 213D-213E). In addition, measure spaces which

are not semi-finite do not really belong to measure theory, but rather to the more general study of -algebras and-ideals. The most important classifications, in terms of the behaviour of a measure space, seem to me to be -finite,localizable and strictly localizable; these are the critical features which cannot be forced by elementary constructions.

If you know anything about Borel subsets of the real line, the argument of part (a) of the proof of 211P must lookvery clumsy. But better proofs rely on ideas which we shall not need until Volume 4, and the proof here is based ona construction which we have to understand for other reasons.

212 Complete spaces

In the next two sections of this chapter I give brief accounts of the theory of measure spaces possessing certain ofthe properties described in211. I begin with completeness. I give the elementary facts about complete measurespaces in 212A-212B; then I turn to the notion of completion of a measure (212C) and its relationships with the otherconcepts of measure theory introduced so far (212D-212G).

212A Proposition Any measure space constructed by Caratheodorys method is complete.

proof Recall that Caratheodorys method starts from an arbitrary outer measure :PX[0, ] and sets ={E: EX, A= (A E) + (A \ E) for every AX}, =

(113C). In this case, ifBE and E= 0, then B = E= 0 (113A(ii)), so for any AXwe have(A B) + (A \ B) = (A \ B)A(A B) + (A \ B),

and B.

212B Proposition (a) If (X, , ) is a complete measure space, then any conegligible subset ofX is measurable.(b) Let (X, , ) be a complete measure space, and f a [

,

]-valued function defined on a subset ofX. Iff is

virtually measurable (that is, there is a conegligible set EXsuch that fE is measurable), then f is measurable.(c) Let (X, , ) be a complete measure space, and fa real-valued function defined on a conegligible subset ofX.

Then the following are equiveridical, that is, if one is true so are the others:(i)fis integrable;(ii) f is measurable and|f| is integrable;(iii)f is measurable and there is an integrable function g such that|f| a.e.g .

proof (a) IfE is conegligible, then X\ E is negligible, therefore measurable, and E is measurable.(b)Let a R. Then there is an H such that

{x: (fE)(x)a}= H dom(fE) = H E dom f.


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212F Complete spaces 19

(c)If = then of course must be complete. If is complete, and E , then there are E, E such thatE E E and (E\ E) = 0. But now E\ E E\ E, so (because (X, , ) is complete) E\ E andE= E (E\ E). As Eis arbitrary, and = and = .

212E The importance of this construction is such that it is worth spelling out some further elementary properties.

Proposition Let (X, , ) be a measure space, and (X,,) its completion.(a) The outer measures , defined from and coincide.(b), give rise to the same negligible and conegligible sets and the same sets of full outer measure.

(c) is the only measure with domain which agrees with on .

(d) A subset ofXbelongs to iff it is expressible as FA whereF and A is -negligible.proof (a) Take any AX. (i) IfAF , then F and F = F, so

A F =F;asF is arbitrary, AA. (ii) IfAE , there is an E such that EE andE= E, so

AE= E;asEis arbitrary, AA.

(b)Now, for AX,A is -negligible

A= 0

A= 0

A is -negligible,

A is -conegligible (X\ A) = 0 (X\ A) = 0 A is -conegligible.

IfA has full outer measure for , F and F A =, then there is an F such that F F and F = F;as F A=, F is -negligible and F is -negligible; as Fis arbitrary, A has full outer measure for . In the otherdirection, of course, ifA has full outer measure for then

(F A) = (F A) = F =Ffor everyF , so A has full outer measure for .

(c)If is any measure with domain extending , we must have

E

E

E, E= E= E,

so E= E, whenever E, E, EEE and(E\ E) = 0.(d)(i)IfE , takeE, E such that EEE and(E\ E) = 0. ThenE\ EE\ E, soE\ E is

-negligible, and E= E(E\ E) is the symmetric difference of a member of and a negligible set.(ii) If E = FA, where F and A is -negligible, take G such that G = 0 and A G; then

F\ GEF G and ((F G) \ (F\ G)) = G= 0, so E .

212F Now let us consider integration with respect to the completion of a measure.

Proposition Let (X, , ) be a measure space and (X, ,) its completion.

(a) A [, ]-valued functionfdefined on a subset ofX is -measurable iff it is -virtually measurable.(b) Let f be a [

,

]-valued function defined on a subset of X. Then fd = f d if either is defined in[, ]; in particular, f is-integrable iff it is -integrable.

proof (a)(i) Suppose that f i s a [, ]-valued -measurable function. For q Q let Eq be such that{x : f(x) q} = dom f Eq, and choose Eq, Eq such that Eq Eq Eq and (Eq\Eq) = 0. SetH=X\qQ(Eq\ Eq); then H is-conegligible. ForaR set

Ga=qQ,q


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20 Taxonomy of measure spaces 212F

(ii)Iff is-virtually measurable, then there is a -conegligible setHXsuch thatfH is -measurable. Since,fHis also -measurable. AndHis -conegligible, by 212Eb. But this means that f is -virtually measurable,therefore-measurable, by 212Bb.

(b)(i) Let f : D [, ] be a function, where D X. If either off d, f d is defined in [, ], thenf is virtually measurable, and defined almost everywhere, for one of the appropriate measures, and therefore for both(putting (a) above together with 212Bb).

(ii) Now suppose that f is non-negative and integrable either with respect to or with respect to . Let Ebe a conegligible set included in dom f such thatfE is -measurable. For nN, k1 set

Enk ={x: xE, f(x)2nk};then eachEnk belongs to and is of finite measure for both and . (Iff is -integrable,

Enk= Enk2n

f d;

iff is -integrable,

Enk = Enk2n

f d.)

So

fn=4n

k=12nEnk

is both -simple and -simple, and

fnd=

fnd. Observe that, for xE,

fn(x) = 2nk ifk


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212X Complete spaces 21

()If (X, ,) is semi-finite and E=, then E=, so there is an F Esuch that 0< F 0. IfF and F H, letFFbe such that F and (F\ F) = 0. ThenE FE and (F(E F)) = 0, soeither F =(E F) = 0 or (H\ F) = (E\ F) = 0. AsFis arbitrary, His an atom for .

(iii) It follows at once that (X,,) is atomless iff (X, , ) is.

(iv)() On the other hand, if (X, , ) is purely atomic and H > 0, there is an E such that E H andE > 0, and an atom F for such that F E; but F is also an atom for . As H is arbitrary, (X, ,) is purelyatomic.

