Constantin Carath´eodory (1873–1950) - Elsevier.comA COURSE IN REAL ANALYIS 2/E John N. McDonald...

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Constantin Caratheodory(1873–1950)

Constantin Caratheodory was born on September 13,1873, in Berlin, Germany, to Greek parents. His fa-ther, Stephanos, was an Ottoman Greek who hadstudied law in Berlin; his mother, Despina, camefrom a Greek family of businessmen. At the timeof Caratheodory’s birth, his family was in Berlin be-cause his father had been appointed there as FirstSecretary to the Ottoman Legation.

Caratheodory was a student at the Belgian Military Academy from 1891–1895 and then worked as an assistant engineer with the British Asyut Damproject in Egypt. Returning to Germany, Caratheodory entered the Universityof Berlin in 1900 to begin his study of mathematics. In 1902, he transferredto the University of Gottingen, where he received his doctorate in 1904 underHermann Minkowski.

The works of Caratheodory encompassed many disciplines. Among themwere a simplified proof of a central theorem in conformal representation, exten-sions of findings by Picard on function theory, and the amplification of measuretheory begun by Emile Borel and Henri Lebesgue. Regarding the latter disci-pline, two particularly important results in measure theory are the Caratheodorycriterion for measurability and the Caratheodory extension theorem.

Caratheodory wrote many outstanding books. Two of particular relevancehere are Vorlesungen Uber Reelle Funktionen (Lectures on Real Functions) andMass und Integral und Ihre Algebraisierung (Measure and Integral and theirAlgebraisation).

Caratheodory taught at the Universities of Hannover (1909), Breslau (1910–1913), Gottingen (1913–1918), Berlin (1918–1920), all in Germany. In 1920,he went to Anatolia, Greece, to establish the University of Smyrna. After theTurks burned Smyrna in 1922, he went to the University of Athens. In 1924,he returned to the University of Munich where he remained, except for oneyear (1936-1937) at the University of Wisconsin, until his death on Febru-ary 2, 1950.

144


5Elements of Measure Theory

In Chapter 3, we expanded the collection of continuous functions to the collectionof Borel measurable functions, the smallest algebra that contains the continuousfunctions and is closed under pointwise limits. Subsequently, in Chapter 4, weextended the Riemann integral so that it applies to all Borel measurable functionsand, in doing so, we encountered Lebesgue measure, the collection of Lebesguemeasurable functions, and the Lebesgue integral.

We will discover, in this chapter, that the concepts and methods of Chapters 3and 4 lend themselves to considerable generalization with relatively little effortand huge rewards. This generalized theory has extensive applications throughoutmathematics and, as well, to a large variety of fields outside of mathematics.

5.1 MEASURE SPACES

When we examine the definition of the Lebesgue integral carefully, we find that itdepends ultimately on the concept of measure. More precisely, the mathematicalframework requires a set, a σ-algebra of subsets, and a set function that assigns toeach set in the σ-algebra a nonnegative number (its measure). In Chapter 3, thisconsisted, respectively, of R, M, and λ. But we can abstract the mathematicalframework to provide a broader setting for the integral. We begin by consideringthe general concept of measure.

In developing Lebesgue measure, we imposed three conditions; namely, Con-ditions (M1)–(M3) on page 91. The first two conditions are specific to thegeneralization of length; but the third is not. In fact, Condition (M3), thecountable-additivity condition, is the primary property of an abstract measure.

145A Course in Real Analysis. Second edition. Copyright © 1999, 2013, Elsevier Inc. All rights reserved.


146 Chapter 5 Elements of Measure Theory

DEFINITION 5.1 Measure, Measurable Space, Measure Space

Let Ω be a set and A a σ-algebra of subsets of Ω. A measure μ on A is anextended real-valued function satisfying the following conditions:a) μ(A) ≥ 0 for all A ∈ A.b) μ(∅) = 0.c) If A1, A2, . . . are in A, with Ai ∩Aj = ∅ for i �= j, then

μ

(⋃n

An

)=∑n

μ(An).

The pair (Ω,A) is called a measurable space and the triple (Ω,A, μ) is calleda measure space.

Note: We will often refer to members of A as A-measurable sets.We should point out the following fact: If μ satisfies (a) and (c) of Defini-

tion 5.1, then it is a measure (i.e., also satisfies (b)) if and only if there is anA ∈ A such that μ(A) < ∞. We leave the proof of this fact to the reader.

EXAMPLE 5.1 Illustrates Definition 5.1

a) (R,M, λ) is a measure space, the one that we studied in Chapter 3.b) (R,B, λ|B) is a measure space.c) Let (Ω,A, μ) be a measure space. For D ∈ A, define AD = {D ∩A : A ∈ A}

and μD = μ|AD. Then AD is a σ-algebra of subsets of D, μD is a measure

on AD, and, hence, (D,AD, μD) is a measure space.d) Referring to part (c), let Ω = R, A = M, μ = λ, and D = [0, 1]. Then

([0, 1],M[0,1], λ[0,1]) is a measure space. λ[0,1] is called Lebesgue measureon [0, 1]. More generally, if D is a Lebesgue measurable set, then (D,MD, λD)is a measure space and λD is called Lebesgue measure on D.

e) Refer to part (c). By Theorem 3.7 on page 88, if D ∈ B, then BD = B(D).f) Let Ω be a nonempty set and A = P(Ω). Define μ on A by

μ(E) =

{N(E), if E is finite;∞, if E is infinite.

where N(E) denotes the number of elements of E. Then μ is a measure on Aand is called counting measure.

g) Let Ω = N , A = P(N ), and μ be counting measure on A, as defined inpart (f). Then, for instance, μ(N ) = ∞ and μ({1, 3}) = 2. We will see laterthat (N ,P(N ), μ) is the appropriate measure space for the analysis of infiniteseries.

h) Suppose that (Ω,A, μ) is a measure space. If μ(Ω) = 1, then (Ω,A, μ) iscalled a probability space and μ a probability measure. Furthermore,μ is usually replaced by a P (for probability). Two simple examples are asfollows:(i) ([0, 1],M[0,1], λ[0,1]) is a probability space since λ([0, 1]) = 1. It is an

appropriate measure space for analyzing the experiment of selecting anumber at random from the unit interval.


5.1 Measure Spaces 147

(ii) Consider the experiment of tossing a coin twice. The set of possible out-comes for that experiment is Ω = {HH, HT, TH, TT} where, for instance,HT denotes the outcome of a head on the first toss and a tail on the sec-ond toss. Set A = P(Ω) and, for E ∈ A, define P (E) = N(E)/4 where,as before, N(E) denotes the number of elements of E. Then (Ω,A, P )is a probability space — the appropriate measure space to use when thecoin is balanced (i.e., equally likely to come up heads or tails). To il-lustrate: The probability of getting at least one head in two tosses of abalanced coin is P ({HH, HT, TH}) = 3/4.

i) Let Ω be a nonempty set, {xn}n a sequence of distinct elements of Ω, and{an}n a sequence of nonnegative numbers. For E ⊂ Ω, define

μ(E) =∑xn∈E

an, (5.1)

where the notation∑

xn∈E means the sum over all indices n such that xn ∈ E.Then μ is a measure on P(Ω) and, consequently, (Ω,P(Ω), μ) is a measurespace. Here are two special cases:(i) If Ω is countable, {xn}n is an enumeration of Ω, and an = 1 for all n,

then the measure μ defined in (5.1) is counting measure.(ii) If the sequence {xn}n consists of only one element, say x0, and if a0 = 1,

then the measure μ defined in (5.1) takes the form

μ(E) =

{1, if x0 ∈ E;0, if x0 �∈ E.

This measure is denoted by δx0and is called the unit point mass or

Dirac measure concentrated at x0. Note that δx0 is a probabilitymeasure.

j) Let (Ω,A) be a measurable space such that {x} ∈ A for each x ∈ Ω. Ameasure μ on A is called discrete if there is a countable set K ⊂ Ω suchthat μ(Kc) = 0. It is not too difficult to show that if μ is a discrete measure,then we can write μ =

∑x∈K μ({x})δx. See Exercises 5.6 and 5.19 for more

on discrete measures.

The following theorem provides some important properties of measures. Weleave the proof as an exercise for the reader.

