Part A: Integration (Hilary Term, 2014)

Lecturer: Z. Qian (Email: qianz@maths.ox.ac.uk)Mathematical Institute, Oxford

March 12, 2014

Contents

1 Introduction 2

2 Measure spaces 42.1 Some definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 The Lebesgue measure 83.1 The outer measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.2 Null sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Caratheodorys extension theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.4 An example of non measurable sets . . . . . . . . . . . . . . . . . . . . . . . . . . 173.5 Lebesgue-Stieltjes measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.6 Complete measure spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 Measurable functions 194.1 Definition and basic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Almost sure properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.3 Lebesgue sub-spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5 Lebesgue integration 265.1 Simple (measurable) functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2 Integrals of non-negative measurable functions . . . . . . . . . . . . . . . . . . . . 285.3 Integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

5.3.1 Convergence theorems summary . . . . . . . . . . . . . . . . . . . . . . . 365.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.5 Integrals depending on parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6 The Fubini theorem 446.1 Measures on product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446.2 Fubini theorem and Tonelli theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 456.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.4 Change of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

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7 Lp-space and convergencies 517.1 Convergence in measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567.2 Convergence in Lp-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

2

The Synopsis for Pat A Integration published by the Mathematical Institute:Measure spaces. Outer measure, null set, measurable set. The Cantor set. Lebesgue measure

on the real line. Counting measure. Probability measures. Construction of a non-measurableset (non-examinable). Measurable function, simple function, integrable function. Reconciliationwith the integral introduced in Moderations. A simple comparison theorem. Integrability of poly-nomial and exponential functions over suitable intervals. Changes of variable. Fatous Lemma(proof not examinable). Monotone Convergence Theorem (proof not examinable). DominatedConvergence Theorem. Corollaries and applications of the Convergence Theorems (includingterm-by-term integration of series). Theorems of Fubini and Tonelli (proofs not examinable).Differentiation under the integral sign. Change of variables. Brief introduction to Lp spaces.Holder and Minkowski inequalities (proof not examinable).

These notes are written for the lectures delivered in Hilary term 2014, which cover the materialin the synopsis above. In writing up these notes, I have benefited from the notes written byprevious lecturers, in particular the notes produced by A. Etheridge and C. Batty. Their notesare available on the course web page. I have adopted many of their worked examples, and inmany places I have closely followed their notes. Any errors appeared in these notes howeverare completely my own responsibility. Please report errors, typos and etc. you may discover toqianz@maths.ox.ac.uk .

1 Introduction

In Prelims Analysis III Integration, we have established the theory of Riemann integrals forbounded functions on a finite interval. In addition to the computational techniques such as themethods of change of variables, integration by parts and etc., we have proved several fundamentalresults about Riemann integrals. Let us recall two important results. The first is the existenceof Riemann integrals. If f is bounded and continuous on (a, b) (where a and b are two numbers),then f is Riemann integrable. The second is the fundamental theorem in calculus (FTC): if Fhas continuous derivative on [a, b] then

ba

F (x)dx = F (b) F (a).

These results are important and are still good, and form the foundation for further study in your3rd year and 4th year. What you have learned in Prelims Calculus and Analysis courses areimportant not only for the study of advanced theories in mathematics, but also are essential inapplying mathematics to solve practical problems in science.

The theory of Riemann integration is powerful tool in particular for computations, it howeverlacks flexibility in handling limit procedures such as taking limits under integration. Here is asimple and illustrative example. Consider the Dirichlet function f(x) = 1 if x is rational and 1if x is irrational. f is a simple function but perhaps is not an interesting one. While it appears asthe limit of simple functions. List all rational numbers in [0, 1] as r1, r2, , and define fn(x) = 1if x {r1, ,rn} and fn(x) = 1 otherwise. Clearly fn f on [0, 1], clearly each fn is Riemannintegrable on [0, 1] and

10fn(x)dx = 1. Therefore limn

10fn(x)dx = 1, but we can not

take limit limn fn = f first then take Riemann integral of f , as f is not Riemann integrable.

3

It was recognized gradually that most difficulties one has with Riemann integrals are due tothe limitation of the theory of Riemanns integration. There was a need to extend the theory ofintegration to a larger class of functions, so that functions we are interested are integrable.

In order to explain the approach we are going to develop in this course, let us recall quickly themain steps in defining Riemann integrals. The first step is to choose a simple class of functionsto which we can assign integrals. For the theory of Riemann integrals we choose the collectionL of all step functions. Recall that a function is step if =ni=1 ai1Ji , where n is a positiveinteger, J1, , Jn are finite intervals, and where 1A denotes the characteristic function of A:1A(x) = 1 or 0 according to x A or not. The integral of a step function is defined to beI() =

ni=1 ai|Ji|, where |J | denotes the length of an interval J (which is the measure of the

interval J). The reason we choose to consider step functions in the Riemann integration liesin the fact that the measure of an interval, namely its length, makes sense for intervals. Thesecond step is to define lower and upper integrals for a bounded function f on a finite interval(a, b). The lower integral

baf(x)dx is defined to be the supremum of all I() where is step

and f1(a,b), and the upper integral baf(x)dx is the infinimum of all I() where is step

and f1(a,b). Finally, f is Riemann integrable if the lower and upper integrals of f over (a, b)coincide, and its Riemann integral

baf(x)dx is the common upper (and lower) integral.

By definition of supremum and infinimum, if we are able to enlarge the class L of simplefunctions, then the corresponding lower integral

baf(x)dx (which is a supremum) would become

greater, and the corresponding upper integral baf(x)dx (an infinimum) become smaller, so they

have better chance to be equal, thus have better chance to be integrable. This suggests thefollowing approach to extend the theory of integration to a larger class of functions. Take acollection of functions in place of step functions with the following form

=ni=1

ai1Ei

where Ei are certain subsets but not necessary intervals. It is required that we should be able todefine integral for such to proceed the definition of integrals for general functions. Namely theintegral of should be defined as I() =

ni=1 ai|Ei|. The problem to carry out this idea is that

for a general subset E, |E|, the length of E, is not well-defined. The good news is that this is themain difficulty we need to overcome in order to establish a new integration theory. Therefore ourfirst task is to extend the notion of lengths for intervals to a larger class of subsets than intervals,the new notion will be called the Lebesgue measures. As long as the measures of certain subsetsare established, the integration theory can be established solely based on measures, no otherstructures of the real line will be needed, which is somehow an unexpected reward by extendingthe notion of length to measures to general sets beyond intervals.

A very short history of Lebesgues theory of integration. E. Borel, and his student H. Lebesguewere interested in the general construction of functions, and they wanted to extend the conceptsof length, areas and volumes to general sets. They discovered it was not always possible to doso. It was H. Lebesgue who recognized the importance of the countable additivity in handlinglimits under integration. His Ph D thesis Integrale, Longueur, Aire was finished in 1902, andin a book form Lecons sur lintegration et la recherche des fonctions primitives which waspublished in 1904, in Paris, Lebesgues theory of integration was established, and basic limittheorems (Monotone Convergence Theorem, Dominated Convergence Theorem), which are the

4

powerful mathematical tools, were proved. H. Lebesgue applied his new integration theory to thestudy of trigonometric series, and published another monograph in 1906, Lecons sur les seriestrigonometriques (Paris). The theory of Lebesgues integration and its generalization, calledthe theory of measures (a measure is a generalization of the concept of length, area, volume),got prominent when, in 1933, A. Kolmogorov firmly established the foundation of probabil-ity theory by interpreting measure spaces as mathematical models for developing probabilitytheory and statistics. A. Kolmogorov published his finding in the article Grundbergriffe derWahrscheinlichkeitsrechnung, Erg. Mat. 2, no.3 (1933). The English translation in a book form,Foundations of the theory of probability is still in print, and is still worthy of reading even to-day. A. Kolmogorov not only founded the probability theory based on the theory of measures, hemade very important contributions even for the theory of measures. It was him who introducedthe concept of conditional expectations which plays a vital role in both analysis and probability.He also developed basic tools for constructing measures out of marginal distributions.

Around 1940, J. L. Doob systematically developed the theory of martingales. He turned Kol-mogorovs conditional expectation into a powerful tool, and identified a class of random variables(while, random variables are exactly those measurable functions according to A. Kolmogorov)called martingales which is the most important class of measurable functions in probability the-ory. Doob established powerful tools in a collection of Doobs inequalities, convergence theoremsand etc. His results on martingales were organized neatly in his classic Stochastic Processes,published in 1953. Meanwhile his student Paul Halmos wrote a book on the subject of measures,Measure Theory (1950), which was and remains a standard reference on Lebesgues theory ofmeasures and integration.

2 Measure spaces

Our first task is to extend the notion of length to some subsets E of R, called the Lebesguemeasures, denoted by m(E) (which is the length |J | if E = J is an interval). Suppose that theclass of subsets of R to which we are able to assign measures is denoted by M, then we expectthe following conditions should be satisfied:

1. Empty set has its measure zero, that is m() = 0.2. The measure of E M is nonnegative, i.e. m(E) 0 but can be infinity.3. Finite additivity : if E1, , En belong toM, then E1 En M. If they are disjoint,

thenm(E1 En) = m(E1) + +m(En).

The finite additivity is not enough in order to do integration, which is replaced by thecountable additivity.

4. Countable additivity : if E1, , En, belong toM, then n=1En M, and if in additionthey are disjoint, then

m

(n=1

En

)=

n=1

m(En).

5

2.1 Some definitions

Consider the extended real line [,] by adding and to R. That is[,] = {} R {}.

The following conventions are assumed throughout the lectures. For every a R, < a

In the definition of a measure space, there is no other structure required over the space rather than a -algebra.

A measure on (,F) is called a probability (measure) if () = 1. In this case is calleda sample space (of fundamental events), an element A in the -algebra F is called an event, and(A) is called the probability that the event A happens.

Proposition 2.4 Let (,F , ) be a measure space. Then

1) (A) (B) if A B,2) if An F and An (that is An An+1 for all n), then (n=1An) = limn (An),3) if An F , An (that is An An+1 for all n) and (A1) < , then (n=1An) =

limn (An).Proof. 1) If A B then B = A (B Ac), since A and B Ac are disjoint, by additivity

of the measure(B) = (A) + (B Ac) (A).