() And if (X, ,) is purely atomic and E > 0, then there is an H Ewhich is an atom for ; now letF be such that F H and (H\ F) = 0, so that F is an atom for and F E. As E is arbitary, (X, , ) ispurely atomic.

212X Basic exercises >>>(a) Let (X, , ) be a complete measure space. Suppose that A E and thatA + (E\ A) = E


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22 Taxonomy of measure spaces 212Xb

>>>(b)Let and be two measures on a setX, with completions and . Show that the following are equiveridical:(i) the outer measures , defined from and coincide; (ii) E= Ewhenever either is defined and finite; (iii)

fd=

f dwheneverfis a real-valued function such that either integral is defined and finite. (Hint: for (i)(ii),if E 0, Y a set, f : X Y a function and f1 theimage measure onY. Show that ifFis the filter of-conegligible subsets ofX, then the image filter f[[F]] (2A1Ib) isthe filter off1-conegligible subsets ofY .

212Y Further exercises (a)Let Xbe a set and an inner measure on X, that is, a functional fromPXto [0, ]such that

= 0,

(A B)A + B ifA B =,(nN An) = limn Anwhenever AnnNis a non-increasing sequence of subsets ofXand A0


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213B Semi-finite, locally determined and localizable spaces 23

213 Semi-finite, locally determined and localizable spaces

In this section I collect a variety of useful facts concerning these types of measure space. I start with the characteristicproperties of semi-finite spaces (213A-213B), and continue with the complete locally determined spaces (213C) and theconcept of c.l.d. version (213D-213H), the most powerful of the universally available methods of modifying a measurespace into a better-behaved one. I briefly discuss locally determined negligible sets (213I-213L), and measurableenvelopes (213L-213M), and end with results on localizable spaces (213N) and strictly localizable spaces (213O).

213A Lemma Let (X, , ) be a semi-finite measure space. Then

E= sup{F :F , F E, F


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24 Taxonomy of measure spaces 213B

f= sup

g is simple,gf a.e.

g

(if either is finite, and therefore also if either is infinite)

= supg is simple,gf a.e.,F


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213G Semi-finite, locally determined and localizable spaces 25

there is an E f such that H E, so that H= (H E) = H). Now suppose thatH and that H=.There is surely an Ef such that (H E)> 0; but now 0 < (H E)


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26 Taxonomy of measure spaces 213Gf d= limn

fnd= limn

fnd=

f d.

(iii) In general, if

f dis defined in R, we havef d=

f+d

fd=

f+d

fd=

f d,

writingf+ =f 0, f= (f) 0.(c)(i)Let fbe a -simple function. Express it as

ni=0 aiHi where Hi


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213H Semi-finite, locally determined and localizable spaces 27

Working in (X, , ), let F be an essential supremum forE.(i) ??? Suppose, if possible, that there is an H H such that (H\ F)> 0. Then there is an E such that

EH\ F andE= (H\ F)> 0 (213Fc). This Ebelongs toEand(E\ F) = E >0; which is impossible ifF isan essential supremum ofE. XXX

(ii) Thus (H\ F) = 0 for every H H. Now take any G such that (H\ G) = 0 for every H H. LetE0 be such that E0 F\ G and E0 = (F\ G); note that F\ E0F G. IfE E, there is an H HsuchthatEH, so that

(E\ (F\ E0)) (H\ (F G))(H\ F) + (H\ G) = 0.BecauseF is an essential supremum forE in ,

0 = (F\ (F\ E0)) = E0 = (F\ G).This shows thatF is an essential supremum forHin . AsHis arbitrary, (X, ,) is localizable.

(iii) The argument of (i)-(ii) shows in fact that ifH thenHhas an essential supremum F in such that Factually belongs to . TakingH={H}, we see that ifH there is an F such that (HF) = 0.

(c)We already know that EEfor every E, with equality ifE


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28 Taxonomy of measure spaces 213I

213I Locally determined negligible sets The following simple idea is occasionally useful.

Definition A measure space (X, , ) has locally determined negligible sets if for every non-negligible A Xthere is an E such that E


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213O Semi-finite, locally determined and localizable spaces 29

proof Forq Q, f setEfq ={x: xdom f, f(x)q} .

For each q Q, letEq be an essential supremum of{Efq : f} in . Seth(x) = sup{q: qQ, xEq} [, ]

forxX, taking sup =if necessary.Iff, g and q Q, then

Efq\ (X\ (dom g \ Egq)) = Efq dom g \ Egq {x: xdom f dom g, f(x)=g(x)}

is negligible; as fis arbitrary,

Eq dom g \ Egq =Eq\ (X\ (dom g \ Egq))is negligible. AlsoEgq\ Eq is negligible, so Egq(Eq dom g) is negligible. SetHg =

qQ Egq(Eq dom g); then

Hg is negligible. But if x dom g\Hg, then, for every q Q, x Eq x Egq ; it follows that for such x,h(x) = g(x). Thus h= g almost everywhere in dom g; and this is true for every g.

The functionh is not necessarily real-valued. But it is measurable, because

{x: h(x)> a}= {Eq : q Q, q > a} for every real a. So if we modify it by setting

h(x) = h(x) ifh(x) R,= 0 ifh(x) {, },

we shall get a measurable real-valued function h : X R; and for any g , h(x) will be equal to g(x) at leastwhenever h(x) = g(x), which is true for almost everyxdom g. Thus h is a suitable function.

213O There is an interesting and useful criterion for a space to be strictly localizable which I introduce at thispoint, though it will be used rarely in this volume.

Proposition Let (X, , ) be a complete locally determined space.(a) Suppose that there is a disjoint familyE such that () E 0 then there is an E E such that (E F)> 0. Then (X, , ) is strictly localizable,

E is conegligible, andE {X\ E}is a decomposition ofX.

(b) Suppose thatXiiIis a partition ofXinto measurable sets of finite measure such that whenever E andE >0 there is an iIsuch that(E Xi)> 0. Then (X, , ) is strictly localizable, andXiiIis a decompositionofX.

proof (a)(i)The first thing to note is that ifF andF 0}.