THEOREM 5.1

Suppose that (Ω,A, μ) is a measure space and that A and B are A-measurablesets. Then the following properties hold:

a) If μ(A) < ∞ and A ⊂ B, then μ(B \A) = μ(B) − μ(A).

b) A ⊂ B ⇒ μ(A) ≤ μ(B). (monotonicity)

c) If {En}∞n=1 ⊂ A with E1 ⊃ E2 ⊃ · · · and μ(E1) < ∞, then

μ

( ∞⋂n=1

En

)= lim

n→∞μ(En).



d) If {En}∞n=1 ⊂ A with E1 ⊂ E2 ⊂ · · · , then

μ

( ∞⋃n=1

En

)= lim

n→∞μ(En).

e) If {En}n ⊂ A, then

μ

(⋃n

En

)≤∑n

μ(En).

This property is called countable subadditivity.

Almost Everywhere and Complete Measure Spaces

Recall from Section 4.4 that a property holds Lebesgue almost everywhere (λ-ae)if it holds except on a set of Lebesgue measure zero. That concept can begeneralized to apply to any measure space.

DEFINITION 5.2 Almost Everywhere

A property is said to hold μ almost everywhere, or μ-ae for short, if it holdsexcept on a set of μ-measure zero, that is, except on a set N with μ(N) = 0.

Note: Several terms are used synonymously for “almost everywhere.” Here are afew: almost always, for almost all x ∈ Ω, and, in probability theory, almostsurely, with probability one, and almost certainly.

Proposition 3.4 on page 106 implies that subsets of Lebesgue measurable setsof Lebesgue measure zero are also Lebesgue measurable sets. On the otherhand, Exercise 4.48 on page 142 indicates that there exist subsets of Borel setsof Lebesgue measure zero that are not Borel sets.

Those two facts have relevance to almost-everywhere (ae) properties of mea-surable functions. For instance, by Proposition 4.7 on page 140, if f is Lebesguemeasurable and g = f λ-ae, then g is Lebesgue measurable. However, as Exer-cise 4.49 on page 142 shows, that result is not true for Borel measurable functions.

We now see that it is important to know whether subsets of sets of measurezero are measurable sets. Hence, we make the following definition.

DEFINITION 5.3 Complete Measure Space

A measure space (Ω,A, μ) is said to be complete if all subsets of A-measurablesets of μ-measure zero are also A-measurable; in other words, if A ∈ A andμ(A) = 0, then B ∈ A for all B ⊂ A.

Thus, (R,M, λ) is a complete measure space, whereas (R,B, λ|B) is not acomplete measure space.


5.1 Measure Spaces 149

The following theorem shows that any measure space can be extended to acomplete measure space. We leave the proof of the theorem as an exercise forthe reader.

THEOREM 5.2

Let (Ω,A, μ) be a measure space. Denote by A the collection of all sets of theform B ∪A where B ∈ A and A ⊂ C for some C ∈ A with μ(C) = 0. For suchsets, define μ(B ∪A) = μ(B). Then A is a σ-algebra, μ is a measure on A,and (Ω,A, μ) is a complete measure space. Furthermore, A ⊂ A and μ|A = μ.

(Ω,A, μ) is called the completion of (Ω,A, μ).

It can be shown that the measure space (R,M, λ) is the completion of themeasure space (R,B, λ|B). See Exercise 5.16.

Exercises for Section 5.1Note: A ★ denotes an exercise that will be subsequently referenced.

5.1 Suppose that (Ω,A, μ) is a measure space and that D is an A-measurable set. DefineAD = {D ∩A : A ∈ A} and μD = μ|AD

. Show that (D,AD, μD) is a measure space.

5.2 Let Ω be a nonempty set and A = P(Ω). Define μ on A by

μ(E) ={N(E), if E is finite;∞, if E is infinite.

where N(E) denotes the number of elements of E. Prove that μ is a measure on A.

5.3 Consider the experiment of selecting a number at random from the interval [−1, 1].a) Construct an appropriate probability space for this experiment.b) Determine the probability that the number selected exceeds 0.5.c) Determine the probability that the number selected is rational.

5.4 Let (Ω,A) be a measurable space, μ and ν measures on A, and α > 0. Define setfunctions μ + ν and αμ on A by

(μ + ν)(A) = μ(A) + ν(A), (αμ)(A) = αμ(A).

a) Show that μ + ν is a measure on A.b) Show that αμ is a measure on A.

5.5 Let (Ω,A) be a measurable space, {μn}∞n=1 a sequence of measures on A, and {αn}∞n=1

a sequence of nonnegative real numbers. Define∑∞

n=1αnμn on A by( ∞∑

n=1

αnμn

)(A) =

∞∑n=1

αnμn(A).

Prove that∑∞

n=1αnμn is a measure on A.

★5.6 Refer to Example 5.1(j). Let (Ω,A) be a measurable space and suppose that {x} ∈ Afor each x ∈ Ω. Show that a measure μ on A is discrete if and only if there is a countablesubset K of Ω such that μ =

∑x∈K

μ({x})δx.

5.7 Suppose that a balanced coin is tossed three times.a) Construct a probability space for this experiment in which each possible outcome

is equally likely.



b) Determine the probability of obtaining exactly two heads.c) Express the probability measure P as a finite linear combination of Dirac measures.

5.8 Let Ω be a nonempty set, {xn}n a sequence of distinct elements of Ω, and {an}n asequence of nonnegative real numbers. For E ⊂ Ω, define

μ(E) =∑xn∈E

an.

a) Show that μ is a measure on P(Ω).b) Interpret the ans in terms of the measure μ.c) Express μ as a linear combination of Dirac measures.

5.9 Suppose that two balanced dice are thrown.a) Construct a probability space for this experiment in which each possible outcome

is equally likely.b) Use part (a) to determine the probability that the sum of the dice is seven or 11.c) Construct a probability space for this experiment in which the outcomes consist of

the possible sums of the two dice.d) Use part (c) to determine the probability that the sum of the dice is seven or 11.

5.10 Prove Theorem 5.1.

5.11 Let (Ω,A) be a measurable space. A measure μ on A is called a finite measureif μ(Ω) < ∞. A measure space (Ω,A, μ) is called a finite measure space if μ is afinite measure. For a finite measure space, prove the following:a) If A and B are A-measurable sets, then

μ(A ∪B) = μ(A) + μ(B) − μ(A ∩B).

b) Generalize part (a) to an arbitrary finite number of A-measurable sets.

5.12 Let {En}∞n=1 be a sequence of A-measurable sets. Prove that

μ

( ∞⋃n=1

( ∞⋂k=n

Ek

))≤ lim inf

n→∞μ(En).

5.13 Let {En}∞n=1 be a sequence of A-measurable sets with μ(⋃∞

n=1En

)< ∞. Prove that

μ

( ∞⋂n=1

( ∞⋃k=n

Ek

))≥ lim sup

n→∞μ(En).

★5.14 Let (Ω,A, μ) be a measure space and {En}∞n=1 a sequence of A-measurable sets. DefineE = {x : x ∈ En for infinitely many n }.a) Prove that E =

⋂∞n=1

(⋃∞k=n

Ek

).

b) Prove that∑∞

n=1μ(En) < ∞ ⇒ μ(E) = 0.

5.15 Prove Theorem 5.2.

5.16 Prove that (R,M, λ) is the completion of (R,B, λ|B). Use the following steps:

a) Verify that B ⊂ M by employing Exercise 3.32 on page 107.b) Show that B ⊃ M by applying Exercise 3.44 on page 108.c) Prove that λ = λ|B. Hint: Use the fact established in parts (a) and (b) that M = B.

★5.17 Let (Ω,A, μ) be a measure space. Suppose that (Ω,F , ν) is a complete measure spacewith F ⊃ A and ν|A = μ. Prove that F ⊃ A and that ν|A = μ. Conclude that (Ω,A, μ)

is the smallest complete measure space that contains (Ω,A, μ).


5.2 Measurable Functions 151

5.18 Let f be a nonnegative M-measurable function. Define μf on M by μf (E) =∫Ef dλ.

Prove that μf is a measure on M.