2) Let E1 = A1 and En = AnAn1 for n 2. Then En are disjoint and n=1An = n=1En.Therefore, by the countable additivity,

(n=1

An

)=

(n=1

En

)=

k=1

(Ek).

On the other hand, since An , nk=1Ek = An, thusk=1

(Ek) = limn

nk=1

(Ek) = limn

(nk=1Ek) = limn

(An)

which yields 2).3) Let Bn = A1 \ An. Then Bn . Since A1 = Bn An, Bn and An are disjoint, so

(A1) = (Bn) + (An)

hence (Bn) = (A1) (An) as (An) (A1)

2.2 Examples

This section I collect some problems for you to think about. Hints or even outlines of solutionsare given for your references.

Example 1. (Limiting sets.) Let {An : n = 1, 2, } be a sequence of subsets of the space .Consider

limn

An = {x : x belongs to infinitely many An}and

limnAn = {x : x does not belong to only finitely many An}= {x : x belongs to eventually all An}.

For example x limnAn means that, there is a sub-sequence n1 < n2 < < nk < suchthat x k=1Ank (where the sub-sequence may depend on x). Similarly, x limnAn meansthat there is a natural number N (which may depend on x) such that x nNAn. We havethe following facts.

1. limnAn limnAn.2. limnAn =

n=1

k=nAk and limnAn =

n=1

k=nAk.

3. If An+1 An for all n (in this case we say that An ), then limnAn = limnAn =n=1An.

4. If An+1 An for all n (in this case we say that An ), then limnAn = limnAn =n=1An.

Example 2. Recall de Morgans law: (A)c = Ac and (A)c = Ac. If Fis a -algebra on , and An F for n = 1, 2, , then

1. n=1An F , and2. both limnAn and limnAn belong to F .

Example 3. Suppose that {F : } is a family of -algebras on , where is a set, then

F = {A : A F for all }

is once again a -algebra over .

Example 4. Let C be a non-empty collection of some subsets of . Then

{C} ={F : F is a -algebra on and F C}

is a -algebra on . Show that if a -algebra H contains C, then {C} H. {C} is called thesmallest -algebra containing C, or the -algebra generated by C.

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Example 5. Suppose that and S are two spaces, and that G is a -algebra on S. Letf : S be a mapping. Prove that

F = {f1(B) : B G}is a -algebra on , here we recall that

f1(B) = {x : f(x) B} if B S.

Example 6. Let (,F) be a measurable space, and let E be a subset. Let

FE = {E A : where A F} .

Prove that FE is a -algebra on E, and therefore (E,FE) is a measurable space.If in addition E F , then FE F . If is a measure on (,F) then the restriction of on

FE is a measure.

3 The Lebesgue measure

In this section we construct 1) the -algebra MLeb of Lebesgue measurable subsets of R, and 2)the Lebesgue measure m :MLeb [0,].

3.1 The outer measure

In this section, by saying a subset we mean a subset of R unless otherwise specified.Recall that |J | denotes the length of an interval J . |J | = if J is unbounded.If A R, then the outer measure of A, denoted by m(A), is defined by

m(A) = inf

{i=1

|Ji| : Ji are intervals andi=1

Ji A}. (3.1)

It follows immediately from definition that m() = 0 and that m(A) 0 for any subset A. IfA B then m(A) m(B).

Lemma 3.1 [Countable sub-additivity] If {An : n = 1, 2, } is a sequence of subsets, then

m

(n=1

An

)

n=1

m(An). (3.2)

Proof. If

n=1m(An) = then (3.2) holds. Suppose

n=1m

(An) 0 there is a countable cover {J (n)i : i =1, 2, } of An, where J (n)i are intervals, such that

i=1

|J (n)i | m(An) +

2n

9

Then {J (n)i : i, n = 1, 2, } forms a countable cover of n=1An so that

m

(n=1

An

)

n=1

i=1

|J (n)i | n=1

(m(An) +

2n

)

=n=1

m(An) + .

Since > 0 is arbitrary, (3.2) follows immediately.We may demand the inf in (3.1) runs over all possible countable coverings of open intervals,

or closed, or all half open intervals only, which does not alter the inf. Let us argue for openinterval coverings only.

Suppose that m(A) 0, there is a countable cover {Ji : i = 1, 2, }of A with the property that

m(A) i=1

|Ji| 2.

Suppose that Ji is an interval with ends ai bi (i = 1, 2, ), replace Ji be a slightly enlargedopen interval Oi = (ai 2i+2 , bi + 2i+2 ), then {Oi : i = 1, 2, } is also a countable cover of A,and

|Oi| = |Ji|+ 2i+1

for i = 1, 2, .Hence

i=1

|Ji| =i=1

|Oi| i=1

2i+1=

i=1

|Oi| 2

and therefore

m(A) i=1

|Oi| .

Thus, according to definition,

m(A) = inf

{i=1

|Ji| : Ji are open intervals andi=1

Ji A}.

Similarly, suppose H denotes the collection of all intervals with form (a, b] [intervals open atthe left and closed at the right end], then

m(A) = inf

{i=1

|Ji| : Ji H such thati=1

Ji A}.

Lemma 3.2 If A = ni=1Ji, where Ji are intervals and are disjoint, then m(A) =n

i=1 |Ji|.

Proof. Since {Ji} is a finite cover of A, and Ji are intervals, so that

m(A) ni=1

|Ji|.

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We next prove that, if J = [a, b] is a closed interval, then m(J) = b a. According todefinition, m(J) b a. On the other hand, for every > 0, there is a countable cover{Ik : k = 1, 2, } of J , where Ik are open intervals, such that

m(J) k=1

|Ik| .

Since [a, b] is closed and bounded, there is a finite subcover {Ik1 , , Ikn} of [a, b] from theopen cover {Ik : k = 1, 2, } of [a, b] [This is a result from Analysis II]. We may assume thatIki = (ai, bi) (i = 1, , n), and arrange them in the order

a1 a2 an.

Then we must have bk ak+1 (for any k = 1, 2, , n 1), a1 a and bn b, so thatk=1

|Ik| ni=1

|Jki | =ni=1

(bi ai) b a.

Therefore m(J) (b a) . Since is arbitrary, m(J) b a.If J = (a, b), then for every (0, (b a)/2)

m((a, b)) m([a , b ]) = b a 2.

Letting 0 to obtain m((a, b)) = b a. Therefore, for any finite interval with end pointsa < b, we have

b a m(J) m((a, b)) = b a.Thus m(J) = |J | for an interval J .

Let us now consider the general case. We may assume that all intervals Ji are finite otherwisethere is nothing to prove. For every > 0 there is a sequence Ik of intervals such that k=1Ik Aand

m(A) k=1

|Ik| m(A) + .

Note that Ik Ji Ik is an interval (may be empty), and {Ik Ji : i = 1, , n} are disjointsub-intervals of Ik for fixed k, so that

ni=1

|Ik Ji| |Ik|.

On the other hand

|Ji| = m(Ji) = m (k=1Ik Ji)

k=1

m(Ik Ji) =k=1

|Ik Ji|

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and therefore

ni=1

|Ji| ni=1

k=1

|Ik Ji| =k=1

ni=1

|Ik Ji|

k=1

|Ik| m(A) + .

Since > 0 is arbitrary, we may conclude that

ni=1

|Ji| m(A).

Hence m(A) =n

i=1 |Ji|.Although m(E) is well defined for every subset E in R, m is not countably additive, thus

m is not a measure on P(R) (the -algebra of all subsets of R). In fact, P(R) is too big for mto be countably additive. The main technical step in the construction of the Lebesgue measureis to identify the -algebra MLeb on which m is countably additive. MLeb should be sufficientlarge and should include all intervals. This will be achieved in the celebrated Caratheodorysextension theorem.

3.2 Null sets

Let us look at some subsets which are small with respect to the outer measure m.

Definition 3.3 A subset A R is called a null set if it has zero outer measure, i.e. m(A) = 0.

According to definition (3.1), A is a null set if and only if for every > 0 there is a sequenceof intervals Ji (i = 1, 2, ), such that i=1Ji A and

i=1 |Ji| < .

Lemma 3.4 If {An : n = 1, 2, } is a sequence of null subsets, so is n=1An.

This follows immediately from Lemma 3.1. In fact, if m(An) = 0, then by the countablesub-additivity of the outer measure

m

(n=1

An

)

n=1

m(An) = 0

hence we must have m(n=1An) = 0. n=1An is null.

Example 3.5 If A is a countable subset, then A is a null set.

There are null sets which are not countable! Here is an example.

The Cantor ternary set. Consider the closed interval J(1)0 = [0, 1].

1) Divide J(1)0 equally into three sub-intervals, the middle open interval I

(1)1 = (

13, 23) is removed

from J(1)0 , the remaining two closed sub-intervals are J

(1)1 = [0,

13] and J

(1)2 =

[23, 1], each has

length 13.

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2) Repeat step 1) for each closed interval J(1)1 and J

(1)2 : divide each equally into three sub-

intervals and remove the middle open ones. From J(1)1 we remove I

(2)1 =

(132, 232

), and from J

(1)2

remove

I(2)2 =

(2

3+

1

32,2

3+

2

32

)=

2

3+

(1

32,2

32

).

The remaining 22 closed intervals are denoted by J(2)1 , J

(2)2 , J

(2)3 , J

(2)4 each has length

132.

3) Repeat the procedure for each J(2)i , remove the middle open intervals I

(2)i (i = 1, , 22)

and the remaining 23 closed intervals are denoted by J(3)i for i = 1, , 23, each has length 133 .

Repeating this process, for each n, we have 2n1 disjoint open intervals I(n)i (i = 1, , 2n1)

with equal length 13n, and 2n disjoint closed intervals J

(n)i (i = 1, 2, , 2n) with length 13n (where

n = 1, 2, ).For each n, I

(n)i , J

(n)k (i = 1, , 2n1 and k = 1, , 2n) are disjoint sub-intervals of [0, 1],

and (nk=1 2

k1

i=1 I(k)i

)(2nk=1J (n)k

)= [0, 1].

Cantors ternary set C is defined to be the subset of [0, 1] by removing all middle intervals

I(n)i . That is

C = [0, 1] \(

n=1

2n1i=1

I(n)i

).

Lemma 3.6 The Cantor ternary set C is a null set.