BecauseEis disjoint, we must have#(En)2nF

for every n N, so that everyEn is finite andE, being the union of a sequence of countable sets, is countable. SetE =

E and F = F\ E, so that both E and F belong to . IfE E, then E E so (E F) = = 0; ifE E \E, then(E F) = (E F) = 0. Thus(E F) = 0 for everyE E. By the hypothesis () onE,F = 0,so (F\ E) = 0, as required. QQQ

(ii) Now suppose that H X is such that H E for every E E. In this caseH . PPP Let F besuch thatF


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30 Taxonomy of measure spaces 213O

(iii) We find also that H =

EE(HE) for every H . PPP () BecauseE is disjoint, we must haveEE(H E)H for every finiteE E, so

EE(H E) = sup{

EE(H E) :E Eis finite} H.() For the reverse inequality, consider first the case H


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213Yb Semi-finite, locally determined and localizable spaces 31

(g)Let (X, , ) be any measure space. Set

sfA= sup{(A E) : E, E (h) Let (X, , ) be a strictly localizable measure space with a decompositionXiiI. Show that A =iI

(A Xi) for everyAX.

>>>(i) Let (X, , ) be a complete locally determined measure space, and let A X be such that max((EA), (E\A)) < Ewhenever E and 0 < E >(j) Let (X, , ) be a measure space, its c.l.d. version, and the measure defined by Caratheodorys methodfrom . (i) Show that the following are equiveridical: () has locally determined negligible sets; () and havethe same negligible sets; () = . (ii) Show that in this case is semi-finite.

(k)Let (X, , ) be a measure space. Show that the following are equiveridical: (i) (X, , ) has locally determinednegligible sets; (ii) the completion and c.l.d. version of have the same sets of finite measure; (iii) and havethe same integrable functions; (iv) = ; (v) the outer measure sfof 213Xg is equal to

.

(l) Let us say that a measure space (X, , ) has the measurable envelope property if every subset ofXhas ameasurable envelope. (i) Show that a semi-finite space with the measurable envelope property has locally determinednegligible sets. (ii) Show that a complete semi-finite space with the measurable envelope property is locally determined.

(m) Let (X, , ) be a semi-finite measure space, and suppose that it satisfies the conclusion of Theorem 213N.Show that it is localizable. (Hint: givenE , setF ={F : F , E F is negligible for every E E}. Let bethe set of functions f from subsets ofX to{0, 1}such that f1[{1}] E andf1[{0}] F.)

(n)Let (X, , ) be a measure space. Show that its c.l.d. version is strictly localizable iff there is a disjoint familyE such thatE


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32 Taxonomy of measure spaces 213Yc

(c)Set X= N, and for AX setA= #(A)2 ifA is finite,ifA is infinite.

Show that satisfies the conditions of 113Yg/212Ya, but that the measure defined from by the method of 113Yg isnot semi-finite.

(d) Let (X, , ) be a complete locally determined measure space. Suppose thatDXand that f : D R is afunction. Show that the following are equiveridical: (i) f is measurable; (ii)

{x: xD E, f(x)a} + {x: xD E, f(x)b} Ewhenevera < b in R, E and E a {x: xD E, f(x)b},and use 213Xi above.)

(e) Let (X, , ) be a complete locally determined measure space and suppose thatE is such that E


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214C Subspaces 33

It is worth noting that while the measure obtained by Caratheodorys method directly from the outer measure

defined from may fail to be semi-finite, even when is (213Yb), a simple modification of (213Xg) yields the c.l.d.version of, which can also be obtained from an appropriate inner measure (213Xe). The measure is of courserelated in other ways to ; see 213Xd.

214 Subspaces

In131 I described a construction for subspace measures on measurable subsets. It is now time to give the gener-alization to subspace measures on arbitrary subsets of a measure space. The relationship between this constructionand the properties listed in211 is not quite as straightforward as one might imagine, and in this section I try to givea full account of what can be expected of subspaces in general. I think that for the present volume only (i) generalsubspaces of-finite spaces and (ii) measurable subspaces of general measure spaces will be needed in any essentialway, and these do not give any difficulty; but in later volumes we shall need the full theory.

I begin with a general construction for subspace measures (214A-214C), with an account of integration with respectto a subspace measure (214E-214G); these (with 131E-131H) give a solid foundation for the concept of integrationover a subset (214D). I present this work in its full natural generality, which will eventually be essential, but even forLebesgue measure alone it is important to be aware of the ideas here. I continue with answers to some obvious questionsconcerning subspace measures and the properties of measure spaces so far considered, both for general subspaces (214I)

and for measurable subspaces (214K), and I mention a basic construction for assembling measure spaces side-by-side,the direct sums of 214L-214M. At the end of the section I discuss a measure extension problem (214O-214P).

214A PropositionLet (X, , ) be a measure space, andYany subset ofX. Let be the outer measure definedfrom (132A-132B), and set Y ={E Y : E }; let Y be the restriction of to Y. Then (Y, Y, Y) is ameasure space.

proof (a) I have noted in 121A that Y is a -algebra of subsets ofY .

(b)Of courseYF [0, ] for every F Y.(c)Y= = 0.(d) IfFnnNis a disjoint sequence in Ywith unionF, then chooseEn,En,E such thatFn= YEn,FnEn

andYFn= En

for each n, F

EandYF =E (using 132Aa repeatedly). Set Gn= En

En

E\m


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34 Taxonomy of measure spaces 214C

(b)(i) IfA is Y-negligible, there is a set F Y such that A F and YF = 0; now A F = 0 so A is-negligible, by 132Ad. (ii) IfA is -negligible, there is an E such thatAEandE= 0; now AE Y Yand Y(E Y) = 0, so A is Y-negligible.

(c)IfAX is-conegligible, thenA Y isY-conegligible, becauseY\ A= Y (X\ A) is-negligible, thereforeY-negligible. IfAY is Y-conegligible, then A (X\ Y) is -conegligible because X\ (A (X\ Y)) = Y\ A isY-negligible, therefore -negligible.

(d)Let AY. (i) IfAE, thenAEY Y, soYAY(EY)E; asEis arbitrary, YAA.(ii) IfAF Y, there is an E such that F Eand YF =F =E; now AE so AE= YF. AsFis arbitrary, AYA.

(e)That Z= (Y)Zis immediate from the definition of Y, etc.; now

(Y)Z=YZ=

Z=Z

by (d).

(f) This is elementary, because E Y and (E Y) = (E Y) for every E.

214D Integration over subsets: Definition Let (X, , ) be a measure space, Y a subset of X and f a[, ]-valued function defined on a subset ofX. By Yf(or Yf(x)(dx), etc.) I mean (fY)dY, if this exists in[, ], following the definitions of 214A-214B, 133A and 135F, and taking dom(fY) = Ydom f, (fY)(x) = f(x)forx

Y

dom f. (Compare 131D.)

214E PropositionLet (X, , ) be a measure space, YX, andfa [, ]-valued function defined on a subsetdom f ofX.