5.19 Let (Ω,A, μ) be a measure space such that {x} ∈ A for each x ∈ Ω. An element x ∈ Ωis said to be an atom of μ if μ({x}) > 0. Assume now that μ is a finite measure, thatis, μ(Ω) < ∞. Prove the following facts.a) μ has only countably many atoms.b) μ can be expressed uniquely as the sum of two measures, μc and μd, where μc has

no atoms and μd is discrete. Moreover, we have that μd =∑

x∈Kμ({x})δx, where

K is the set of atoms of μ.

5.2 MEASURABLE FUNCTIONS

The next step in developing the abstract Lebesgue integral is to introduce theconcept of measurability for functions defined on an abstract space. In additionto real-valued functions, we will also consider complex-valued and extended real-valued functions. We begin with real-valued functions.

Real-Valued Measurable Functions

Let (Ω,A) be a measurable space and f : Ω → R. We want to specify when f ismeasurable. In previous chapters, we discussed two kinds of measurable func-tions: Borel measurable functions and Lebesgue measurable functions. Recallthat a real-valued function f is Borel measurable if and only if f−1(O) ∈ B foreach open set O ⊂ R and it is Lebesgue measurable if and only if f−1(O) ∈ M foreach open set O ⊂ R. Hence, it is quite natural to make the following definition.

DEFINITION 5.4 Real-Valued Measurable Function

Let (Ω,A) be a measurable space. A real-valued function f on Ω is said to be anA-measurable function if the inverse image of each open subset of R under fis an A-measurable set, that is, if f−1(O) ∈ A for all open sets O ⊂ R.


a) Let Ω = R. Then, as we know from Chapters 3 and 4, the Borel measur-able functions are the B-measurable functions, and the Lebesgue measurablefunctions are the M-measurable functions.

b) Let (Ω,A) be a measurable space, D ∈ A, and AD = {D ∩A : A ∈ A}. Thena function f :D → R is AD-measurable if and only if for each open subset Oof R, f−1(O) is of the form D ∩A for some A ∈ A.

c) Every real-valued function on a nonempty set Ω is P(Ω)-measurable. An im-portant special case: If Ω = N , then A is usually taken to be P(N ); hence,all functions f :N → R are A-measurable. But functions on N are infinite se-quences. Consequently, in this case, the A-measurable functions are preciselythe infinite sequences.



The following proposition provides some useful equivalent conditions for afunction to be A-measurable. To prove the proposition, we proceed in a similarmanner as we did in the proof of Lemma 3.5 on page 85.

PROPOSITION 5.1

Let (Ω,A) be a measurable space and f a real-valued function on Ω. Then thefollowing statements are equivalent:a) f is A-measurable.b) For each a ∈ R, f−1

((−∞, a)

) ∈ A.

c) For each a ∈ R, f−1((a,∞)

) ∈ A.

d) For each a ∈ R, f−1((−∞, a]

) ∈ A.

e) For each a ∈ R, f−1([a,∞)

) ∈ A.

Theorem 5.3, which we prove next, gives several important properties of real-valued A-measurable functions. Theorem 4.1 on page 113 is a special case.

THEOREM 5.3

Let (Ω,A) be a measurable space. The collection of all real-valued A-measurablefunctions forms an algebra. In other words, if f and g are A-measurable andα ∈ R, thena) f + g is A-measurable.b) αf is A-measurable.c) f · g is A-measurable.

PROOF a) By Proposition 5.1, to prove that f + g is A-measurable, it suffices to showthat {x : f(x) + g(x) > a } ∈ A for each a ∈ R. Now,

{x : f(x) + g(x) > a } = {x : f(x) > a− g(x) }=⋃r∈Q

{x : f(x) > r > a− g(x) }

=⋃r∈Q

(f−1

((r,∞)

) ∩ g−1((a− r,∞)

)).

This last union is an A-measurable set since f and g are A-measurablefunctions, A is a σ-algebra, and Q is countable. Consequently, f + g is anA-measurable function.

b) If α = 0, then αf ≡ 0, which is A-measurable (why?). So, assume α �= 0and let O be any open set in R. Then α−1O = {α−1y : y ∈ O } is open.Therefore, because f is A-measurable, (αf)−1(O) = f−1(α−1O) ∈ A. Thisfact proves that αf is A-measurable.

c) First we show that if f is A-measurable, then so is f2. If a < 0, then(f2)−1((a,∞)) = Ω ∈ A. If a ≥ 0, then we have

(f2)−1((a,∞)

)= {x : f(x)2 > a } = {x : f(x) >

√a } ∪ {x : f(x) < −√

a }= f−1

((√a,∞)

) ∪ f−1((−∞,−√

a)).

This last union is an A-measurable set because f is A-measurable. Hence,f2 is an A-measurable function whenever f is.



Now, for any two functions f and g, we can write

f · g =1

4

((f + g)2 − (f − g)2

).

Applying parts (a) and (b) of this theorem and the fact that the squareof an A-measurable function is A-measurable, we conclude that f · g is anA-measurable function.

We should emphasize that the measurability (or nonmeasurability) of a func-tion depends only on the σ-algebra, A, of subsets of Ω; that is, it has nothingto do with a measure. Nonetheless, if (Ω,A, P ) is a probability space, then theA-measurable functions are called random variables.

Thus, an A-measurable function is a random variable only when consideredin the context of a probability space. By the way, in probability theory, randomvariables are usually denoted by uppercase italicized English-alphabet lettersthat are near the end of the alphabet (e.g., X, Y , and Z) instead of the moreusual f , g, and h.

EXAMPLE 5.3 Illustrates Random Variables

Let (Ω,A, P ) be the probability space from subpart (ii) of Example 5.1(h)on page 147. Define X(HH) = 2, X(HT) = X(TH) = 1, and X(TT) = 0. ThenX: Ω → R is a random variable. It indicates the number of heads obtained whena balanced coin is tossed twice.

Our next result is a generalization of Proposition 4.7 on page 140 to an ar-bitrary complete measure space. Its proof is essentially identical to that ofProposition 4.7.

PROPOSITION 5.2

Suppose that (Ω,A, μ) is a complete measure space. If f is A-measurable andg = f μ-ae, then g is A-measurable.

Complex-Valued Measurable Functions

In applying real analysis, we often encounter complex-valued functions. Thisoccurs, for instance, in Fourier analysis. We will denote the set of all complexnumbers by C. Here now is the definition of measurability for complex-valuedfunctions.

DEFINITION 5.5 Complex-Valued Measurable Function

Let (Ω,A) be a measurable space. A complex-valued function f on Ω is said tobe an A-measurable function if the inverse image of each open subset of Cunder f is an A-measurable set, that is, if f−1(O) ∈ A for all open sets O ⊂ C.

The following theorem provides a useful characterization of measurability forcomplex-valued functions. We leave the proof of the theorem as an exercise forthe reader.



THEOREM 5.4

A complex-valued function f on Ω is A-measurable if and only if both its realpart �f and its imaginary part �f are (real-valued) A-measurable functions.

EXAMPLE 5.4 Illustrates Complex-Valued Measurable Functions

a) If f is a real-valued A-measurable function on Ω, then it is also a complex-valued A-measurable function.

b) Let Ω = R and A = B. Define f :R → C by f(x) = eix. The real and imagi-nary parts of f(x) are cosx and sinx, respectively. Since those two functionsare continuous, they are B-measurable. Consequently, by Theorem 5.4, f isa complex-valued B-measurable function.

c) If g and h are real-valued A-measurable functions, then, by Theorem 5.4, thecomplex-valued function f = g + ih is also A-measurable.

d) Let {an}∞n=1 be a sequence of complex numbers and define f :N → C byf(n) = an. Then f is a complex-valued P(N )-measurable function.

Theorem 5.3 holds also for complex-valued A-measurable functions. Thatis, the collection of complex-valued A-measurable functions forms a (complex)algebra. See Exercise 5.32.

Extended Real-Valued Measurable Functions

In addition to real- and complex-valued functions, we frequently must deal withextended real-valued functions, in other words, functions that take valuesin R∗ = R∪ {−∞,∞}. This is especially so when considering suprema, infima,and limits. For instance, define

fn(x) =n√2π

e−(nx)2/2

for x ∈ R and n ∈ N . Then, as n → ∞, fn(x) → 0 if x �= 0 and fn(0) → ∞.Consequently, the sequence {fn}∞n=1 of real-valued functions converges pointwiseto the extended real-valued function f , where

f(x) =

{0, if x �= 0;∞, if x = 0.