Proof. By the construction, for every n, J(n)i (i = 1, , 2n) is a finite cover of C, so that

m(C) 2ni=1

|J (n)i | = 2n1

3n 0

as n. Therefore m(C) = 0, and C is a null subset by definition.Ternary expansions For decimal expansion of x = 0.x1x2 (0, 1] we mean that

x =n=1

xn10n

.

Similarly x (0, 1] can be written in its ternary expansion

x =n=1

an3n

= 0.a1a2

where an = 0, 1, 2, and we use the convention that it should not end with ak = 2 for all k N .Then, in terms of ternary expansion

(1

3,2

3) = (0, 1, 0.2)

and x I(1)1 = (13 , 23) if and only if x = 0.a1a2 with a1 = 1. Similarly, I(n)j contains preciselythe numbers whose ternary expansion x = 0.a1a2 where an = 1.

We leave the reader as an exercise to show that 1) C is closed, 2) C is uncountable, and 3)describe C in terms of ternary expansions for real numbers.

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3.3 Caratheodorys extension theorem

This is a major theorem in the construction of Lebesgues measure, admittedly the most difficultone as it looks lack of initiative at the first instance. It is a general theorem, and its proof is notexaminable. We will use this theorem to identify some important measurable subsets which weshould be familiar with.

Definition 3.7 A subset E is (Lebesgue) measurable if E satisfies the Caratheodory condition

m(F ) = m(F E) +m(F Ec) for any subset F . (3.3)

The collection of all (Lebesgue) measurable subsets is denoted by MLeb.

Since F \ E = F Ec, (3.3) may be written as

m(F ) = m(F E) +m(F \ E) for any subset F . (3.4)

Since F = (F E) (F Ec), by sub-additivity of m

m(F ) m(F E) +m(F Ec)

for any E and F , the Caratheodory condition (3.3) is equivalent to the inequality

m(F ) m(F E) +m(F Ec) (3.5)

for any subset F .E is measurable if and only if (3.5) holds for any subset F .

Theorem 3.8 MLeb is a -algebra on R, and m is a measure on MLeb.

Proof. (The proof is not examinable.) Clearly the empty set MLeb, and R MLeb.Also, according to (3.4), E MLeb if and only if Ec MLeb.

Let A,B MLeb. We show that AB MLeb so thatMLeb is an algebra. Since (AB)c =Ac Bc, we need to show that

m(F ) = m(F (A B)) +m(F (Ac Bc))

for any subset F .Since A is measurable, so

m(F ) = m(F A) +m(F Ac), (3.6)

and since B is measurable, applying (3.4) to F A (in the place of F ) and B we obtain

m(F A) = m(F A B) +m(F A Bc). (3.7)

Substitute (3.7) into (3.6) to obtain

m(F ) = m(F (A B)) +m(F A Bc) +m(F Ac). (3.8)

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Use again (3.4) to A (which is measurable) and F (Ac Bc) to obtain

m(F (Ac Bc)) = m(F (Ac Bc) A) +m(F (Ac Bc) Ac)= m(F Bc A) +m(F Ac),

here we have used the elementary equalities (Ac Bc) A = Bc A and (Ac Bc) Ac = Ac.Together with (3.8) we deduce that

m(F ) = m(F (A B)) +m(F (Ac Bc))

for any F , so that A B MLeb. It follows from de Morgan law that

A B = (Ac Bc)c MLeb.

Thus MLeb is an algebra.We next to show thatMLeb is a -algebra. To this end, consider En MLeb, n = 1, 2, . We

show that E = n=1En MLeb. Without losing generality we may assume that En are disjoint,otherwise we may consider An instead, where A1 = E1 MLeb and An = En \ (j

which implies that j=1Ej MLeb. Since we must have

m(F ) m(F (j=1Ej)) +m(F (j=1Ej)c)for any F and Ej, the inequalities in (3.9) must be equalities. Hence

m(F ) =j=1

m(F Ej) +m(F (j=1Ej)c)

for any subset F . In particular, by applying the equality above to F = k=1Ek we obtain

m(j=1Ej) =j=1

m((k=1Ek) Ej) =j=1

m(Ej).

That is, m is countably additive, so that m is a measure on MLeb.The restriction m on the -algebra MLeb will be denoted by m. The measure space

(R,MLeb,m) constructed in the theorem above is called the Lebesgue measure space.In the proof above, we have only used the facts that m is non-negative, m is monotonic:

A B then m(A) m(B), and that m is sub-additive. Therefore, the Caratheodory theoremis a quite general statement which allows us to construct a measure out of an outer measure.

Next we want to see how large the -algebra MLeb is.Proposition 3.9 If A is a null set, then A MLeb.

Proof. For every subset F , we have m(F A) m(A) = 0 so thatm(F ) m(F Ac) = m(F A) +m(F Ac).

By definition, A is measurable.

Proposition 3.10 If J is a finite interval, then J MLeb and m(J) = |J |.Proof. According to Lemma 3.6, m(J) = |J |. We want to show that J MLeb. That is, J

satisfies the Caratheodorys condition:

m(F ) = m(F J) +m(F J c)for any F . Since by the subadditivity

m(F ) m(F J) +m(F J c)we therefore need to show that

m(F ) m(F J) +m(F J c).The inequality is trivial if m(F ) =, so let us assume that m(F ) 0 thereare intervals Jk (k = 1, 2, ) such that kJk F and

m(F ) k=1

|Jk| m(F ) + .

16

Since Jk J is an intervals, and Jk J c is a union of at most two intervals, so that, accordingto Lemma 3.2

|Jk| = |Jk J |+m(Jk J c)= m(Jk J) +m(Jk J c)

for every k. Therefore

m(F ) k=1

|Jk| =k=1

(m(Jk J) +m(Jk J c))

=k=1

m(Jk J) +k=1

m(Jk J c)

m(F J) +m(F J c) .Since > 0 is arbitrary, we must have

m(F ) m(F J) +m(F J c).Thus J MLeb.Proposition 3.11 Let MBor be the smallest -algebra containing all finite intervals. ThenMBor MLeb.

Proof. Since MLeb is a -algebra, and any finite interval belongs to MLeb, so by definitionMBor MLeb.

MBor is called the Borel -algebra on R, and an element inMBor is called a Borel measurablesubset or simply a Borel subset. Therefore any Borel subset is Lebesgue measurable.

Corollary 3.12 Open sets and closed sets are measurable.

Proof. If U is an open set in R, then U can be written as a union of at most countable openintervals, so that U MLeb. Since MLeb is a -algebra, any closed set is also measurable.

Actually we can say a bit more about the measure of an open set. We can show that anyopen set U of R can be written as a union of at most countable many disjoint open intervals

U = j=1(aj, bj).By the countable additivity we have

m(U) =j=1

(bj aj) .

Therefore, Lebesgue measure m is really an extension of the notion of length to measurable sets.We should point out thatMBor is strictly smaller thanMLeb. However we have the following

fact

Proposition 3.13 If E MLeb then there are A,B MBor such that A E B and bothE \ A and B \ E are null sets.

The proof will be set as an exercise.

17

3.4 An example of non measurable sets

While not every subset in R is measurable, that is, the -algebra MLeb does not include everysubset of R. On the other hand, it is not very easy to produce an example of non measurablesets, which is a good news indeed. In fact we need to invoke the axiom of choice to produce aset which is not in MLeb.

If E is a subset of R, and a is a real number, then a + A = {a + x : x A} is a translationof A. It is elementary that if J is an interval, then so is a + J , and |a + J | = |J |. According todefinition of the outer measure m we may deduce that m(a + A) = m(A). It follows that, ifE MLeb then so is a+ E MLeb and m(a+ E) = m(E). That is to say that m is a measurewhich is invariant under translations.

Consider the unit interval [0, 1]. We say two numbers x, y [0, 1] are equivalent if x y Q,we divide [0, 1] into equivalent classes, and we choice exactly one number from each equivalentclass to form a subset A [0, 1]. We claim that A is not measurable, i.e. A / MLeb. Infact, list the rational numbers in [0, 1] by r1, r2, , by definition i=1(ri + A) [0, 1] andi=1(ri + A) [0, 2]. Thus

1 m(

i=1

(ri + A)

) 2.

Note that ri + A are disjoint, and m(ri + A) = m

(A) for every i. If A were measurable, thenso are ri + A. According to the countable additivity, we would have

m

(i=1

(ri + A)

)=

i=1

m(ri + A) =i=1

m(A)

so that

1 i=1

m(A) 2

which is impossible as

i=1m(A) = 0 or

i=1m

(A) = (according to m(A) = 0 or not).Thus A must be not Lebesgue measurable.

3.5 Lebesgue-Stieltjes measures

By demanding the length of an interval J to be its length |J | we have constructed the Lebesguemeasure m which is the extension of the concept of length. If we give a different scale of thelength for intervals, then we may construct different measures. Let us outline how to constructLebesgue-Stieltjes measures on R, the method is however the same as that for the Lebesguemeasure.

Let : R R be a right continuous increasing function. Here is right continuous meansthat for every a, the right limit limta (t) = (a+) (which must exist as is increasing AnalysisII) coincides with (a). The left limit limta (t) = (a) exists. The increment (a) (a) iscalled the jump of at a. Define for an interval (a, b] its length by

m((a, b]) = (b) (a).Let R be the collection of all subsets with the following form

(a1, b1] (aN , bN ]

18

for some N and a1 < b1 a2 < < bN , i.e. subsets of finite unions of intervals which open tothe left and closed to the right. Then R is an algebra but not a -algebra. We define the outermeasure associated with by

m(E) = inf

{n=1

m(Jn) : Jn = (an, bn] andn=1

Jn A}. (3.10)

Then m(E) 0, m() = 0, if E F then m(E) m(F ), and we have the countablesub-additivity

m

(n=1

En

)

n=1

m(En).

Moreover, if E = (a1, b1] (aN , bN ] R is a disjoint finite union, then

m(E) =Ni=1

((bi) (ai)) .

We say that a subset E is measurable with respect to m if the Caratheodory condition issatisfied:

m(F ) = m(F E) +m(F Ec) F . (3.11)

The collection of all m-measurable subsets is a -algebra on R, denoted by MLeb,, and mrestricted on MLeb, is a measure, denoted by m.

It is still true that MBor MLeb,, and it can be shown that

m({a}) = m(

n=1

(a n1, a])= lim

nm((a n1, a])

= limn

((a) (a n1))

= (a) (a).Hence m({a}) = 0 if and only if the jump (a) (a) = 0. Moreover

m([a, b]) = m({a}) +m((a, b])= (a) (a) + (b) (a)= (b) (a) and etc.