(a) Iff is-integrable then fY isY-integrable, andYf

f iffis non-negative.

(b) If dom f Y and f is Y -integrable, then there is a -integrable function f on X, extending f, such thatF

f=FYf for everyF .

proof (a)(i)Iff is-simple, it is expressible asn

i=0 aiEi, whereE0, . . . , E n, a0, . . . , an Rand Ei


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214F Subspaces 35

F

af+ bg= a

F

f+ b

F

g

=a

FY

f+ b

FY

g=

FY

af+ bg

for everyF , so af+ bg is an enveloping extension ofaf+ bg.(iii) Putting (i) and (ii) together, we see that every Y-simple function fhas an enveloping extension.

(iv)Now suppose thatfnnN is a non-decreasing sequence of non-negative Y-simple functions converging Y-almost everywhere to a Y-integrable function f. For each n N let fn be an enveloping extension of fn. Thenfna.e. fn+1. PPPIfF then

Ffn=

FYfn

FYfn+1 =

F

fn+1.

So fna.e. fn+1, by 131Ha. QQQ Alsolimn

F

fn= limnFYfn=

FYf

for every F . Taking F = X to begin with, B.Levis theorem tells us that h = limn fn is defined (as areal-valued function) -almost everywhere; now letting F vary, we have

Fh =

FYf for every F , because

hF= limn fnF F-a.e. (I seem to be using 214Cb here.) Now hY =f Y-a.e., by 214Cb again. If we define fby setting

f(x) = f(x) for xdom f, h(x) for xdom h \ dom f, 0 for other xX,then f is defined everywhere in X and is equal to h -almost everywhere; so that if F , fF will be equal tohF F-almost everywhere, and

Ff=

F

h =FYf.

As Fis arbitrary, f is an enveloping extension off.

(v) Thus every non-negative Y-integrable function has an enveloping extension. Using (ii) again, every Y-integrable function has an enveloping extension, as claimed.

214F PropositionLet (X, , ) be a measure space, Ya subset ofX, andf a [, ]-valued function such thatX

f is defined in [, ]. IfeitherY is of full outer measure in X orfis zero almost everywhere in X\ Y, then

Yfis defined and equal to X

f.

proof (a) Suppose first that f is non-negative, -measurable and defined everywhere in X. In this casefY is Y-measurable. SetFnk ={x: xX, f(x)2nk} fork, nN, fn =

4nk=12

nFnk for nN, so thatfnnN is anon-decreasing sequence of real-valued measurable functions converging everywhere to f, and

X

f= limnX

fn.For each nN and k1,

Y(Fnk Y) = (Fnk Y) = Fnkeither because Fnk\ Y is negligible or because Xis a measurable envelope ofY . So

Y

f = limn

Y

fn= limn

4nk=1

2nY(Fnk Y)

= limn

4n

k=1 2nFnk = limnX fn= X f.(b) Now suppose that f is non-negative, defined almost everywhere in X and -virtually measurable. In this case

there is a conegligible measurable set E dom f such that fE is measurable. Set f(x) = f(x) for x E, 0 forxX\ E; then fsatisfies the conditions of (a) and f= f -a.e. Accordingly fY = fY Y-a.e. (214Cc), and

Yf=

Y

f=X

f=X

f.

(c)Finally, for the general case, we can apply (b) to the positive and negative parts f+, f offto getY

f=Y

f+ Y

f=X

f+ X

f=X

f.


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36 Taxonomy of measure spaces 214G

214G Corollary Let (X, , ) be a measure space, Ya subset ofX, andE a measurable envelope ofY. Iffis a [, ]-valued function such that Ef is defined in [, ], then Yfis defined and equal to Ef.proofBy 214Ce, we can identify the subspace measure Ywith the subspace measure (E)Yinduced by the subspacemeasure on E. Now, regarded as a subspace ofE, Yis of full outer measure, so 214F gives the result.

214H Subspaces and Caratheodorys method The following easy technical results will occasionally be useful.

LemmaLet Xbe a set, Y

Xa subset, and an outer measure on X.

(a)Y =PYis an outer measure on Y .(b) Let, be the measures on X, Y defined by Caratheodorys method from the outer measures , Y, and , T

their domains; let Ybe the subspace measure on Y induced by, and Y its domain. Then(i) Y T and F YF for every F Y;(ii) ifY then = Y;(iii) if = (that is, is regular) then extends Y;(iv) if = andY


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214I Subspaces 37

(d) If (X, , ) is complete and locally determined, then (Y, Y, Y) is complete and semi-finite.(e) If (X, , ) is complete, locally determined and localizable then so is ( Y, Y, Y).

proof (a)(i)Suppose that (X, , ) is complete. IfAUY andYU= 0, there is anE such that U=EYandE= YU= 0; now AEso A and A = A Y Y.

(ii) YY =Y X, so Yis totally finite if is.

(iii) IfXnnN is a sequence of sets of finite measure for which covers X, thenXn YnN is a sequence ofsets of finite measure for Ywhich covers Y. So (Y, Y, Y) is-finite if (X, , ) is.

(iv) Suppose thatXiiIis a decomposition ofX for. ThenXi YiIis a decomposition ofY for Y. PPPBecauseY(Xi Y)Xi 0. Then there is an E such that E 0. PPP???Otherwise,E U is-negligible wheneverE


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38 Taxonomy of measure spaces 214I

there is a U U such that E = (EU); now (EU\W) = 0, so E = (EUW) (EH) andE\ H is negligible. As E is arbitrary, His an essential upper bound forEandG \ His negligible; but this means thatG Y\ W is negligible. As Wis arbitrary, G Y is an essential supremum forU. QQQ

AsUis arbitrary, Y is localizable.

214J Upper and lower integrals The following elementary facts are sometimes useful.

Proposition Let (X, , ) be a measure space, A a subset ofXand fa real-valued function defined almost everywherein X. Then

(a) ifeitherf is non-negativeorA has full outer measure in X, (fA)dA f d;(b) ifA has full outer measure in X,

f d (fA)dA.

proof (a)(i) Suppose that f is non-negative. If

f d =, the result is trivial. Otherwise, there is a -integrablefunctiong such that f g -a.e. and f d= g d, by 133Ja. Now fAgA A-a.e., by 214Cb, and (gA) dAis defined and less than or equal to

g d, by 214Ea; so

(fA)dA

(gA)dA

g d=

f d.