Thus, we next consider measurability for extended real-valued functions. Re-call that, by definition, a real-valued function f is A-measurable if f−1(O) ∈ Afor all open sets O ⊂ R. Also, by definition, a complex-valued function f isA-measurable if f−1(O) ∈ A for all open sets O ⊂ C. Hence, once we iden-tify the open sets of R∗, we have a natural way to define extended real-valuedA-measurable functions.

DEFINITION 5.6 Open Subsets of the Extended Real Numbers

A subset of R∗ is said to be open if it can be expressed as a union of intervalsof the form (a, b), [−∞, b), and (a,∞], where a, b ∈ R.



DEFINITION 5.7 Extended Real-Valued Measurable Function

Let (Ω,A) be a measurable space. An extended real-valued function f on Ωis said to be an A-measurable function if the inverse image of each opensubset of R∗ under f is an A-measurable set, that is, if f−1(O) ∈ A for all opensets O ⊂ R∗.

The next proposition provides the analogue of Proposition 5.1 for extendedreal-valued functions. Its proof is left as an exercise.

PROPOSITION 5.3

Let (Ω,A) be a measurable space and f an extended real-valued function on Ω.Then the following statements are equivalent:a) f is A-measurable.b) For each a ∈ R, f−1

([−∞, a)

) ∈ A.

c) For each a ∈ R, f−1((a,∞]

) ∈ A.

d) For each a ∈ R, f−1([−∞, a]

) ∈ A.

e) For each a ∈ R, f−1([a,∞]

) ∈ A.

Theorem 5.3 shows that the collection of real-valued A-measurable functionsforms an algebra. In the case of extended real-valued functions, if we adopt theconvention that ∞−∞ is some fixed extended real number, then the collectionof extended real-valued A-measurable functions is closed under addition, scalarmultiplication, and multiplication. See Exercises 5.39 and 5.40.

The following theorem establishes that the collection of extended real-valuedA-measurable functions is closed under maxima, minima, suprema, infima, andpointwise limits. Note that Theorem 4.2 on page 113 is an immediate conse-quence.

THEOREM 5.5

Suppose that f and g are extended real-valued A-measurable functions and that{fn}∞n=1 is a sequence of extended real-valued A-measurable functions. Then

a) f ∨ g and f ∧ g are A-measurable.

b) supn fn and infn fn are A-measurable.

c) lim supn→∞ fn and lim infn→∞ fn are A-measurable.

d) If {fn}∞n=1 converges pointwise, then limn→∞ fn is A-measurable.

PROOF a) Let h = f ∨ g and a ∈ R. Then h−1((a,∞]

)= f−1

((a,∞]

) ∪ g−1((a,∞]

).

This union is in A because f and g are A-measurable functions. Thus, f ∨ g isA-measurable. Similarly, f ∧ g is A-measurable.

b) Let h = supn fn and a ∈ R. Then h−1((a,∞]

)=⋃∞

n=1 f−1n

((a,∞]

). This

union is in A because each fn is an A-measurable function. Hence, we seethat supn fn is A-measurable. Similarly, infn fn is A-measurable.



c) Noting that lim supn→∞ fn = infn supk≥n fk, it follows from part (b) thatlim supn→∞ fn is A-measurable. Using an entirely similar argument, we findthat lim infn→∞ fn is A-measurable.

d) If {fn}∞n=1 converges pointwise, then we have limn→∞ fn = lim supn→∞ fn.So, limn→∞ fn is A-measurable by part (c).

A common application of Theorem 5.5 occurs when {fn}∞n=1 is a sequenceof real-valued A-measurable functions, but where at least one of the functionsinfn fn, supn fn, lim infn→∞ fn, lim supn→∞ fn, and limn→∞ fn is an extendedreal-valued A-measurable function.

EXAMPLE 5.5 Illustrates Theorem 5.5

a) Let (Ω,A, μ) = (R,M, λ). For x ∈ R and n ∈ N , define

fn(x) =n√2π

e−(nx)2/2.

Then {fn}∞n=1 is a sequence of real-valued M-measurable functions. More-over, fn → f pointwise, where

f(x) =

{0, if x �= 0;∞, if x = 0.

By Theorem 5.5(d), f is an extended real-valued A-measurable function, afact that we can easily verify directly.

b) Let f be an extended real-valued A-measurable function. By Theorem 5.5(a),|f | is A-measurable since |f | = f ∨ −f .

Theorem 5.5(d) shows that if a sequence {fn}∞n=1 of A-measurable functionsconverges pointwise to a function f , then f is an A-measurable function. Whatif the convergence is only almost everywhere? In general, we cannot concludethat f is A-measurable; however, for complete measure spaces we can.

PROPOSITION 5.4

Let (Ω,A, μ) be a complete measure space. Suppose that {fn}∞n=1 is a se-quence of complex-valued or extended real-valued A-measurable functions andthat fn → f μ-ae. Then f is an A-measurable function.

PROOF The proof is essentially identical to that of Proposition 4.8 on page 140 and isleft to the reader.

Exercises for Section 5.2

5.20 Prove Proposition 5.1 on page 152.

5.21 Let (Ω,A) be a measurable space and f a real-valued function on Ω. Prove that f isA-measurable if and only if f−1(B) ∈ A for each B ∈ B.

5.22 Suppose that (Ω,A) is a measurable space and that f : Ω → R is an A-measurablefunction. Further suppose that g:R → R is a Borel measurable function. Prove thatg ◦ f is A-measurable.



5.23 Let D ∈ B. Show that C(D), the collection of Borel measurable functions on D, isprecisely the collection of BD-measurable functions.

5.24 Let (Ω,A) be a measurable space, D ∈ A, and AD = {D ∩A : A ∈ A}.a) If f : Ω → R is A-measurable, show that f|D is AD-measurable.b) Suppose that g:D → R is AD-measurable. Define f : Ω → R by

f(x) =

{g(x), x ∈ D;

0, x /∈ D.

Prove that f is A-measurable. (This result shows that every AD-measurable func-tion can be extended to an A-measurable function.)


5.26 Provide an example to show that the hypothesis of completeness cannot be omittedfrom Proposition 5.2.

5.27 If O is an open subset of R and α is a nonzero real number, show that α−1O is anopen subset of R.

5.28 Prove Theorem 5.4 on page 154. Hint: Use the fact that each open set in C is acountable union of open rectangles. [An open rectangle in C is a set of the form{u + iv ∈ C : a < u < b, c < v < d }.]

5.29 Show that every real-valued A-measurable function is a complex-valued A-measurablefunction.

5.30 The collection B2 of Borel sets of C is defined to be the smallest σ-algebra of subsetsof C that contains all the open subsets of C. Show that f : Ω → C is A-measurable ifand only if f−1(B) ∈ A for all B ∈ B2.

5.31 Let (Ω,A, P ) be a probability space, X a random variable on Ω, and t a fixed realnumber. Define g: Ω → C by g = eitX ; that is, for each x ∈ Ω, g(x) = eitX(x). Provethat g is A-measurable. Is g a random variable? Explain your answer.

★5.32 Prove that the collection of complex-valued A-measurable functions forms a complexalgebra. That is, if f and g are complex-valued A-measurable functions and α ∈ C,show that f + g, αf , and f · g are complex-valued A-measurable functions.

5.33 Show that each open subset of R is also an open subset of R∗.

5.34 Prove Proposition 5.3 on page 155. Hint: Show that each open set in R∗ can be writtenas a countable union of intervals of the form (a, b), [−∞, b), and (a,∞], where a, b ∈ R.

5.35 Show that f : Ω → R∗ is A-measurable if and only if (i) f−1({−∞}) and f−1({∞}) arein A and (ii) f−1(B) ∈ A for all B ∈ B.

5.36 Show that every real-valued A-measurable function is also an extended real-valuedA-measurable function.

5.37 Show that a set O ⊂ R is open in R if and only if there is an open subset U of R∗ suchthat O = R∩ U .

5.38 Suppose that f and g are extended real-valued A-measurable functions. Prove that thefollowing three sets are A-measurable:a) {x : f(x) > g(x) }.b) {x : f(x) ≥ g(x) }.c) {x : f(x) = g(x) }.