3.6 Complete measure spaces

As a consequence of Caratheodorys extension theorem, one may build a new measure space froma given measure space, called the completion of a measure space.

Suppose that (,F , ) is a measure space, and suppose that it is not trivial, that is, there isat least one subset E F such that 0 < (E) < . Then, we can define an outer measure associated with the given measure , defined namely by

(A) = inf

{j=1

(Ei) : Ei F s.t. i=1 Ei A}.

19

possesses the familiar properties for outer measures: 1) () = 0, (A) 0 for any subset,2) (A) (B) if A B, 3) is countably sub-additive:

(i=1Ai) i=1

(Ai).

Let F denote the collection of all subset E which satisfies the Caratheodorys condition:(F ) = (F E) + (F Ec) for any F .

Theorem 3.14 1) F is a -algebra on , and restricted on F is a measure.2) F F and for every A F , (A) = (A).3) If (A) = 0 then A F. That is, F contains all null set with respect to .

The -algebra F is called the completion of F with respect to the measure .Thanks to Theorem 3.14, given a measure space (,F , ), we naturally extend it (in many

applications, without further comments) to its completion (,F, ), which contains all nullsubsets.

Definition 3.15 A measure space (,F , ) is called complete, if for every A F with (A) =0, then any subset of A also belongs to F .

According to definition, if (,F , ) is a measure space, then its completion (,F, ) is acomplete space.

If E is Lebesgue measurable, then the Lebesgue space (E,MLeb,m) is complete.

4 Measurable functions

In this section we identify a class of functions, called measurable functions, for which we aim todefine integrals.

4.1 Definition and basic properties

The concept of measurability can be developed independent of a measure, we therefore developthis concept for a general measurable space (,F).

Definition 4.1 Let f : R be a function on . Then f is F-measurable if f1(I) F forevery interval I.

In probability theory, an F -measurable function on is called a random variable.

Remark 4.2 (about notations) f1(I) = {x : f(x) I}, and it is often written as {f I}.For example,

{f > a} = {x : f(x) > a} = f1((a,)).If f and g are two functions on , then

{f 6= g} = {x : f(x) 6= g(x)} and etc.

20

The proof of the following Proposition is a question in your assignment.

Proposition 4.3 Let B = Bi be any one of the following collections of subsets in R.1) B1: all intervals.2) B2: all intervals of form (a,), a real numbers.3) B3: all intervals of form (, a), a real numbers.4) B4: all intervals of form (, a], a real numbers.5) B5: all closed intervals [a, b], a, b real numbers.6) B6: all open intervals (a, b), a, b real numbers.7) B7: all open intervals (a, b], a, b real numbers.8) B8: all open intervals [a, b), a, b real numbers.9) B9: all open subsets.Let {B} denote the smallest -algebra containing B. ThenMBor = {Bi} (for i = 1, , 9).

Proof. (Outline) (2-7) {B2} = {B7): it follows from the relations: (a,) \ (b,) = (a, b]and (a,) = n=1(a, n].

(2-3) {B2} = {B3}:

(, a) = n=1(, a1

n] = n=1(a

1

n,)c

and

(a,) = n=1[a1

n,) = n=1(, a

1

n)c.

(3-4) {B3} = {B4}: (, a] = (a,)c and (a,) = (, a]c.(4-5) {B4} = {B7}: (b, a] = (, a] \ (, b] and

(, a] = n=1(n, a].

(5-7) {B7} = {B5}:[a, b] = n=1(a

1

n, b]

and

(a, b] = n=1[a1

n, b].

(6-7) {B7) = {B6}:(a, b) = n=1(a, b

1

n]

and

(a, b] = n=1(a, b+1

n).

(6-8) {B6) = {B8}:[a, b) = n=1(a

1

n, b)

and

(a, b) = n=1[a+1

n, b).

(6-9) {B9} = {B6}: since any open set is a countable union of open intervals.

21

Proposition 4.4 Suppose that f : R is a function. Then

Ff ={B R : f1(B) F}

is a -algebra.

Proof. Clearly both and belong to Ff . Since

f1(Bc) = { : f() Bc}= { : f() / B} = \ f1(B)c

therefore, if B Ff then Bc Ff . Suppose Bi Ff then

f1(i=1Bi) = i=1f1(Bi) Ff .

Hence Ff is a -algebra.

Theorem 4.5 Let f : R be a real function on . Then f is F-measurable if and only iff1(G) F for every G MBor.

Proof. Let Ff be defined as in Proposition 4.4. Suppose f is measurable, then any intervalI Ff . SinceMBor is the smallest -algebra containing all intervals, and since Ff is a -algebra,thus MBor Ff , so that for any G MBor, f1(G) F . The sufficiency part follows fromdefinition.

Proposition 4.6 f : R is F-measurable if one of the following properties holds:1) For every a R, {f > a} F .2) For every a R, {f < a} F .3) For every a R, {f a} F .

Proof. It follows directly from By Proposition 4.3 and Proposition (4.4).

Proposition 4.7 Let A . Then 1A is F-measurable if and only if A F .

Proof. For any interval I we have 11A (I) = A, Ac, or , so that IA is measurable if and

only if A F .A function f : R R which isMLeb-measurable is called a Lebesgue measurable function, or

simply called a measurable function if no confusion may arise.Recall that a function h : R R is continuous if for every open set U , h1(U) is open, in

particular, h1(U) MBor for any open set U , and therefore, according to Proposition (4.3)h1(G) MBor for any G MBor. Hence a continuous function h is MBor-measurable.

A function f : R R which is MBor-measurable, is called Borel measurable.Since MBor MLeb, a Borel measurable function is Lebesgue measurable.

Proposition 4.8 If f : R is F-measurable, and h : R R is Borel measurable, then h fis F-measurable. In particular, if f is F-measurable, h : R R is continuous, then h f isF-measurable.

22

Proof. In fact, for any G MBor, h1(G) MBor, thus (h f)1(G) = f1(h1(G)) F ,so h f is F -measurable.Proposition 4.9 Let a be a real number, and f, g be two F-measurable functions on . Then

1) af is measurable.2) f + g and f g are F-measurable.3) fg and f

g(if g 6= 0) are F-measurable.

4)f g and f g are F-measurable.5) |f |, f+ and f are F-measurable.Proof. There is nothing to prove if a = 0. If a > 0 then for any b we have

{af > b} = {f > ba} F

and if a < 0 then

{af > b} = {f < ba} F

so that af is measurable. In particular g is measurable if g is measurable. Thus we only needto show that f + g is measurable. For any b we have

{f + g > b} =qQ

{f > q and g > b q}

=qQ

{f > q} {g > b q}.

Since f, g are measurable so that {f > q} F and {g > b q} F and therefore{f > q} {g > b q} F q.

Since Q is countable, it follows thatqQ

{f > q} {g > b q} F

and therefore f + g is countable.To show 3) we note that f 2 = h f where h(x) = x2 is a continuous function, so that f 2 is

measurable if f is measurable. Now

fg =1

4

[(f + g)2 (f g)2]

is measurable.Similarly, since h(x) = |x| is continuous, so that |f | = h f is measurable. Hence

f g = 12[(f + g) + |f g|]

and

f g = 12[(f + g) |f g|]

are measurable. In particular f+ = f 0 and f = (f) 0 are measurable.

23

Definition 4.10 A function f : [,] is F-measurable, if {f = }, {f = } aremeasurable, and for any interval I, f1(I) is measurable.

Next we prove that the class of F -measurable functions on a measurable space (,F) is closedunder limit operations.

Suppose that {fn} is a sequence of F -measurable functions on . For each n considergn(x) = sup{fn(x), fn+1(x), }

andhn(x) = inf{fn(x), fn+1(x), }.

Then (gn) is a [point-wise] decreasing sequence and (hn) is an increasing sequence, hence gn gand hn h, where h may take value and g may take value . g and h are called theupper limit and lower limit of (fn), denoted by limnfn (or lim sup fn) and limnfn (or bylim inf fn) respectively.

For any a we have

{gn a} =

m=n

{fm a}

which implies that gn is F -measurable. On the other handhn = inf{fn(x),fn+1(x), }

so that hn is F -measurable. Finally sincelimnfn = inf {gn : n 1}

thus limnfn is F -measurable, and limnfn = limn(fn) is also F -measurable.

Proposition 4.11 If (fn) is a sequence of F-measurable functions, then both limnfn andlimnfn are F-measurable. In particular, if f = limn fn exists, then f is F-measurable.

4.2 Almost sure properties

In this sub-section we consider the Lebesgue space (R,MLeb,m) only. Suppose P is a propertydepending on x R. We say the property P holds almost everywhere (or almost surely), if{x : P (x) doesnt hold} is a null set. For example, if f and g are two functions, then we sayf = g almost everywhere if {x : f(x) 6= g(x)} is a null set. Another example, suppose that (fn)is a sequence of functions, then we say fn converges to f almost everywhere, if {x : fn(x) doesnot converge to f(x)} is a null set.

Proposition 4.12 If f is Lebesgue measurable, and g = f almost everywhere, then g is alsoLebesgue measurable.

This is an exercise in Problem Sheet 2.

Proposition 4.13 Suppose that (fn) is a sequence of Lebesgue measurable functions, and thatfn f almost everywhere, then f is measurable.

24

4.3 Lebesgue sub-spaces

Suppose that E R is Lebesgue measurable, then

MLeb(E) = {E A : where A MLeb}= {A : where A E and A MLeb}

is a -algebra on E. On the other handMLeb(E) MLeb so that the restriction of the Lebesguemeasure m, denoted by m|E or by m if no confusion is possible, is obviously a measure on(E,MLeb(E)), called the Lebesgue measure on E. The triple (E,MLeb(E),m) is therefore ameasure space, called a Lebesgue subspace. A function f defined on E which is MLeb(E)-measurable is called Lebesgue measurable on E. MLeb(E) will be simply denoted (by abusingnotations) by MLeb if no confusion may arise.

4.4 Examples

We give in this part further properties about measurable functions in terms of examples.Example 1. If g : R R is monotone, then g is Borel measurable. In fact if I is an interval,

then g1(I) is a union of at most countable many intervals, so it belongs to MBor. Hence g isMBor-measurable.