(ii) Now suppose that A has full outer measure in X. If g is such that f g -a.e. and g d is defined in[, ], then fA gA A-a.e. and

(gA)dA =

g d, by 214F. So

(fA)dA

g d. As g is arbitrary,

(fA)dA

f d.

(b)Apply (a) tof, and use 133J(b-iv).

214K Measurable subspaces: PropositionLet (X, , ) be a measure space.(a) Let E and let Ebe the subspace measure, with E its domain. If (X, , ) is complete, or totally finite,

or-finite, or strictly localizable, or semi-finite, or localizable, or locally determined, or atomless, or purely atomic, sois (E, E, E).

(b) Suppose thatXiiIis a partition ofX into measurable sets (not necessarily of finite measure) such that ={E: EX, E Xi for every iI},

E=

iI(E Xi) for every E.Then (X, , ) is complete, or strictly localizable, or semi-finite, or localizable, or locally determined, or atomless, orpurely atomic, iff (Xi, Xi , Xi) has that property for every i

I.

proof I really think that if you have read attentively up to this point, you ought to find this easy. If you are in anydoubt, this makes a very suitable set of sixteen exercises to do.

214L Direct sums Let(Xi, i, i)iI be any indexed family of measure spaces. SetX =iI(Xi {i}); for

EX, iI setEi={x: (x, i)E}. Write ={E: EX, Eii for everyiI},

E=

iIiEi for every E.Then it is easy to check that (X, , ) is a measure space; I will call it the direct sum of the family(Xi, i, i)iI.Note that if (X, , ) is any strictly localizable measure space, with decompositionXiiI, then we have a naturalisomorphism between (X, , ) and the direct sum (X, , ) =

iI(Xi, Xi , Xi) of the subspace measures, if we

match (x, i)X with xX for every iIandxXi.For some of the elementary properties (to put it plainly, I know of no properties which are not elementary) of directsums, see 214M and 214Xh-214Xk.

214M Proposition Let(Xi, i, i)iIbe a family of measure spaces, with direct sum ( X, , ). Let f be areal-valued function defined on a subset ofX. For each iI, set fi(x) = f(x, i) whenever (x, i)dom f.

(a)f is measurable ifffi is measurable for every iI.(b) Iff is non-negative, then

fd=

iI

fidi if either is defined in [0, ].

proof (a) ForaR, set Fa ={(x, i) : (x, i)dom f, f(x, i)a}. (i) Iff is measurable, iI andaR, then thereis an E such that Fa= E dom f; now


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*214O Subspaces 39

{x: fi(x)a}= dom fi {x: (x, i)E}belongs to the subspace -algebra on dom fi induced by i. As a is arbitrary, fi is measurable. (ii) If every fi ismeasurable and a R, then for each i I there is an Ei i such that{x : (x, i) Fa} = Eidom f; settingE ={(x, i) : i I, x Ei}, Fa = dom fE belongs to the subspace -algebra on dom f. As a is arbitrary, f ismeasurable.

(b)(i)Suppose first that f is measurable and defined everywhere. Set Fnk={(x, i) : (x, i)X, f(x, i)2nk}fork, nN, gn =

4n

k=12nFnk for nN, Fnki ={x: (x, i)Fnk} for k , nN and iI, gni(x) =gn(x, i) for iI,

xXi. Then f d = lim

n

gnd= sup

nN

4nk=1

2nFnk

= supnN

4nk=1

iI

2nFnki=iI

supnN

4nk=1

2nFnki

=iI

supnN

gnidi=

iI

fidi.

(ii) Generally, if

fd is defined, there are a measurable g : X [0, [ and a conegligible measurable set E

dom fsuch thatg = fonE. NowEi =

{x: (x, i)

Xi

}belongs to ifor eachi, and iIi(Xi \Ei) = (X\E) = 0,so Ei isi-conegligible for every i. Settinggi(x) = g(x, i) for xXi, (i) tells us that

iI

fidi=

iI

gidi=

g d=

f d.

(iii) On the other hand, if

fidi is defined for every iI, then for eachiIwe can find a measurable functiongi : Xi[0, [ and a i-conegligible measurable set Eidom fi such that gi = fi on Ei. Settingg (x, i) =gi(x) foriI, xXi, (a) tells us that g is measurable, while g = f on{(x, i) : iI , xEi}, which is conegligible (by thecalculation in (ii) just above); so

f d=

g d=

iI

gidi =

iI

fidi,

again using (i) for the middle step.

214N Corollary Let (X, , ) be a measure space with a decomposition

Xi

iI. Iff is a real-valued function

defined on a subset ofX, then(a)f is measurable ifffXi is measurable for every iI,(b) iff0, then f= iIXi fif either is defined in [0, ].

proof Apply 214M to the direct sum of(Xi, Xi , Xi)iI, identified with (X, , ) as in 214L.

*214O I make space here for a general theorem which puts rather heavy demands on the reader. So I ought to saythat I advise skipping it on first reading. It will not be quoted in this volume, in the full form here I do not expectto use it anywhere in this treatise, only the special case of 214Xm is at all often applied, and the proof depends on aconcept (ideal of sets) and a technique (transfinite induction, part (d) of the proof of 214P) which are used nowhereelse in this volume. However, extension of measures is one of the central themes of Volume 4, and this result mayhelp to make sense of some of the patterns which will appear there.

LemmaLet (X, , ) be a measure space, andI

an ideal of subsets ofX, that is, a family of subsets ofX such that I, IJ I for all I, J I, and I I whenever I J I. Then there is a measure on X such that I dom , E= E+ supII(E I) for everyE, and I= 0 for every I I.proof (a)Let be the set of those F Xsuch that there are E and a countableJ Isuch thatEF J.Then is a -algebra of subsets ofXincluding I. PPP because EE for everyE.I becauseI {I} for every I I. In particular, . IfF , let E andJ Ibe such thatJis countable andFEJ; then (X\F)(X\E)J so X\F . IfFnnN is a sequence in with union F, then foreach nN choose En,Jn I such thatJn is countable and EnFn

Jn; then E= nN En belongs to ,J = nN Jn is a countable subset ofIandEF J, so F . Thus is a -algebra.QQQ

(b)ForF set


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40 Taxonomy of measure spaces *214O

F= sup{E: E, EF, (E I) = 0 for every I I}.Then is a measure. PPP The only subset of is, so = 0. LetFnnN be a disjoint sequence in with union F;set u =

n=0 Fn. (i) IfE , E F and (E I) = 0 for every I I, then for each n set En = E Fn. As

(En I) = 0 for every I I, EnFn for eachn. NowEnnN is disjoint and has union E, soE=

n=0 En

n=0 Fn= u.