5.39 Let f and g be extended real-valued A-measurable functions and let β ∈ R∗. Set

E = {x : f(x) = ∞, g(x) = −∞} ∪ {x : f(x) = −∞, g(x) = ∞}.For x ∈ E, define (f + g)(x) = β; otherwise, define (f + g)(x) = f(x) + g(x), as usual.Prove that f + g is A-measurable.



5.40 With the convention established in the preceding exercise, prove that the collection ofextended real-valued A-measurable functions is closed under scalar multiplication andmultiplication.

5.41 Suppose that {fn}∞n=1 is a sequence of extended real-valued A-measurable functions.Verify that {x : limn→∞ fn(x) exists } is an A-measurable set.

5.42 Suppose {fn}∞n=1 is a sequence of complex-valued A-measurable functions that con-verges pointwise to a complex-valued function f . Prove that f is A-measurable.

5.43 Construct a sequence {fn}∞n=1 of A-measurable functions that converges almost every-where to a function f that is not A-measurable. Hint: Take (Ω,A, μ) = (R,B, λ|B)and do something with a non-Borel measurable subset of the Cantor set.


★5.45 Suppose that {fn}∞n=1 is a sequence of complex-valued A-measurable functions. Define

f(x) =

{lim

n→∞fn(x), if lim

n→∞fn(x) exists;

0, otherwise.

Prove that f is A-measurable.

5.46 Suppose that {fn}∞n=1 is a sequence of complex-valued A-measurable functions andthat fn → g μ-ae. Prove that there exists an A-measurable function f such thatfn → f μ-ae. Note: g need not be A-measurable unless, of course, (Ω,A, μ) is complete.

5.47 Suppose that E is an open subset of C and that g is a real-valued continuous functionon E. Further suppose that f is a complex-valued A-measurable function on Ω withthe range of f being a subset of E. Prove that g ◦ f is a real-valued A-measurablefunction on Ω. Repeat the proof if E is a closed subset of C.

5.48 Suppose that f : Ω → C is A-measurable. Verify that f can be written in the “polar”form, f = ReiΘ, where R: Ω → [0,∞) and Θ: Ω → R are A-measurable functions.

5.3 THE ABSTRACT LEBESGUE INTEGRAL FOR NONNEGATIVE FUNCTIONS

Now that we have discussed measure spaces and measurable functions, we canproceed to develop the abstract Lebesgue integral, that is, the Lebesgue integralon an arbitrary measure space (Ω,A, μ). As we will see, the development of theabstract Lebesgue integral is almost identical to that of the Lebesgue integralon the real line, that is, on (R,M, λ), given in Chapter 4. Consequently, manyof the proofs will be left to the reader.

Following the procedure used in Chapter 4, we will first define the abstractLebesgue integral of a simple function, then of a nonnegative A-measurable func-tion, and then of a real-valued A-measurable function. In addition, we will alsodefine the abstract Lebesgue integral of extended real-valued and complex-valuedA-measurable functions. Nonnegative functions will be considered in this sectionand general functions in the next.

The Lebesgue Integral of a Nonnegative Simple Function

Let (Ω,A, μ) be a measure space. An A-measurable function on Ω is called asimple function if it takes on only finitely many values. More precisely, wehave the following definition.


5.3 The Abstract Lebesgue Integral for Nonnegative Functions 159

DEFINITION 5.8 Simple Function and Canonical Representation

An A-measurable function s is said to be a simple function if its range isa finite set. Let a1, a2, . . . , an denote the distinct nonzero values of s andset Ak = {x : s(x) = ak }, 1 ≤ k ≤ n. Then

s =

n∑k=1

akχAk.

This is called the canonical representation of s.

We leave it as an exercise for the reader to show that the sets A1, A2, . . . , An,appearing in the canonical representation of an A-measurable simple function,are A-measurable and pairwise disjoint.


a) The Lebesgue measurable simple functions introduced in Chapter 4 are M-measurable simple functions in the sense of Definition 5.8.

b) If Ω is a finite set, then every A-measurable function is simple.

We now give the definition of the abstract Lebesgue integral of a nonneg-ative A-measurable simple function. It is a straightforward generalization ofthe definition presented in Chapter 4 for the Lebesgue integral of a nonnegativeLebesgue measurable simple function.

DEFINITION 5.9 Integral of a Nonnegative Simple Function

Let (Ω,A, μ) be a measure space and s a nonnegative A-measurable simple func-tion on Ω with canonical representation s =

∑nk=1 akχAk

. Then the (abstract)Lebesgue integral of s over Ω with respect to μ is defined by∫

Ω

s(x) dμ(x) =

n∑k=1

akμ(Ak).

If E ∈ A, then the (abstract) Lebesgue integral of s over E with respectto μ is defined by ∫

E

s(x) dμ(x) =

∫Ω

χE(x)s(x) dμ(x).

Note: The notations∫Es dμ and

∫Es(x)μ(dx) are commonly used in place

of∫Es(x) dμ(x).

The next proposition shows how we can obtain the abstract Lebesgue integralof a nonnegative simple function from a possibly noncanonical representation.The proof is identical to that of Proposition 4.2 on page 114.



PROPOSITION 5.5

Let s be a nonnegative A-measurable simple function that can be expressed inthe form s =

∑mk=1 bkχBk

, where this representation is not necessarily canonicalbut Bk ∈ A for 1 ≤ k ≤ m and Bi ∩Bj = ∅ for i �= j. Then∫

Ω

s(x) dμ(x) =

m∑k=1

bkμ(Bk).

More generally, we have∫E

s(x) dμ(x) =

m∑k=1

bkμ(Bk ∩ E)

for each E ∈ A.

The following fact is proved in precisely the same way as Lemma 4.1 onpage 116.

PROPOSITION 5.6

Suppose that s and t are nonnegative A-measurable simple functions and thatα, β ≥ 0. Then αs + βt is a nonnegative A-measurable simple function and∫

E

(αs + βt) dμ = α

∫E

s dμ + β

∫E

t dμ

for each E ∈ A.

The Lebesgue Integral of a Nonnegative A-measurable Function

The next thing on the agenda is the definition of the abstract Lebesgue integralfor a nonnegative extended real-valued A-measurable function. Proposition 5.7,whose proof is left to the reader as an exercise, provides the motivation for thatdefinition.

PROPOSITION 5.7

a) Suppose that f is a nonnegative extended real-valued A-measurable functionon Ω. Then there is a nondecreasing sequence of nonnegative A-measurablesimple functions that converges pointwise to f . In other words, there is asequence {sn}∞n=1 of nonnegative A-measurable simple functions such that,for all x ∈ Ω, s1(x) ≤ s2(x) ≤ · · · and limn→∞ sn(x) = f(x).

b) If {sn}∞n=1 is a sequence of nonnegative A-measurable simple functions thatconverges pointwise on Ω to a function f , then f is a nonnegative extendedreal-valued A-measurable function.

Proposition 5.7 shows that the functions that can be approximated by non-negative A-measurable simple functions are precisely the nonnegative extendedreal-valued A-measurable functions. Thus, we make the following definition.



DEFINITION 5.10 Lebesgue Integral of a Nonnegative Function

Let f be a nonnegative extended real-valued A-measurable function on Ω. Thenthe (abstract) Lebesgue integral of f over Ω with respect to μ is defined by∫

Ω

f(x) dμ(x) = sups

∫Ω

s(x) dμ(x),

where the supremum is taken over all nonnegative A-measurable simple functionsthat are dominated by f . If E ∈ A, then the (abstract) Lebesgue integral of fover E with respect to μ is defined by∫

E

f(x) dμ(x) =

∫Ω

χE(x)f(x) dμ(x).

Note: The abstract Lebesgue integral of a nonnegative M-measurable functionwith respect to λ is identical to its Lebesgue integral, as defined in Chapter 4.

Some of the more important properties of the abstract Lebesgue integral fornonnegative extended real-valued A-measurable functions are provided in Propo-sition 5.8. The proof is left as an exercise for the reader.