Example 2. Let (,F) be a measurable space. Suppose that fn : R are F -measurable.Recall that {fn} converges (to a finite number) at x if and only if {fn(x)} is a Cauchysequence. Hence

{x : (fn(x)) converges to a finite number}

=k=1

N=1

m,n=N

{|fn fm| < 1

k

}

and

{x : (fn(x)) doesnt converge to a finite number}

=k=1

N=1

m,n=N

{|fn fm| 1

k

}

are both measurable.

Example 3. Furthermore, if f : R is F -measurable, then

{fn f} =k=1

N=1

n=N

{|fn f | < 1

k

}

and

{fn 9 f} =k=1

N=1

n=N

{|fn f | 1

k

}

are measurable as well.

25

Example 4. Let E R, and let f, fn : E R (n = 1, 2, ) be Lebesgue measurablefunctions on E. Then fn f almost everywhere on E (with respect to the Lebesgue measure),i.e.

m ({x E : fn(x) does not converge to f(x)}) = 0[where

{fn doesnt converge to f} = {x E : fn(x)9 f(x)}and similar notations apply to other sets], if and only if for every > 0

m

(

N=1

n=N

{|fn f | })= 0 . (4.1)

If in addition m(E) n. Thenfn f = 1(0,) everywhere. For every (0, 1) we have

{|fn f | } = (n,)

so thatm ({|fn f | }) =.

Thus (4.2) doesnt hold. On the other hand

N=1

n=N

{|fn f | } =

N=1

(N,) = .

[This is a simple example that fn f everywhere, but fn does not converge to f in measure,see Definition 7.5 below].

Example 6. (Egorovs theorem) Let E R be measurable such that m(E) < . Supposethat fn, f : E R are measurable, and that fn f almost surely on E. Then for every > 0there is a measurable E E such that m(E \ E) < and fn f uniformly on E.

Proof. Since fn f almost surely on E and m(E) < , according to (4.2) in Example 4,for every > 0 and for any k = 1, 2, (applying (4.1) to = 1

k) there is nk such that

m

(j=n

{|fj f | 1

k

})

2kfor n nk.

Let

Ek =

j=nk

{|fj f | < 1

k

}=

{|fj f | < 1

k: j nk

}

26

and E =

k=1Ek. Then m(Eck) 2k and

m(E \ E) = m(

k=1

Eck

)

k=1

m(Eck) k=1

2k= .

We claim that fn f uniformly on E. In fact, x E if and only if x Ek for all k = 1, 2, ,and if and only if

|fj(x) f(x)| < 1k

j nk.Therefore

supxE

|fj(x) f(x)| 1k, j nk

and fn f uniformly on E.

5 Lebesgue integration

In this section we study Lebesgues theory of integration. First of all we need to define the classof simple (measurable) functions, and we define integrals for simple and non-negative functions,then define integrals for non-negative measurable functions. Finally we define Lebesgue integralsfor measurable functions.

Although we only deal with the Lebesgue space (R,MLeb,m), but the approach applies to ageneral measure space with only notational modifications. In particular what we are doing hereapply to Lebesgue subspaces, and to Lebesgue-Stieltjes measures m.

5.1 Simple (measurable) functions

Let (,F) be a measurable space. Then we say a function : R is simple, or called asimple (F -measurable) function if

=ki=1

ci1Ei

for some positive integer k, some reals ci and some F -measurable subsets Ei. If Ei are disjoint,then we say that f is in a standard form. Clearly a simple function is F -measurable.

Lemma 5.1 Suppose =k

i=1 ci1Ei is a simple function, where Ei are disjoint. Then isnon-negative if and only if ci 0 for all i = 1, , k.

This is obvious by definition. The collection of all non-negative, simple functions is denotedby S+(,F) or by S+ if no confusion may arise.

Theorem 5.2 Suppose that f is a measurable function taking values in [0,]. Then there is anincreasing sequence of simple functions (fn) such that fn f .

27

Proof. For each n we define

fn =22n1k=0

k

2n1E

(n)k

+ 2n1An ,

where

E(n)k =

{x :

k

2n f(x) < k + 1

2n

}and An = {x : f(x) 2n}

which are measurable. Then 0 fn f .Therefore we have the following simple structure theorem for measurable functions in terms

of simple measurable functions.

Corollary 5.3 A function f on taking values in [,] is measurable if and only if thereis a sequence of simple measurable functions fn such that fn f .

Proof. We note that f = f+f is measurable if and only if both f+ and f are measurable.Apply the previous theorem to f+ and f.

Let us now work with the Lebesgue space (R,MLeb,m).Suppose that =

ki=1 ci1Ei is a simple (Lebesgue measurable) function, where ci 0, thus

S+. Its (Lebesgue) integral is defined byR

=

R

dm =

(x)dx ki=1

cim(Ei)

where the convention that 0 = 0 has been used. The definition of R does not depend on

the representation of =k

i=1 ci1Ei as long as ci 0 for all i = 1, , k. That is, if

=ki=1

ci1Ei =Nj=1

aj1Aj

where ci 0 and aj 0 for all i, j, thenki=1

cim(Ei) =Nj=1

ajm(Aj).

This previous equality follows from the additivity of m.If =

ki=1 ci1Ei is a simple (measurable) function where ci > 0, then

R

Proof. We may choose measurable sets Ei (i = 1, , k) which are disjoint such that

=ki=1

ci1Ei , =ki=1

di1Ei

so that

+ =ki=1

(ci + di)1Ei

and therefore R

(+ ) =ki=1

(ci + di)m(Ei) =

R1+

R

,

which proves 1). If then ci di for all i, henceR

=ki=1

cim(Ei) ki=1

dim(Ei) =

R

which is 3).

5.2 Integrals of non-negative measurable functions

We are now in a position to define Lebesgue integrals for non-negative measurable functions.Suppose that f : R [0,] is measurable, then the (Lebesgue) integral of f is defined by

R

f = sup

{R

: S+ s.t. f}. (5.1)

A non-negative measurable function f is integrable ifRf

In particular this definition applies to Lebesgue sub-space (E,MLeb(E),m) where E R ismeasurable, and to define Lebesgue integral

Ef . Similarly, we say a non-negative measurable

function f on E is (Lebesgue) integrable over E ifEf

Proof. (The proof is not examinable. We use the proof given in W. Rudin: Real AndComplex Analysis, Third Edition, page 21 ). Since fn f , f is measurable and takes values in[0,]. Moreover fn fn+1 f so that

fn

fn+1

f . It follows that limn

fn

exists (but may be ), and limnfn

f . We next prove the reversed inequality that

limnfn

f .

Suppose that S+ and that f . Let (0, 1) be any fixed number. ConsiderEn = {fn }.

Since fn , En and n=1En = R (or the space depends on which measure space you areconsidering). In fact, for any x, if f(x) = 0 then x E1. If f(x) > 0, then (x) < f(x) (as f and < 1), since fn f , there is N such that fn(x) (x) for alln N (though ingeneral N depends on x) so that x En for n N . Thus in any case x n=1En, and thereforen=1En = R. Since

fn fn1En 1Enit follows that

fn fn1En

1En =

1En . (5.3)

Suppose =k

i=1 ci1Ai where ci 0 and Ai are disjoint. Then

1En =ki=1

ci1Ai1En =ki=1

ci1AiEn

so that 1En =

ki=1

cim(Ai En).

Since Ai En Ai as n, thus [Proposition 2.4, item 2)]m(Ai En) m(Ai En) as n.

Therefore 1En

ki=1

cim(Ai) =

as n .

Hence, by letting n in (5.3) we obtain

limn

fn

S+ s.t. f

which implies that

limn

fn

f

for all (0, 1). Thus we must have limnfn =

f .

Let us draw several useful consequences which follow directly from MCT.

Corollary 5.9 Suppose that f is measurable and takes values in [0,], then there is sequenceof simple non-negative functions n S+ such that n f and

f = limn

n.

31

Proof. This follows from Theorem 5.2 and MCT directly.

Corollary 5.10 If f, g are measurable and non-negative, then(f + g) =

f +

g.

Proof. Choose n, n S+ such that n f and n g. Then n + n f + g, so that(f + g) = lim

n

(n + n) = lim

n

(n +

n

)

= limn

n + lim

n

n =

f +

g.

Theorem 5.11 (MCT for series of non-negative measurable functions, B. Levi) Let fn be mea-surable and take values in [0,], n = 1, 2, , and f =n=1 fn. Then

f =n=1

fn.

Therefore, f is integrable if and only if

n=1

fn

Corollary 5.13 Suppose that h : R [0,] is Lebesgue measurable. Define (E) = Eh for

any E MLeb. Then is a measure on (R,MLeb). d is denoted by hdm, and h is called thedensity of the measure with respect to the Lebesgue measure. Suppose that f : R [0,] isLebesgue measurable. Then

fd =

fhdm.

As another consequence, we prove following result called Fatous lemma.

Theorem 5.14 (Fatous lemma) Let E R be measurable. Suppose that fn : E [0,] aremeasurable (n = 1, 2, ), then

E

limnfn limnE

fn.

Proof. (The proof is not examinable) Let gn(x) = infin fi(x) for n = 1, 2, , so gi fn forall i n, and gn limnfn. Apply MCT to (gn) we have

limnfn = limn

gn. While

gn fi for all i n, so that

gn infin

fi and therefore

limnfn = limn

gn lim

ninfin

fi = limn

fn.

5.3 Integrable functions

Suppose that E is a measurable subset, and f : E [,] is a measurable function. Thus|f | is measurable and takes values in [0,]. If

E|f |

is a null set. Then g is measurable too, and both N and E \N are measurable, henceE

g+ =

E\N

g+ +

N

g+ =

E\N

g+ =

E\N

f+ =

E

f+

and similarly,Ef =

Eg. Therefore f L1(E) if and only if g L1(E), and in this case

Ef =

Eg. Therefore, the definition of Lebesgue integrals on a measurable set E applies to

measurable functions which may be well defined on E only almost surely.Suppose that E and F are two measurable sets, that F E and that E \F is null. Suppose

that f : E [0,] be measurable, thenE

f =

F

f +

E\F

f =

F

f

In particular, if a < b, then (a,b)

f =

[a,b)

f =

(a,b]

f =

[a,b]

f .

Therefore baf(x)dx denotes the Lebesgue integral of f on an interval with endpoints a < b.