As E is arbitrary, F u. (ii) Take any < u. Forn N, set n = Fn 2n1 min(1, u ) ifFn is finite, otherwise. For each n, we can find an En

such that En

Fn,

(En

I) = 0 for every I

I, and En

n. Set

E= nN En; then E F and E I= nN En I is -negligible for every I I, so F E= n=0 En.As is arbitrary, F u. (iii) AsFnnN is arbitrary, is a measure. QQQ

(c) Now take any E and set u = supII(EI). If u = then we certainly have E = = E+ u.Otherwise, letInnN be a sequence inI such that limn (EIn) = u; replacing In by

mn Im for each n

if necessary, we may suppose thatInnN is non-decreasing. SetA = EnN In; because E In has finite outer

measure for each n, A can be covered by a sequence of sets of finite measure, and has a measurable envelope H forincluded inE (132Ee). Observe that

H= A= supnN (E In) = u

by 132Ae.SetG = E\ H. Then(G I) = 0 for every I I. PPPFor any nNthere is an F such that F E (In I)

and F u; in which case

(G I) + (E In)(F\ H) + (F H)u.As n is arbitrary, (G I) = 0. QQQ Accordingly

u + EH+ G= E.On the other hand, ifF is such that F Eand(F I) = 0 for every I I, then

(E In)(E\ F) + (F In) = (E\ F)for everyn, so

u + F (E\ F) + F =E;as Fis arbitrary, u + EE.

(d) IfJ I, F , F J and (F I) = 0 for every I I, then F J= F is -negligible; as F is arbitrary,J= 0. Thus has all the required properties.

*214P Theorem Let (X, , ) be a measure space, andA a family of subsets ofXwhich is well-ordered by therelation. Then there is an extension of to a measure on Xsuch that(E A) is defined and equal to(E A)wheneverE and A A.proof (a) Addingand X toAif necessary, we may suppose thatAhasas its least member and Xas its greatestmember. By 2A1Dg,A is isomorphic, as ordered set, to some ordinal; sinceAhas a greatest member, this ordinal is asuccessor, expressible as + 1; let A :+ 1 Abe the order-isomorphism, so thatA is a non-decreasingfamily of subsets ofX, A0 = and A=X.

(b) For each ordinal , write for the subspace measure on A, for its domain andI for


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214Xg Subspaces 41

E Afor each, so E. If, then for each either < and

A A = A Iorand A A = A belongs to . SoA for every.

(d) Finally, (EA) = (EA) whenever E and . PPP??? Otherwise, because the ordinal + 1 iswell-ordered, there is a least such that (EA)=(E A). AsA0 =we surely have (E A0) = (E A0)and >0. Note that if > , then(E

A) = 0; so

(E A) = (E A A) = (E A).Now

(E A) = (E A) + sup(d)Let (X, , ) be a measure space, and Ya subset ofX such that the subspace measure Y is semi-finite. (i)

Show that a set F Y is an atom for Y iff it is of the form E Y where E an atom for . (ii) Show that if isatomless or purely atomic, so is Y.

(e)Let (X, , ) be a localizable measure space, and Yany subset ofX. Show that the c.l.d. version of the subspacemeasure on Y is localizable.

(f)Let (X, , ) be a measure space with locally determined negligible sets, and Ya subset ofX, with its subspacemeasureY. Show that Y has locally determined negligible sets.

>>>(g) Let (X, , ) be a measure space. Show that (X, , ) has locally determined negligible sets iff the subspacemeasureY is semi-finite for every Y X.


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42 Taxonomy of measure spaces 214Xh

>>>(h)Let(Xi, i, i)iIbe a family of measure spaces, with direct sum (X, , ) (214L). Set Xi =Xi {i} Xfor each iI. Show thatXi, with the subspace measure, is isomorphic to (Xi, i, i). Under what circumstances isXiiIa decomposition ofX? Show that is complete, or strictly localizable, or localizable, or locally determined,or semi-finite, or atomless, or purely atomic iff every i is. Show that a measure space is strictly localizable iff it isisomorphic to a direct sum of totally finite spaces.

>>>(i) Let(Xi, i, i)iIbe a family of measure spaces, and (X, , ) their direct sum. Show that the completionof (X, , ) can be identified with the direct sum of the completions of the (Xi, i, i), and that the c.l.d. version of

(X, , ) can be identified with the direct sum of the c.l.d. versions of the (Xi, i, i).(j)Let(Xi, i, i)iIbe a family of measure spaces. Show that their direct sum has locally determined negligible

sets iff every i has.

(k)Let(Xi, i, i)iIbe a family of measure spaces, and (X, , ) their direct sum. Show that (X, , ) has themeasurable envelope property (213Xl) iff every (Xi, i, i) has.

(l) Let (X, , ) be a measure space, Y a subset ofX, and f : X [0, ] a function such that Yf is defined in[0, ]. Show that Yf= f Y d.

>>>(m)Write out a direct proof of 214P in the special case in which A={A}. (Hint: for E, F ,((E A) (F\ A)) = (E A) + sup{G: G, GF\ A}.)

>>>(n) Let (X, , ) be a measure space andA a finite family of subsets ofX. Show that there is a measure on X,extending, which measures every member ofA.

214Y Further exercises (a)Let (X, , ) be a measure space andA a subset ofXsuch that the subspace measureon A is semi-finite. Set = sup{E : E , E A}. Show that if A then there is a measure on X,extending, such that A = .

(b) Let (X, , ) be a measure space andAnnZ a double-ended sequence of subsets ofX such that Am Anwhenevermn in Z. Show that there is a measure on X, extending , which measures every An. (Hint: use 214Ptwice.)

(c)Let Xbe a set andA

a family of subsets ofX. Show that the following are equiveridical: (i) for every measureon Xthere is a measure onXextending and measuring every member ofA; (ii) for every totally finite measure on Xthere is a measure on Xextending and measuring every member ofA. (Hint: 213Xa.)

214 Notes and commentsI take the first part of the section, down to 214H, slowly and carefully, because while noneof the arguments are deep (214Eb is the longest) the patterns formed by the results are not always easy to predict.There is a counter-example to a tempting extension of 214H/214Xb in 216Xb.

The message of the second part of the section (214I-214L) is that subspaces inherit many, but not all, of the propertiesof a measure space; and in particular there is a difficulty with semi-finiteness, unless we have locally determinednegligible sets (214Xg). (I give an example in 216Xa.) Of course 213Hb shows that if we start with a localizable space,we can convert it into a complete locally determined localizable space without doing great violence to the structure ofthe space, so the difficulty is ordinarily superable.