PROPOSITION 5.8

Let f and g be nonnegative extended real-valued A-measurable functions on Ω,α ≥ 0, and E ∈ A. Then

a) f ≤ g μ-ae ⇒ ∫Ef dμ ≤ ∫

Eg dμ.

b) B ⊂ E and B ∈ A ⇒ ∫Bf dμ ≤ ∫

Ef dμ.

c) f(x) = 0 for all x ∈ E ⇒ ∫Ef dμ = 0.

d) μ(E) = 0 ⇒ ∫Ef dμ = 0.

e)∫Eαf dμ = α

∫Ef dμ.

Convergence Properties of the Abstract Lebesgue Integral forNonnegative A-measurable Functions

We now present two major convergence theorems for the abstract Lebesgue in-tegral of nonnegative extended real-valued A-measurable functions — the mono-tone convergence theorem (MCT) and Fatou’s lemma. The proofs are similar tothose given in Section 4.2 (page 121 onward).

The MCT is stated first. Observe that it applies to extended real-valuedA-measurable functions as well as to real-valued A-measurable functions.

THEOREM 5.6 Monotone Convergence Theorem (MCT)

Suppose that {fn}∞n=1 is a monotone nondecreasing sequence of nonnegativeextended real-valued A-measurable functions. Then∫

E

limn→∞ fn dμ = lim

n→∞

∫E

fn dμ

for each E ∈ A.



COROLLARY 5.1

Let f , g, f1, f2, . . . be nonnegative extended real-valued A-measurable functionsand let E ∈ A. Then

a)∫E

(f + g) dμ =∫Ef dμ +

∫Eg dμ.

b)∫E

∑∞n=1 fn dμ =

∑∞n=1

∫Efn dμ.

c) If {En}n ⊂ A are pairwise disjoint, then∫⋃

nEn

f dμ =∑

n

∫En

f dμ.

Proposition 5.8(e) and Corollary 5.1(a) together imply that if f and g arenonnegative extended real-valued A-measurable functions and α, β ≥ 0, then∫

Ω

(αf + βg) dμ = α

∫Ω

f dμ + β

∫Ω

g dμ. (5.2)

Equation (5.2), Proposition 5.7, and the MCT are frequently used together for“bootstrapping arguments.” That is, suppose we want to prove that a certainLebesgue-integral property holds for all nonnegative A-measurable functions.To bootstrap, we employ three steps: First we show that the property holdsfor characteristic functions of A-measurable sets; next we apply (5.2) to con-clude that the property holds for nonnegative simple functions; and then we useProposition 5.7(a) and the MCT to deduce that the property holds for all non-negative A-measurable functions. Exercises 5.60 and 5.61 provide illustrationsof bootstrapping.

Next we state Fatou’s lemma. This version of Fatou’s lemma not only gen-eralizes to arbitrary measure spaces the version presented in Theorem 4.6 onpage 126 but its hypotheses are less restrictive. Specifically, it does not imposeany convergence conditions on {fn}∞n=1.

THEOREM 5.7 Fatou’s Lemma

Let {fn}∞n=1 be a sequence of nonnegative extended real-valued A-measurablefunctions. Then ∫

E

lim infn→∞ fn dμ ≤ lim inf

n→∞

∫E

fn dμ

for each E ∈ A.

EXAMPLE 5.7 Illustrates the Abstract Lebesgue Integral

a) Let (Ω,A, μ) be a measure space and f a nonnegative extended real-valuedA-measurable function on Ω. Suppose that x0 ∈ Ω and that {x0} ∈ A. Weclaim that ∫

{x0}f dμ = f(x0)μ({x0}). (5.3)

To see this, note that χ{x0}f is the simple function f(x0)χ{x0} and, hence,by Definition 5.9 on page 159,∫

{x0}f dμ =

∫Ω

χ{x0}f dμ =

∫Ω

f(x0)χ{x0} dμ = f(x0)μ({x0}).



More generally, let C = {xn}n be a countable subset of Ω with {xn} ∈ Afor each n. Then, by Corollary 5.1(c) and (5.3),∫

C

f dμ =

∫⋃n{xn}

f dμ =∑n

∫{xn}

f dμ =∑n

f(xn)μ({xn}). (5.4)

b) Let μ be counting measure on P(N ). Then, as we learned in Example 5.2(c),a nonnegative real-valued P(N )-measurable function f on N is a nonnegativeinfinite sequence {an}∞n=1, where we have let an = f(n). Thus, by (5.4),∫

Nf dμ =

∞∑n=1

f(n)μ({n}) =

∞∑n=1

an.

Hence, we can apply abstract measure theory to study infinite series.c) Let (Ω,A, P ) be a probability space and X a nonnegative random variable.

Then the abstract Lebesgue integral of X over Ω with respect to P is calledthe mean (expectation, expected value) of X. The mean of X is denotedby E(X). Thus,

E(X) =

∫Ω

X dP.

For instance, consider the random experiment of tossing a balanced cointwice. An appropriate probability space for that experiment is (Ω,A, P ),where Ω = {HH, HT, TH, TT}, A = P(Ω) and, for E ∈ A, P (E) = N(E)/4.Let X denote the number of heads obtained. Then, by (5.4), the mean of Xequals

E(X) =

∫Ω

X dP = X(HH)P ({HH}) + X(HT)P ({HT})

+ X(TH)P ({TH}) + X(TT)P ({TT})

= 2 · 1

4+ 1 · 1

4+ 1 · 1

4+ 0 · 1

4= 1,

which is intuitively what it should be.d) Let Ω be a set, {xn}n a sequence of distinct elements of Ω, and {bn}n a

sequence of nonnegative real numbers. For E ⊂ Ω, define

μ(E) =∑xn∈E

bn.

Then μ is a measure on P(Ω). Let f be a nonnegative function on Ω andset C = {xn}n. Then, by Corollary 5.1(c), Proposition 5.8(d) on page 161,and (5.4), ∫

Ω

f dμ =

∫C

f dμ +

∫Cc

f dμ =

∫C

f dμ + 0

=∑n

f(xn)μ({xn}) =∑n

f(xn)bn.(5.5)

We will employ (5.5) frequently.



e) Let Ω be a set, A = P(Ω), and μ counting measure on A. If f is a nonnegativefunction on Ω, then ∫

Ω

f dμ =∑x∈Ω

f(x),

where∑

x∈Ω f(x) = sup{∑

x∈F f(x) : F finite, F ⊂ Ω}. The verification of

this fact is left to the reader.


5.49 Establish that the sets appearing in the canonical representation of an A-measurablesimple function are A-measurable and pairwise disjoint.

5.50 Prove Proposition 5.7 on page 160. Hint: Refer to Proposition 4.3 on page 116.

5.51 Suppose that f is a nonnegative extended real-valued A-measurable function on Ω,c > 0, and Ac = {x : f(x) ≥ c }. Prove that

μ(Ac) ≤ 1

c

∫Ω

f dμ.

★5.52 Let f be a nonnegative extended real-valued A-measurable function on Ω and E ∈ A.Prove that

∫Ef dμ = 0 if and only if f = 0 μ-ae on E.

★5.53 Suppose that f is a nonnegative extended real-valued A-measurable function on Ω andthat

∫Ωf dμ < ∞. Show that f is finite μ-ae.

5.54 Prove Proposition 5.8 on page 161. Hint: Refer to Proposition 4.4 on page 118.

5.55 Prove the MCT, Theorem 5.6 on page 161.

5.56 Show that for a fixed E ∈ A, the conclusion of the MCT remains valid if the hypothesesare satisfied only on E.

5.57 Prove Corollary 5.1 on page 162.

5.58 Suppose that f is a nonnegative extended real-valued A-measurable function on Ω.Also, suppose that {En}∞n=1 ⊂ A with E1 ⊂ E2 ⊂ · · · . Prove that∫⋃∞

n=1En

f dμ = limn→∞

∫En

f dμ.

5.59 Prove Fatou’s lemma, Theorem 5.7 on page 162.

5.60 Suppose that (Ω,A, μ) is a measure space, D ∈ A, and f is a nonnegative extended real-valued A-measurable function on Ω. Let (D,AD, μD) be as defined in Example 5.1(c)on page 146. Show that ∫

D

f dμ =

∫D

f|D dμD.