Theorem 5.15 Let E be a measurable subset. Then L1(E) is a linear space, and the integraloperation f

Ef is linear.

Proof. 1) follows from definition and Corollary (5.10). In fact, if f, g L1(E) then Ef+ 1, and f is integrable on (0, 1) if and only if p > 1. Thereforef is integrable on (0,) if and only if p > 1 and q p > 1.

Example 4. f(x) = sinxx. By means of contour integral, we have evaluated the (improper

Riemann) integral 0

sin x

xdx = lim

R

R0

sin x

xdx =

pi

2

38

[for example, on page 115, 2.7 Example, in J. B. Conway: Functions of one complex variable I].The function sinx

xis continuous and bounded on (0, R] for any R > 0. Therefore it is (Riemann)

integrable on (0, R]. The existence of the limit limR R0

sinxxdx doesnt necessarily imply that

f is integrable on (0,). In fact, for any n = 1, 2, , consider gn = |f |1En |f | whereEn = (0, npi). According to MCT

0

|f | = limn

0

gn = limn

npi0

| sin x|x

dx.

Since

npi0

| sin x|x

dx =n1k=0

(k+1)kpi

| sin x|x

dx n1k=0

(k+1)kpi

| sin x|(k + 1)pi

dx

=n1k=0

1

(k + 1)pi

pi0

sin xdx = 2n1k=0

1

(k + 1)pi

so that0|f | =. By definition, f is not integrable on (0,).

Example 5. Let us consider sinxx

on [1, R] for R > 1, which is (Riemann) integrable, and

limR

R1

sin x

xdx =

pi

2 10

sin x

xdx.

On the other hand, we may perform integration by parts to obtain R1

sin x

xdx = cos x

x

R1 R1

cosx

x2dx .

Since | cosx|x2

1x2

and cosxx2

is measurable, so cosxx2

is integrable on [1,). Apply DCT to gn =cosxx2

1[1,n]. Since |gn(x)| 1x2 , and gn(x) cosxx2 , thus the Lebesgue integral 1

cosx

x2dx = lim

n

1

gn = limn

n1

cos x

x2dx

= limn

( n1

sin x

xdx+

cosn

n cos 1

)

=pi

2 10

sin x

xdx cos 1.

In the following exercises, E R is a Lebesgue measurable set, and m is the Lebesguemeasure. L1(E) is the linear space of Lebesgue integrable functions on E.

Exercise 1. Use definition to prove carefully the following claims.a) If m(E) = 0, then

Ef = 0 for any non-negative function f .

b) Hence, prove that if m(E) = 0, thenEf = 0 for any function f .

Hint. First prove thatE = 0 for any simple measurable function .

39

Exercise 2. a) Suppose f : E [0,]. Prove that {f > 0} = n En where En = {f 1n}.Hence prove that if

Ef = 0 then m({f 1

n}) = 0 for any n = 1, 2, , and conclude that

m({f > 0}) = 0. Hence prove that, if f L1(E) then f is finite almost surely.Hint. Prove for every > 0, we have

m({f }) 1

E

f

for non-negative measurable function f . This is called the Markov inequality.b) If f is measurable, and

E|f | = 0, prove that f = 0 almost everywhere on E. Further

prove that if f, g L1(E) and E|f g| = 0, then f = g almost everywhere on E.

Exercise 3. If f L1(E) prove that E

f

E

|f | .

Hint : WriteEf =

Ef+

Ef and

E|f | =

Ef+ +

Ef.

Exercise 4. Let f L1(E).a) Prove that

limn

En

f = 0.

where En = {x E : |f(x)| n}.Hint. Since

En f En |f | so without losing generality, we can assume that f is non-

negative. Let fn = f1E\En . Since f is finite almost everywhere, we have fn f almost everywhereon E, hence by MCT, we have

E\En

f E

f

so thatEnf =

Ef

E\Enf 0.

b) Prove that for any > 0 there is > 0 such thatA

f

< for any measurable A E such that m(A) < .

Hint. Since

Af

A|f | so we may assume that f is non-negative in order to prove b).

Given > 0, choose N such that{fn}

f < 2for all n N , and use the following (in-)equalities

A

f =

A{f

Exercise 5. (MCT, another version). a) Suppose fn L1(E) and fn f almost surely onE. If

Efn is bounded above, i.e. there is a constant K such that

Efn K for all n, then

f L1(E) and E

f = limn

E

fn .

Hint. Let A = {fn f}. Then A is measurable and E \ A is a null set. Apply MCT tofn f1 f f1 on A to conclude that f f1 L1(A) and therefore f L1(E).

b) If fn are measurable on E and

n=1

E

|fn| 0.Define

() =

0

x1exdx for > 0.

For n = 1, 2, , consider fn(x) = x1ex1[ 1n,n]. Then fn f so, according to MCT

0

x1exdx = limn

n1n

x1exdx [Riemann integral]

On the other hand, using integration by parts we have n1n

x1exdx = n

1n

x1dex

= x1exn1n

+ ( 1) n

1n

x(1)1exdx,

41

so that, by MCT, 0

x1exdx = ( 1) 0

x(1)1exdx

if 1 > 0. Hence( + 1) = () for > 0.

Suppose = [ 1] + r where r [1, 0), then() = ( 1) ( 1 [ 1])(r).

In particular, if n is positive integer then

(n+ 1) =

0

xnexdx = n(n) = = n!(1)

= n!

0

exdx = n! .

Exercise 7. If > 0 then x1ex is integrable on (0, 1]. Moreover

x1ex =n=0

(1)nn!

xn+1.

Let hn(x) =(1)n

n!xn+1 which is integrable on [0, 1] for any n = 0, 1, 2, . Moreover 1

0

|hn| = 1n!

10

xn+1dx =1

n!(n+ )

so thatn=0

10

|hn| =n=0

1

n!(n+ )

5.5 Integrals depending on parameters

Let E be a measurable function, and G R. Suppose for every t G, ft : E [,] isintegrable, so that we may form a function F : G R by taking F (t) =

Eft for every t G.

We seek for sufficient conditions to ensure F is continuous on G. Let us begin with a simpleexample.

Example 1. Consider ft(x) = tet2x2 for t R and x R which is continuous in (t, x). Then

f0 = 0 so that

F (0) =

f0(x)dx = 0

while if t 6= 0 we have

F (t) =

tet2x2dx =

ex2

dx =2pi

and therefore F (t) =

tet2x2dx is not continuous at t = 0.

Theorem 5.22 Let E be measurable, and J R be an interval. For t J , ft : E [,]is measurable. Suppose

1) for any t0 J , ft ft0 almost surely on E, and2) there is g L1(E) such that |ft(x)| g(x) almost surely on E for all t J .Then ft L1(E) and F (t) =

Eft is continuous on J .

Proof. By 2), ft L1(E) for every t J , so F (t) =Eft is well defined real function on J .

To see the continuity at t0 J , consider any sequence tn J such that tn t0, apply DCT toftn we may deduce that

F (tn) =

E

ftn E

ft0 = F (t0),

i.e. F (tn) F (t0) for any tn t0, by definition of continuity, F is continuous at t0.It is important to notice that the control function g in 2) is independent of the parameter t,

as required in the DCT.We may apply the theorem above to ft+hft

hto obtain the following

Theorem 5.23 Let E be a measurable set, and J R be an interval. For each t J , ft : E Ris measurable, and the following conditions are satisfied:

1) for every t J , ft L1(E), and define F (t) =Eft for t J ,

2) for every x E, the partial derivative

tft(x) = lim

h0

ft+h(x) ft(x)h

exists for every t J (here the limit runs over h 0 such that t+ h J), and3) there is a control function g L1(E) such that tft(x)

g(x)almost surely on E for all t J .

Then F is differentiable on J and

F (t) =

E

tft .

43

Proof. Suppose t J we want to show F (t) = E

tft. We may assume that there is > 0

such that [t, t+ ) J (similarly consider the case that (t , t] J). Consider

gh(x) =ft+h(x) ft(x)

hfor x E and h (0, ).

By mean value theorem

|gh(x)| g(x) almost surely for all h (0, ).

By applying DCT to ghn where hn (0, ) and hn 0, to obtainF (t+ hn) F (t)

hn=

E

ghn E

tft

so that F exists and F (t) =E

tft, which completes the proof.

Example 2. The Gamma function () =0x1exdx is continuous on (0,). In fact for

every any 0 > 0, choose J = (a, b) such that 0 < a < 0 < b, and consider f(x) = x1ex for

x (0,) and (a, b). Let

g(x) = xa11(0,1] + xb1ex1(1,).

Then g L1(0,) and |f(x)| g(x) for all x (0,) and (a, b). According to Theorem5.22, () is continuous at 0. Since 0 > 0 is arbitrary, so is continuous on (0,).

If f : E C is a complex function on a measurable function, then f L1(E) if and only ifRef and Imf are integrable on E. In this case the Lebesgue integral

E

f =

E

Ref + i

E

Imf .

It is easy to see that a measurable complex function f on E is integrable if and only ifE|f |

so that gy L1(R). Moreover ygy(x) = eiyxf(x) and ygy(x) = |f(x)| y and x.

By Theorem 5.23 we have

d

dy

gy(x)dx =

d

dy

eiyx 1ix

f(x)dx

=

eiyxf(x)dx = f(y).

That is

f(y) =d

dy

eiyx 1ix

f(x)dx.

6 The Fubini theorem

We may construct the Lebesgue measure on Rd in a similar way as for R. Another way is to usethe Lebesgue measure m on R and construct a product measure on R2 and repeat the definitionof product measures to define the Lebesgue measure on Rd for general d.

6.1 Measures on product spaces

In this part we explain the idea how to construct the Lebesgue measure on R2 as the productmeasure of the Lebesgue measure on R. The same idea actually applies to the general setting ofmeasure spaces only with notational modification. In particular, what we are going to outlineapplies to Lebesgue sub-spaces.

Recall the definition of a Cartesian product space X Y . If X and Y are two spaces (i.e.two abstract sets. For example subsets of R), then the set of all ordered pairs (x, y) where x Xand y Y is called the (Cartesian) product of X and Y , denoted by X Y . For example,R R = R2, and [0, 1] [0, 1] = [0, 1]2 is the unit square in R2.