By far the most important case of 214P is when

A=

{A

}is a singleton, so that the argument simplifies dramatically

(214Xm). In439 of Volume 4 I will return to the problem of extending a measure to a given larger -algebra in theabsence of any helpful auxiliary structure. That section will mostly offer counter-examples, in particular showing thatthere is no general theorem extending 214Xn from finite families to countable families, and that the special conditionsin 214P and 214Yb are there for good reasons. But in552 of Volume 5 I will present some positive results dependenton special axioms beyond those of ZFC.


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215B -finite spaces and the principle of exhaustion 43

215 -finite spaces and the principle of exhaustion

I interpolate a short section to deal with some useful facts which might get lost if buried in one of the longer sectionsof this chapter. The great majority of the applications of measure theory involve -finite spaces, to the point thatmany authors skim over any others. I myself prefer to signal the importance of such concepts by explicitly stating justwhich theorems apply only to the restricted class of spaces. But undoubtedly some facts about-finite spaces need tobe grasped early on. In 215B I give a list of properties characterizing -finite spaces. Some of these make better sensein the light of the principle of exhaustion (215A). I take the opportunity to include a fundamental fact about atomlessmeasure spaces (215D).

215A The principle of exhaustion The following is an example of the use of one of the most important methodsin measure theory.

Lemma Let (X, , ) be any measure space andE a non-empty set such that supnN Fn is finite for everynon-decreasing sequenceFnnN inE.

(a) There is a non-decreasing sequenceFnnN inE such that, for every E, eitherthere is an nN such thatE Fn is not included in any member ofE or, settingF =

nN Fn,

limn (E\ Fn) = (E\ F) = 0.In particular, ifE E andEF, then E\ F is negligible.

(b) IfE is upwards-directed, then there is a non-decreasing sequenceFnnN inEsuch that, setting F =nN Fn,

F= supEEEandE\ F is negligible for every E E, so thatF is an essential supremum ofEin in the sense of211G.(c) If the union of any non-decreasing sequence inEbelongs toE, then there is anF Esuch thatE\Fis negligiblewhenever E E andF E.proof (a) ChooseFnnN,EnnN andunnN inductively, as follows. Take F0 to be any member ofE. GivenFn E, setEn ={E : Fn E E} and un = sup{E : E En} in [0, ], and choose Fn+1 En such thatFn+1min(n, un 2n); continue.

Observe that this construction yields a non-decreasing sequenceFnnN inE. SinceEn+1 En for everyn,unnNis non-increasing, and has a limit u in [0, ]. Since min(n, u 2n)Fn+1un for everyn, limn Fn= u. Ourhypothesis onEnow tells us that u is finite.

IfE is such that for every n N there is an En E such thatE FnEn, then En En, soFn(E Fn)Enun

for everyn, and limn (E

Fn) = u. But this means that

(E\ F)limn (E\ Fn) = limn (E Fn) Fn= 0,as stated. In particular, this is so ifE E and EF.

(b) TakeFnnN from (a). IfE E, then (becauseEis upwards-directed) E Fn is included in some member ofE for every n N; so we must have the second alternative of (a), and E\ F is negligible. It follows that

supEEEF= limn FnsupEEE,soF= supEEE.

IfG is any measurable set such that E\ Fis negligible for every E E, then Fn\ G is negligible for every n, sothatF\ G is negligible; thus F is an essential supremum forE.

(c) Again takeFnnN from (a), and set F =

nN En. Our hypothesis now is that F E, so has both theproperties declared.

215B -finite spaces are so important that I think it is worth spelling out the following facts.

Proposition Let (X, , ) be a semi-finite measure space. WriteNfor the family of-negligible sets and f for thefamily of measurable sets of finite measure. Then the following are equiveridical:

(i) (X, , ) is -finite;(ii) every disjoint family in f \ N is countable;(iii) every disjoint family in \ N is countable;(iv) for everyE there is a countable setE0 Esuch that E\

E0 is negligible for every E E;(v) for every non-empty upwards-directedE there is a non-decreasing sequenceFnnN inE such that E\nN Fn is negligible for every E E;


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44 Taxonomy of measure spaces 215B

(vi) for every non-emptyE , there is a non-decreasing sequenceFnnN inEsuch thatE\nN Fnis negligible

wheneverE E andEFn for everyn N;(vii)eitherX= 0 orthere is a probability measure onXwith the same domain and the same negligible sets as

;(viii) there is a measurable integrable function f :X]0, 1];(ix) either X= 0 or there is a measurable function f :X]0, [ such that f d= 1.

proof (i)(vii) and (viii) IfX= 0, (vii) is trivial and we can take f = X in (viii). Otherwise, letEnnN bea disjoint sequence in f covering X. Then it is easy to see that there is a sequence

n

n

N of strictly positive real

numbers such thatn=0 nEn = 1. Set E= n=0 n(E En) for E ; then is a probability measure withdomain and the same negligible sets as . Alsof=

n=0min(1, n)En is a strictly positive measurable integrable

function.

(vii)(vi) and (v)Assume (vii), and letEbe a non-empty family of measurable sets. IfX= 0 then (vi) and (v)are certainly true. Otherwise, letbe a probability measure with domain and the same negligible sets as . SincesupEE E 1 is finite, we can apply 215Aa to find a non-decreasing sequence FnnN inE such that E\

nN Fn

is negligible whenever E E includes nN Fn; and ifEis upwards-directed, E\nN Fn will be negligible for everyE E, as in 215Ab.

(vi)(iv)Assume (vi), and letEbe any subset of . SetH={

E0 :E0 E is countable}.

By (vi), there is a sequenceHnnN inHsuch thatH\nN Hn is negligible whenever H Hand HHnfor everyn N. Now we can express each Hn as En, whereEn E is countable; settingE0 =nN En,E0 is countable. IfE E, then EnN Hn =({E} E0) belongs toH and includes every Hn, so that E\ E0 = E\nN Hn isnegligible. SoE0 has the property we need, and (iv) is true.

(iv)(iii) Assume (iv). IfEis a disjoint family in \ N, take a countableE0 E such that E\ E0 is negligible

for everyE E. ThenE= E\ E0 is negligible for everyE E \E 0; but this just means thatE \ E0 is empty, so thatE=E0 is countable.