Hint: Use bootstrapping.

★5.61 Let (Ω,A, μ) be a measure space and g a nonnegative A-measurable function on Ω.For E ∈ A, define

ν(E) =

∫E

g dμ.

a) Show that ν is a measure on A.


5.4 The General Abstract Lebesgue Integral 165

b) Show that ∫Ω

f dν =

∫Ω

fg dμ

for each nonnegative A-measurable function f . Hint: Bootstrap.

5.62 Let {amn}∞m,n=1 be a double sequence of nonnegative numbers. Prove that∞∑

n=1

∞∑m=1

amn =

∞∑m=1

∞∑n=1

amn.

Hint: Refer to Example 5.7(b) on page 163.

5.63 Let f : Ω → [0, 1] be an A-measurable function.

a) Prove that limn→∞∫Ωf

1n dμ = μ

(f−1((0, 1]

)).

b) If μ(Ω) < ∞, prove that limn→∞∫Ωfn dμ = μ

(f−1({1})

).

5.4 THE GENERAL ABSTRACT LEBESGUE INTEGRAL

In the previous section, we discussed the abstract Lebesgue integral for non-negative extended real-valued A-measurable functions. We will now expand thedefinition of the abstract Lebesgue integral so that it applies to A-measurablefunctions that are not necessarily nonnegative. We begin with extended real-valued functions.

Lebesgue Integral of an Extended Real-Valued Function

Let (Ω,A, μ) be a measure space. To define the abstract Lebesgue integral ofan extended real-valued A-measurable function f on Ω, we follow the procedureused in Section 4.3 for defining the Lebesgue integral of a real-valued Lebesguemeasurable function on R.

DEFINITION 5.11 Integral of an Extended Real-Valued Function

Let f be an extended real-valued A-measurable function on Ω and E ∈ A. Thenthe (abstract) Lebesgue integral of f over E with respect to μ is defined by∫

E

f dμ =

∫E

f+ dμ−∫E

f− dμ (5.6)

provided that the right-hand side makes sense; that is, at least one of the integralson the right-hand side of (5.6) is finite. Here f+ = f ∨ 0 and f− = −(f ∧ 0)denote the positive and negative parts of f , respectively. In addition, we saythat f is Lebesgue integrable over E if both integrals on the right-hand sideof (5.6) are finite or, equivalently, if∫

E

|f | dμ =

∫E

f+ dμ +

∫E

f− dμ < ∞. (5.7)

If f is Lebesgue integrable over Ω, then we say that f is Lebesgue integrable.

We should mention that if f is Lebesgue integrable (over Ω), then it isLebesgue integrable over every E ∈ A. Here are some examples.




a) Let (Ω,A, μ) = (R,M, λ) and f(x) = x. Then

f+(x) =

{x, x ≥ 0;0, x < 0.

and f−(x) =

{0, x ≥ 0;−x, x < 0.

(i) If E = R, then∫R f+ dλ =

∫R f− dλ = ∞. Hence, the integral

∫R f dλ

is not defined.(ii) If E = [−1, 2], then

∫Ef+ dλ = 2 and

∫Ef− dλ = 1/2 and, consequently,∫

Ef dλ = 2 − 1/2 = 3/2. And, because

∫E|f | dλ = 2 + 1/2 = 5/2 < ∞,

we see that f is Lebesgue integrable over [−1, 2].(iii) If E = (−∞, 1), then

∫Ef+ dλ = 1/2 and

∫Ef− dλ = ∞ and, conse-

quently,∫Ef dλ = 1/2 −∞ = −∞. But, as

∫E|f | dλ = 1/2 + ∞ = ∞,

we see that f is not Lebesgue integrable over (−∞, 1).b) Let (Ω,A, μ) = (N ,P(N ), μ), where μ is counting measure on P(N ). Then

real-valued A-measurable functions are simply infinite sequences of real num-bers. Referring to Example 5.7(b) on page 163, we see that a sequence of realnumbers {an}∞n=1 is Lebesgue integrable (over N ) if and only if

∞∑n=1

|an| < ∞, (5.8)

that is, if and only if the series is absolutely convergent. For instance, the se-quence {(−1)n/np}∞n=1 is Lebesgue integrable if and only if p > 1. Note that,although

∑∞n=1(−1)n/n converges, {(−1)n/n}∞n=1 is not Lebesgue integrable

as the series is not absolutely convergent.

Lebesgue Integral of a Complex-Valued Function

Next we define the abstract Lebesgue integral for complex-valued A-measurablefunctions. First we present some preliminaries.

DEFINITION 5.12 Modulus of a Complex-Valued Function

Let f be a complex-valued function on Ω. Then the modulus of f , denotedby |f |, is defined to be the real-valued function

|f | =√

(�f)2 + (�f)2.

In other words, |f |(x) = |f(x)|, where |f(x)| denotes the modulus of the complexnumber f(x).

The following two propositions will be required. We leave the proofs as exer-cises for the reader.



PROPOSITION 5.9

Let f be a complex-valued function on Ω. Thena) |f | ≤ |�f | + |�f |.b) |�f | ≤ |f | and |�f | ≤ |f |.c) |f | is A-measurable if f is.

PROPOSITION 5.10

Let f be a complex-valued A-measurable function on Ω and E ∈ A. Then |f | isLebesgue integrable over E if and only if both �f and �f are.

In view of Proposition 5.10 and the fact that f = �f + i�f , it is reasonableto make the following definition.

DEFINITION 5.13 Integral of a Complex-Valued Function

Let f be a complex-valued A-measurable function on Ω and E ∈ A. We saythat f is Lebesgue integrable over E with respect to μ if |f | is Lebesgueintegrable over E with respect to μ; that is,∫

E

|f | dμ < ∞.

In that case, the (abstract) Lebesgue integral of f over E with respectto μ is defined by ∫

E

f dμ =

∫E

(�f) dμ + i

∫E

(�f) dμ.

If f is Lebesgue integrable over Ω, then we say that f is Lebesgue integrable.

For a measure space (Ω,A, μ), the collection of all complex-valued Lebesgueintegrable functions is denoted by L1(Ω,A, μ). When no confusion will arise,we write L1(μ) for L1(Ω,A, μ).


a) Let (Ω,A, μ) = (R,M, λ) and consider the complex-valued function f definedby f(x) = eix/(1 + x2). Then it is easy to see that f is measurable. Moreover,�f(x) = cosx/(1 + x2), �f(x) = sinx/(1 + x2), and |f(x)| = 1/(1 + x2). ByExercise 4.20 on page 127 and Theorem 4.9 on page 135,∫

R|f(x)| dλ(x) =

∫R

(1 + x2)−1 dλ(x)

= limn→∞

∫[−n,n]

(1 + x2)−1 dλ(x) = limn→∞

∫ n

−n

dx

(1 + x2)

= limn→∞

(arctan(n) − arctan(−n)

)= π < ∞.

Therefore, f ∈ L1(λ).



b) Let (Ω,A, μ) = (N ,P(N ), μ), where μ is counting measure on P(N ). Thencomplex-valued A-measurable functions are simply infinite sequences of com-plex numbers. Referring to Example 5.7(b) on page 163, we see that a se-quence of complex numbers {an}∞n=1 is in L1(μ) if and only if the series∑∞

n=1 an converges absolutely. We point out here that the notations 1

or 1(N ) are generally used in place of L1(N ,P(N ), μ).c) Let (Ω,A) be a measurable space. A measure μ on A is said to be a finite

measure if μ(Ω) < ∞. If μ is a finite measure, then (Ω,A, μ) is called afinite measure space. For a finite measure space, each bounded complex-valued A-measurable function f is in L1(μ). Indeed, if |f | ≤ M , then byProposition 5.8(a) on page 161,∫

Ω

|f | dμ ≤∫

Ω

M dμ = Mμ(Ω) < ∞.

Note that boundedness is a sufficient but not necessary condition for integra-bility. For instance, let (Ω,A, μ) =

((0, 1),M(0,1), λ(0,1)

)and f(x) = x−1/2.

Then f is not bounded on (0, 1) but is in L1(λ(0,1)

).

d) If (Ω,A, P ) is a probability space, then the integrable functions, that is,members of L1(P ), are called random variables with finite mean or finiteexpectation.