Let us consider R2 as the product space of R with itself. Suppose A R and B R aretwo measurable subsets, then AB R2 is called a measurable rectangle type set, its measurenaturally is defined to be

(A B) = m(A)m(B)here we use the convention that 0 = 0 = 0. The collection of all subset E in R2 with thefollowing form

E =ki=1

Ai Bi (6.1)

where Ai, Bi are measurable subsets, is denoted by R. If E R with representation (6.1) and ifin addition, Ai Bi (i = 1, , k) are disjoint, then we define

m(E) =ki=1

m(Ai)m(Bi). (6.2)

45

Then m(E) is well defined, i.e. (6.2) does not depend on the representation (6.1). As forconstructing Lebesgue measure m on the real line R, we then define the outer measure by

m(G) = inf

{i=1

m(Ai)m(Bi) : Ai, Bi MLeb s.t.i=1

Ai Bi G}

for any subset G R2. We have the following properties which are completely similar to theouter measures m.

1) m(G) 0 and m(E) = m(E) if E R.2) m(G) m(Q) if G Q.3) Countable sub-additivity:

m(i=1Gi) i=1

m(Gi) .

We say a subset E R2 is (Lebesgue) measurable if

m(F ) = m(F E) +m(F Ec)

for any F R2. Then, the collection of all (Lebesgue) measurable subsets is a -algebra onR2, denoted again by MLeb, and m is a measure on the measurable space (R2,MLeb), which isdenoted by m. Of course we have R MLeb, that is, any measurable rectangle set is Lebesguemeasurable.

If E R2 is (Lebesgue) measurable, then

MLeb(E) = {E F : F MLeb}

is a -algebra on E, and MLeb(E) MLeb. Thus the restriction of the Lebesgue measure m onMLeb(E) is a measure on (E,MLeb(E)).

The Lebesgues theory of integration is applicable to the measure space (R2,MLeb,m) andthe measure space (E,MLeb(E),m) if E R2 is measurable.

The integral of a non-negative measurable function / integrable function f on E MLebis denoted by

Ef(x, y)dxdy or by

Ef . If f is integrable, then

Ef(x, y)dxdy is called the

Lebesgue integral, also called a double integral.Suppose E R2 is bounded and closed, and f : E R is continuous, then the double

integral (in the sense of Riemann) E

f(x, y)dxdy (6.3)

exists. In this case f L1(E) and the Lebesgue integral Ef coincides with the (Riemann)

double integral (6.3).

6.2 Fubini theorem and Tonelli theorem

Let us begin with the following simple observation.

46

Lemma 6.1 Suppose that is a non-negative simple function on R2:

=ki=1

ci1AiBi , Ai, Bi MLeb, ci 0,

then1) for any y R, y S+ where y(x) = (x, y), and F S+ as well, where F (y) =

Ry,

and2) we have

R

F =

R2

that is R

(R

(x, y)dx

)dy =

R

(R

(x, y)dy

)dx =

R2(x, y)dxdy.

Proof. According to the definition of the Lebesgue measure on R2, we haveR2(x, y)dxdy =

ki=1

cim(Ai Bi) =ki=1

cim(Ai)m(Bi).

On the other hand, for each y,

y = (, y) =ki=1

ci1Bi(y)1Ai

which is a non-negative measurable simple function on R, and

F (y) =

R

y =

R

(x, y)dx =ki=1

ci1Bi(y)m(Ai)

=ki=1

cim(Ai)1Bi(y)

which is a non-negative simple function on R, and its integralR

F =

R

(R

(x, y)dx

)dy =

ki=1

cim(Ai)m(Bi)

=

R2(x, y)dxdy.

Proposition 6.2 Let f : R2 [0,) be measurable. Then1) x f(x, y) is measurable for almost all y,2) y

Rf(x, y)dx is (define for almost all y) non-negative and measurable, and

3) we have the equalityR2f(x, y)dxdy =

R

(R

f(x, y)dx

)dy =

R

(R

f(x, y)dy

)dx.

47

The proof is a careful applications of DCT together with the definition of the outer measurem. We omit the detail of its proof.

Theorem 6.3 (The Fubini theorem) Let A,B MLeb [so AB MLeb(R2)] and f L2(AB). Then

1) for almost all y B, fy L1(B), where fy(x) = f(y, x) for x A, so F (y) =Afy is well

defined for almost all y B, and2) F defined in 1) is integrable on B (so in particular, F is Lebesgue measurable), and

B

F =

AB

f .

Therefore B

(A

f(x, y)dx

)dy =

A

(B

f(x, y)dy

)dx =

AB

f(x, y)dxdy.

Proof. Apply Proposition 6.2 to f+ and f.

Theorem 6.4 (The Tonelli theorem) Suppose that f : R2 R is measurable, and suppose eitherof the repeated integrals exists and is finite:

R

(R

|f(x, y)|dx)dy,

R

(R

|f(x, y)|dy)dx.

Then f L1(R2), so that Fubinis theorem is applicable to both f and |f |.

Proof. According to Proposition 6.2, under the assumptions,R2|f | < , so that f

L1(R2).

6.3 Examples

Let us give several examples to demonstrate how to use Fubini and Tonellis theorems. I tooksome examples from Battys notes written for 2013H.

Example 1. Let E = [1, 1] [1, 1] and f(x, y) = xy(x2+y2)2

if (x, y) 6= (0, 0) and f(0, 0) = 0.f is measurable on E, and for any y 6= 0, x xy

(x2+y2)2is continuous on [1, 1] so integrable.

Moreover 11

xy

(x2 + y2)2dx = 0 for any y 6= 0

and therefore 11

( 11

xy

(x2 + y2)2dx

)dy = 0.

Similarly 11

( 11

xy

(x2 + y2)2dy

)dx = 0.

48

so two repeated integrals exist and they coincide, but f is not integrable. In fact, f is notintegrable on (0, 1] (0, 1] E, since

10

xy

(x2 + y2)2dy =

1

2x x

2(x2 + 1)

which is not integrable on (0, 1], so by Fubinis theorem, f can not be integrable on (01]2, neitheron [1, 1]2.

Example 2. The repeated integral 10

( x0

1 yx ydy

)dx

may be written as

I =

10

( 10

1 yx y1{y

so that 0

( 0

yey2(1+x2)dy

)dx =

0

1

2

1

1 + x2dx =

pi

4.

Thus f is integrable on (0,) (0,). By Fubinis theorem, the other related integral of f onE is pi

4as well. Thus

pi

4=

0

( 0

yey2(1+x2)dx

)dy =

0

(ey

2

0

ey2x2)ydx

)dy

=

0

(ey

2

0

ex2

dx

)dy =

( 0

ex2

dx

)2and therefore

0

ex2

dx =

pi

2.

Another way to calculate the integral I =0ex

2dx can be described as the following.

Consider I2 and write it as

I2 =

0

ey2

dy

0

ex2

dx =

0

(ey

2

0

ex2

dx

)dy

=

0

( 0

e(x2+y2)dx

)dy.

According to Proposition 6.2 0

( 0

e(x2+y2)dx

)dy =

E

e(x2+y2) = lim

n

E

e(x2+y2)1{x2+y2n}

where the last equality follows from DCT. On the other hand, the integralEe(x

2+y2)1{x2+y2n2}coincides with the integral which can be calculated in terms of polar coordinates, so that

E

e(x2+y2)1{x2+y2n2} =

0

Outline of the proof. [Not examinable] By MCT, we only need to show the statement is validfor f = 1E where E (c, d) is a measurable subset. While it is true if E = J (c, d) is aninterval. Hence, by definition, we can show [in this step, you have to apply the definition of theouter measure m carefully] that

m(E) =

dc

1E((t))(t)dx

for any measurable subset E.Example 4. Consider the Fourier transform of ex

2. Note that

ex2 1

1 + x2

so that g(x) = ex2is integrable on R. For s R consider f(x) = eisxg(x). Then f is

measurable and |f | g, thus f is integrable. Moreover

f(x) = eisxex2

=n=0

(is)nn!

xnex2

.

Now xnex2is integrable, so that if n = 2m+ 1 is odd then, by DCT we have

x2m+1ex2

dx = limN

x2m+1ex

2

1[N,N ]dx

= limN

NN

x2m+1ex2

dx = 0.

If n = 2m is even,

x2mex2

dx = 2

0

x2mex2

dx =

0

zm12 ezdz

[we have made change of variable z = x2 to obtain the last equality], so that

x2mex2

dx = (m+1

2) =

0

zm12 ezdz

=

(m 1

2

)(m 1 1

2

) 1

2

(1

2

)

=(2m 1)!!

2m

(1

2

).

Hence

n=0

(is)nn! xnex2 =

m=0

s2m

(2m)!

(2m 1)!!2m

(1

2

)

=

m=0

s2m

m!

1

4m

(1

2

)= e

s2

4

(1

2

)< ,

51

and therefore, according to MCT (series version)

f(s) =

eisxex2

dx =n=0

(is)nn!

xnex2

dx

=

m=0

(is)2m(2m)!

(2m 1)!!2m

(1

2

)= e

s2

4

(1

2

).

On the other hand

(1

2

)=

0

z12 ezdz =

0

2ex2

dz =pi

so that the Fourier transform of ex2ispie

s2

4 .

6.4 Change of variables

We have the following theorem about change of variables.

Theorem 6.6 Let G be an open subset of R2, and

T : G E = T (G) R2

be one-to-one, and differentiable. Suppose f : E R is measurable. Then f is integrable if andonly if f T | det JT | is integrable on G, where det JT is the Jacobian of T . In that case

E

f =

G

f T | det JT |

Recall that if we write T (u, v) = (x(u, v), y(u, v)) then

JT =

(xu

xv

yu

yv

).

For example, if we use polar coordinate T (r, ) = (x, y), where x = r cos , y = r sin then| det JT | = r. Suppose that E = T (G) and f is measurable. Then f is integrable on E if andonly if

f(r cos , r sin )r

is integrable on G.

7 Lp-space and convergencies

Let (E,F , ) be a measure space. For example E is a measurable subset of R or R2, F =MLeband = m is the Lebesgue measure on E.

Suppose that f : E [,] is measurable, and suppose that p is a positive constant.Then |f |p is measurable. If

E|f |p

is again equivalent to thatE|f |p < . The space of all p-th integrable measurable complex

functions is denoted again by Lp(E) (sorry we use the same notation as for real functions). Lp(E)is a linear space over the complex field. In fact it is obvious that if f Lp(E) then f Lp(E)for any constant . Since

|f + g|p 2p(|f |p + |g|p)f + g Lp(E) for any f, g Lp(E).