(iii)(ii) is trivial.(ii)(i)Assume (ii). LetP be the set of all disjoint subsets of f\N, ordered by. ThenP is a partially ordered

set, not empty (as P), and ifQP is non-empty and totally ordered then it has an upper bound in P. PPP SetE=

Q, the union of all the disjoint families belonging to Q. IfE E then E C for someC Q, so Ef \ N.

IfE, F E andE=F, then there areC,D Q such that E C, F D; now Q is totally ordered, so one ofC,D islarger than the other, and in either caseC D is a member ofQ containing both Eand F. But since any member ofQ is a disjoint collection of sets, E F=. As EandFare arbitrary,Eis a disjoint family of sets and belongs to P.And of courseC E for everyC Q, soE is an upper bound for Q in P. QQQ

By Zorns Lemma (2A1M), P has a maximal elementE say. By (ii),E must be countable, so E . NowH = X\ E is negligible. PPP??? Suppose, if possible, otherwise. Because (X, , ) is semi-finite, there is a set G offinite measure such thatGHandG >0, that is, Gf \ N andG E=for every E E. But this means that{G} Eis a member ofP strictly larger thanE, which is supposed to be impossible. XXXQQQ

LetXnnN be a sequence running overE {H}. ThenXnnN is a cover ofXby a sequence of measurable setsof finite measure, so (X, , ) is-finite.

(v)(i)If (v) is true, then we have a sequenceEnnN in f such thatE\nN Enis negligible for every Ef.

Because is semi-finite,X\nN En must be negligible, soXis covered by a countable family of sets of finite measure

and is -finite.(viii)(ix) If X = 0 this is trivial. Otherwise, if f is a strictly positive measurable integrable function, then

c=

f >0 (122Rc), so 1

cf is a strictly positive measurable function with integral 1.

(ix)(i) If f : X ]0, [ is measurable and integrable,{x : f(x) 2n}nN is a sequence of sets of finitemeasure covering X.

215C Corollary Let (X, , ) be a -finite measure space, and suppose thatE is any non-empty set.(a) There is a non-decreasing sequenceFnnN inE such that, for every E, eitherthere is an n N such that

E Fn is not included in any member ofE orE\nN Fn is negligible.


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215Yb -finite spaces and the principle of exhaustion 45

(b) IfEis upwards-directed, then there is a non-decreasing sequenceFnnN inEsuch thatnN Fn is an essential

supremum ofE in .(c) If the union of any non-decreasing sequence inEbelongs toE, then there is anF Esuch thatE\Fis negligible

whenever E E andF E.proof By 215B, there is a totally finite measure on Xwith the same measurable sets and the same negligible setsas. Since supEE Eis finite, we can apply 215A to to obtain the results.

215D As a further example of the use of the principle of exhaustion, I give a fundamental fact about atomless

measure spaces.

Proposition Let (X, , ) be an atomless measure space. IfE and 0 E 0, there is a measurable setHGsuch that 0< H.(b) LetH be the family of all those H such that H E and H . IfHnnN is any non-decreasing

sequence in

H, then (nN Hn) = limn Hn , sonN Hn H. So 215Ac tells us that there is an F Hsuch that H\ F is negligible whenever H Hand F H. ??? Suppose, if possible, that F < . By (a), there is an

HE\ Fsuch that 0< H F. But in this case H F Hand ((H F) \ F)> 0, which is impossible. XXXSo we have found an appropriate set F.

215X Basic exercises (a) Let (X, , ) be a measure space and a non-empty set of-integrable real-valuedfunctions from X to R. Suppose that supnN

fn is finite for every sequencefnnN in such that fna.e. fn+1

for every n. Show that there is a sequencefnnN in such that fna.e. fn+1 for every n and, for every integrablereal-valued functionf on X, eitherfa.e. supnN fn orthere is an nN such that no member of is greater thanor equal to max(f, fn) almost everywhere.

>>>(b)Let (X, , ) be a measure space. (i) Suppose thatE is a non-empty upwards-directed subset of such thatc= supEEEis finite. Show that E\

nN Fn is negligible whenever E EandFnnN is a sequence inEsuch that

limn

Fn = c. (ii) Let be a non-empty set of integrable functions on X which is upwards-directed in the sense

that for all f, g there is an h such that max(f, g)a.e. h, and suppose that c = supf f is finite. Showthatfa.e.supnN fn whenever f andfnnN is a sequence in such that limn

fn= c.

(c)Use 215A to shorten the proof of 211Ld.

(d) Give an example of a (non-semi-finite) measure space (X, , ) satisfying conditions (ii)-(iv) of 215B, but not(i).

>>>(e) Let (X, , ) be an atomless -finite measure space. Show that for any > 0 there is a disjoint sequenceEnnN of measurable sets with measure at most such thatX=

nN En.

(f)Let (X, , ) be an atomless strictly localizable measure space. Show that for any >0 there is a decompositionXiiI ofXsuch that Xi for everyiI.

215Y Further exercises (a) Let (X, , ) be a -finite measure space andfmnm,nN,fmmN, fmeasurablereal-valued functions defined almost everywhere inXand such thatfmnnNfma.e. for each m andfmmNfa.e. Show that there is a strictly increasing sequencenmmN in N such thatfm,nmmNf a.e. (Compare 134Yb.)

(b) Let (X, , ) be a -finite measure space. LetfnnN be a sequence of measurable real-valued functions suchthat f = limn fn is defined almost everywhere in X. Show that there is a non-decreasing sequenceXkkN ofmeasurable subsets ofXsuch that

kN Xk is conegligible in XandfnnNfuniformly on everyXk, in the sense

that for any > 0 there is an mN such that|fj(x) f(x)| is defined and less than or equal to whenever j m,xXk.

(This is a version of Egorovs theorem.)


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46 Taxonomy of measure spaces 215Yc

(c)Let (X, , ) be a totally finite measure space andfnnN, fmeasurable real-valued functions defined almosteverywhere inX. Show that fnnNfa.e. iff there is a sequence nnNof strictly positive real numbers, convergingto 0, such that

limn (kn{x: xdom fk dom f,|fk(x) f(x)| n}) = 0.

(d) Find a direct proof of (v)(vi) in 215B. (Hint: givenE , use Zorns Lemma to find a maximal totallyorderedE Esuch that EF / N for any distinct E, F E, and apply (v) toE.)

215 Notes and comments The common ground of 215A, 215B(vi), 215C and 215Xa is actually one of the mostfundamental ideas in measure theory. It appears in such various forms that it is often easier to prove an applicationfrom first

fremlin measure theory book 2

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