For the case that p = 1, define

d1(f, g) =

E

|f g|

called the L1-metric. Since E

|f + g| E

|f |+E

|g|

so that the triangle inequality holds

d1(f, g) d1(f, h) + d1(h, g).

Moreover

d1(f, g) =

E

|f g| = 0

if and only if f = g almost everywhere on E. Therefore, if we identify the functions which areequal almost everywhere on E [i.e. we identify an integrable function f as its equivalent class:if f = g almost surely on E, then we regard f and g as the same function on E], then L1(E) isa metric space.

We wish to define the Lp-metric. If f is measurable, then we define

||f ||p =(

E

|f |p) 1

p

which can be infinity in the case that f is not p-th integrable over E. Thus, a measurable functionf Lp(E) if and only if ||f ||p

To prove (7.1), we recall that a function : (a, b) R is convex, if

(s+ (1 )t) (s) + (1 )(t) (7.2)

for all s, t (a, b) and [0, 1]. From Prelims Analysis II, we know that, if the second derivative exists and is non-negative on (a, b) then h is convex on (a, b).

Let us consider the function (t) = tp for t > 0, where p 1, then (t) = p(p 1)tp2 isnonnegative on (0,) so that

(s+ (1 )t)p sp + (1 )tp (7.3)

for all s, t 0 and [0, 1].We are now in a position to prove the Minkowski inequality (7.1).If ||f ||p or ||g||p vanishes, then ||f + g||p = ||f ||p + ||g||p, so (7.1) is an equality in this case.

Suppose that ||f ||p 6= 0 and ||g||p 6= 0. Apply (7.3) with s = |f |||f ||p and t =|g|||g||p

to obtain

(|f |||f ||p + (1 )

|g|||g||p

)p |f |

p

||f ||pp + (1 )|g|p||g||pp

for any [0, 1]. Integrating both sides of the inequality over E we obtainE

(

||f ||p |f |+1 ||g||p |g|

)p

E

|f |p||f ||pp + (1 )

E

|g|p||g||pp = 1. (7.4)

Choose such that

||f ||p =1 ||g||p

i.e.

=||f ||p

||g||p + ||f ||pwhich belongs to [0, 1], and plug the value into (7.4) to deduce that

E

(|f |+ |g|)p (||g||p + ||f ||p)p .

Therefore E

|f + g|p E

(|f |+ |g|)p (||g||p + ||f ||p)p

which yields (7.1).

Proposition 7.2 Let p 1. Then || ||p is a norm on Lp(E). That is, the mapping f ||f ||ppossesses the following three properties:

1) ||f ||p = 0 if and only if f = 0 almost everywhere on E,2) ||f ||p = ||||f ||p if f Lp(E) and is a constant,3) Triangle inequality holds:

||f + g||p ||f ||p + ||g||p.

54

Let us derive another important inequality related to Lp-norms.

Proposition 7.3 (Holders inequality) Let 1 < p, q < be a pair of conjugate numbers in thesense that

1

p+

1

q= 1

(for example, if p = 2, then q = 2, and if p = 3 then q = 32, and etc.), and let f Lp(E) and

g Lq(E). Then fg L1(E), and E

|fg| ||f ||p||g||q .

Proof. (The proof is not examinable). Let us prove the following elementary but very usefulinequality:

st sp

p+tq

q(7.5)

for all s, t 0. If t = 0 then there is nothing to prove. Suppose t > 0. Then the inequality aboveis equivalent to that

s

tq1 1p

sp

tq+

1

q=

1

p

( stq/p

)p+

1

q.

Since qp= q 1, (7.5) is equivalent to [with x = s

tq1]

x 1pxp +

1

qfor all x > 0. (7.6)

Let us prove (7.6). Consider h(x) = x 1pxp 1

q. Then h(0) = 1

q< 0, h(x) , and

h(x) = 1 xp1, thus x = 1 is the global maximum, and h(1) = 0. Therefore h(x) 0 for allx > 0, which proves (7.6) and therefore (7.5).

Let us now prove the Holder inequality. If ||f ||p = 0 or ||g||q = 0, then there is nothing toprove, so assume that both ||f ||p 6= 0 and ||g||q 6= 0. Apply (7.5) with

s =|f |||f ||p and t =

|g|||g||p

and integrate over E to obtainE

|fg|||f ||p||g||p

1

p

E

|f |p||f ||pp +

1

q

E

|g|q||g||qq = 1

which yields the Holder inequality.Example 1. Let E be measurable with m(E) < . Suppose that 1 < p1 < p2, and suppose

that f is measurable. Let p, q be a conjugate pair: 1p+ 1

q= 1, where 1 < p < . Apply Holder

inequality to |f |p1 and g = 1, to obtainE

|f |p1 =E

|f |p11 (

E

|f |p1p) 1

p(

E

1q) 1

q

= m(E)1q

(E

|f |p1p) 1

p

. (7.7)

55

Choose p > 1 such that p1p = p2 which is possible and p =p2p1> 1 by assumption that

p2 > p1. Moreover1p= p1

p2and

1 =1

p+

1

q=p1p2

+1

q

which yields that1

q= 1 p1

p2=p2 p1p2

and q =p2

p2 p1 .

Inserting these values into (7.7) we obtain

E

|f |p1 m(E)p2p1p2

(E

|f |p2) p1

p2

and therefore||f ||p1 m(E)

p2p1p1p2 ||f ||p2 .

Corollary 7.4 1) If m(E)

Proof. If ||f || = 0, then there is nothing to prove as ||f ||p = 0 for all p. Assume that||f || > 0. Then (E) > 0. For every > 0 there is a null set A E such that |f(x)| ||f ||+for all x E \ A, hence

||f ||p (||f || + ) ((E))1p

since ((E))1p 1 as p,

lim sup ||f ||p ||f || + ,and since > 0 is arbitrary, we must have lim sup ||f ||p ||f ||. On the other hand, for every (0, ||f ||), there is a subset B E with (B) > 0 and |f(x)| ||f || for all x B.Therefore

||f ||p (

B

|f |p) 1

p

(||f || )(B)1p ||f ||

as p , which yields that lim inf ||f ||p ||f || . Letting 0 to obtain lim inf ||f ||p ||f ||. This completes the proof.

7.1 Convergence in measure

Recall the following definition.

Definition 7.6 fn f almost everywhere on E, if there is a null subset A E, such thatfn(x) f(x) for every x E \ A.

Suppose that fn, f : E C are measurable, then, x {fn f}, if and only if every > 0,there is N such that x {|fn f | } for all n N , that is, if and only if for every > 0

x

N=1

n=N

{|fn f | }.

Thus, fn f almost everywhere on E if and only if for every > 0,

m

(

N=1

n=N

{|fn f | > })= 0.

In particular, if m(E) < , fn, f : E R are measurable, then fn f almost everywhereon E if and only if for every > 0

limN

m

(

n=N

{|fj f | > })= 0.

Therefore, if m(E) }) 0 as n.

Definition 7.7 Let E R be a measurable subset, and fn, f : E R be measurable. Thenfn f in measure if for every > 0

m ({|fn f | > }) 0 (7.9)as n, where

{|fn f | > } = {x E : |fn(x) f(x)| > }.

57

If m(E) 1

2k

})

1

2j

}

so that

m(E \G) m(

j=k

{|fnj f | >

1

2j

})

j=k

m

({|fnj f | >

1

2j

})

j=k

1

2j

for any k. Since

j=k12j 0 as k , therefore m(E \G) = 0. The proof is complete.

7.2 Convergence in Lp-space

Recall that for p 1, Lp(E) is a vector space, and dp(f, g) = ||f g||p is a metric on Lp(E).

Definition 7.9 Let p 1. We say fn f in Lp-norm if ||fn f ||p 0 as n.

Proposition 7.10 If fn f in Lp-norm, then fn f in measure, and therefore there is asubsequence (fnk) such that fnk f almost surely.

58

Proof. For any > 0 we have

m({|fn f | > }) =E

1{|fnf |>} E

|fn f |pp

1{|fnf |>}

1p

E

|fn f |p = 1p||fn f ||pp

which goes to zero as n. The second claim then follows from Proposition 7.8.Theorem 7.11 Let p 1. Lp(E) with the metric dp(f, g) = ||f g||p is a complete metricspace. That is, if (fn) is a Cauchy sequence in L

p(E): fn Lp(E) and||fn fm||p 0 as n,m

then1) there is a subsequence (fnk) and f Lp(E) such that fnk f almost surely on E, and2) ||fn f ||p 0 as n.Therefore Lp(E) equipped with the norm || ||p is a Banach space.Proof. (The proof is not examinable) 1) The proof is the modification of that of Proposition

7.8. Since ||fn fm||p 0 so for every k = 1, 2, , there is nk such that

||fn fm||pp 1

2k

(1

2k

)pfor any n,m nk.

We can choose nk such that nk is strictly increasing. Then

m

({|fn fm| 1

2k

})=

E

1{|fnfm| 12k}

E

|fn fm|p(12k

)p=

||fn fm||pp(12k

)p 12k

n,m nk. (7.10)

Let

Ek =j=k

{|fnj+1 fnj |

1

2j

}, G =

k=1

Ek

Then if x G, |fni+1(x) fni(x)| 12i for all i k for some k, so that

|fni(x) fnj(x)| i=j

|fni+1(x) fni(x)| i=j

1

2i=

1

2j1i j k

and therefore (fnk(x)) converges to some f(x) for every x G. On the other hand

E \G =k=1

j=k

{|fnj+1 fnj |

1

2j

}

so that, by (7.10)

m(E \G) j=k

m

({|fnj+1 fnj |

1

2j

})

j=k

1

2j=

1

2k1

59

for any k = 1, 2, , hence m(E \G) = 0. Therefore (fnk) converges to f almost surely.We next show that f Lp(E) and ||fn f ||p 0 as n . For every > 0 there is N

such that ||fn fm||p < for all n,m N . ThereforeE

|fn fnk |p < p for any k, n N .

Since |fn fnk | |fn f | almost surely on E, so by Fatous LemmaE

|fn f |p limkE

|fn fnk |p p n N ,

therefore fn f Lp(E) and ||fn f ||p . By definition fn f in Lp(E) and f Lp(E).

60