The following theorem, whose proof is left as an exercise, provides some im-portant properties of Lebesgue integrable functions.

THEOREM 5.8

Suppose that f and g are in L1(Ω,A, μ) and that α ∈ C. Then

a) f + g ∈ L1(μ) and ∫Ω

(f + g) dμ =

∫Ω

f dμ +

∫Ω

g dμ.

b) αf ∈ L1(μ) and ∫Ω

αf dμ = α

∫Ω

f dμ.

c) If f and g are real-valued and f ≤ g on Ω, then∫Ωf dμ ≤ ∫

Ωg dμ.

d)∣∣∫

Ωf dμ

∣∣ ≤ ∫Ω|f | dμ.

e) μ(E) = 0 ⇒ ∫Ef dμ = 0.

f) If A and B are disjoint A-measurable sets, then∫A∪B

f dμ =

∫A

f dμ +

∫B

f dμ.

Remark: Parts (a) and (b) of Theorem 5.8 together imply that if α, β ∈ C andf , g ∈ L1(μ), then ∫

Ω

(αf + βg) dμ = α

∫Ω

f dμ + β

∫Ω

g dμ.

This result is called the linearity property of the abstract Lebesgue integral.



As mentioned in Section 4.4, we often encounter functions that are only de-fined almost everywhere. Because the integral of an A-measurable function isnot affected by its values on a set of measure zero, it is reasonable to make thefollowing definition.

DEFINITION 5.14 Integral of a Function Defined Almost Everywhere

Let (Ω,A, μ) be a measure space. Suppose that f is a function defined μ-ae on Ω;that is, if D is the domain of f , then μ(Dc) = 0. Further suppose that there isan A-measurable function g such that g(x) = f(x) for x ∈ D. Then, for E ∈ A,we define the (abstract) Lebesgue integral of f over E by∫

E

f dμ =

∫E

g dμ,

provided that the integral on the right-hand side exists.

Dominated Convergence Theorem

Theorem 4.8 on page 133 gives the dominated convergence theorem (DCT) forreal-valued functions on the measure space (R,M, λ). Our next theorem gener-alizes that version of the DCT so that it applies to complex-valued functions onan arbitrary measure space (Ω,A, μ). Note that the version of the DCT givenhere has weaker hypotheses than the one presented in Theorem 4.8.

THEOREM 5.9 Dominated Convergence Theorem (DCT)

Let (Ω,A, μ) be a measure space. Suppose that {fn}∞n=1 is a sequence ofcomplex-valued A-measurable functions that converges μ-ae. Further supposethat there is a nonnegative Lebesgue integrable function g such that |fn| ≤ g μ-aefor each n ∈ N . Then ∫

E


n→∞

∫E

fn dμ (5.9)

for each E ∈ A.

PROOF Without loss of generality, we can assume that, for each n ∈ N , |fn| ≤ g every-where on Ω (why?). Define

f(x) =

{limn→∞ fn(x), if lim

n→∞ fn(x) exists;

0, otherwise.

By Exercise 5.45 on page 158, f is A-measurable. Moreover, since {fn}∞n=1

converges μ-ae, fn → f μ-ae. From Definition 5.14,∫E

limn→∞ fn dμ =∫Ef dμ

and, therefore, to prove (5.9) it suffices to prove∫E

f dμ = limn→∞

∫E

fn dμ (5.10)

for each E ∈ A.



First suppose that each fn is real-valued. Then (5.10) can be proved byemploying Fatou’s lemma (Theorem 5.7 on page 162) and the same argumentthat was used in the proof of the DCT for the Lebesgue integral on the real line(Theorem 4.8).

Next, we remove the restriction that each fn is real-valued. Observe that{|f − fn|}∞n=1 is a sequence of real-valued A-measurable functions that convergesto 0 μ-ae. Furthermore, for each n ∈ N , we have |f − fn| ≤ |f | + |fn| ≤ 2g, anintegrable function. Consequently, by Theorem 5.8 and the previous paragraph,as n → ∞, ∣∣∣∣∫

E

f dμ−∫E

fn dμ

∣∣∣∣ ≤ ∫E

|f − fn| dμ →∫E

0 dμ = 0

for each E ∈ A. This completes the proof of the DCT.

Three of the many important corollaries of the DCT are given next. Severalother corollaries are considered in the exercises.

COROLLARY 5.2

Suppose that {fn}∞n=1 is a sequence of complex-valued A-measurable functionssuch that ∞∑

n=1

∫Ω

|fn| dμ < ∞.

Then∑∞

n=1 fn converges μ-ae and∫E

∞∑n=1

fn dμ =

∞∑n=1

∫E

fn dμ

for each E ∈ A.

PROOF From Corollary 5.1(b) on page 162, we know that∫Ω

∞∑n=1

|fn| dμ =

∞∑n=1

∫Ω

|fn| dμ.

By assumption, the sum on the right-hand side of the previous equation is finiteand, hence, so is the integral on the left-hand side. In other words, if we setg =

∑∞n=1 |fn|, then g is Lebesgue integrable. From Exercise 5.53 on page 164, we

conclude that g is finite μ-ae which, in turn, implies that∑∞

n=1 fn converges μ-ae.Set gn =

∑nk=1 fk. Then, for each n ∈ N , |gn| ≤ g and, as we have just

seen, {gn}∞n=1 converges μ-ae (to∑∞

n=1 fn). Therefore, by the DCT and Theo-rem 5.8(a),∫

E

∞∑n=1

fn dμ =

∫E

limn→∞ gn dμ = lim

n→∞

∫E

gn dμ

= limn→∞

∫E

n∑k=1

fk dμ = limn→∞

n∑k=1

∫E

fk dμ =

∞∑n=1

∫E

fn dμ

for each E ∈ A.



COROLLARY 5.3

Let (Ω,A, μ) be a measure space. Suppose that f ∈ L1(μ) and that {En}∞n=1 isa sequence of A-measurable sets with E1 ⊂ E2 ⊂ · · · . Then∫⋃∞

n=1En

f dμ = limn→∞

∫En

f dμ.

PROOF For convenience, let E =⋃∞

n=1 En. It is easy to see that χEnf → χEf pointwise

and that |χEnf | ≤ |f | ∈ L1(μ) for each n ∈ N . Thus, by the DCT,∫E

f dμ =

∫Ω

χEf dμ = limn→∞

∫Ω

χEnf dμ = limn→∞

∫En

f dμ,

as required.

COROLLARY 5.4 Bounded Convergence Theorem

Let (Ω,A, μ) be a finite measure space. Suppose that {fn}∞n=1 is a sequenceof uniformly bounded, complex-valued, A-measurable functions that convergesμ-ae. Then ∫

E


n→∞

∫E

fn dμ

for each E ∈ A.

PROOF By assumption, there is a real number M such that |fn| ≤ M for all n ∈ N .Because (Ω,A, μ) is a finite measure space, the function g(x) ≡ M is Lebesgueintegrable (why?). Applying the DCT completes the proof.

EXAMPLE 5.10 Illustrates the DCT

a) Suppose that, for each n ∈ N , {ank}∞k=1 is a sequence of complex numbersand that limn→∞ ank = ak for each k ∈ N . Further suppose that there isa sequence of nonnegative numbers {bk}∞k=1 such that

∑∞k=1 bk < ∞ and

|ank| ≤ bk for k, n ∈ N . We claim that

limn→∞

∞∑k=1

ank =

∞∑k=1

ak. (5.11)

Indeed, consider the measure space (N ,P(N ), μ), where μ is counting mea-sure. Define fn(k) = ank, f(k) = ak, and g(k) = bk. By assumption, g isintegrable, |fn| ≤ g for all n ∈ N , and fn → f pointwise on N . Thus, bythe DCT,

∫N fn dμ → ∫

N f dμ as n → ∞. However,∫N fn dμ =

∑∞k=1 ank

and∫N f dμ =

∑∞k=1 ak (see Exercise 5.73). Thus, (5.11) holds.

Without a dominating integrable sequence, (5.11) may fail. For instance,take ank = δnk and ak ≡ 0. Then limn→∞ ank = ak for each k ∈ N . But, as∑∞

k=1 ank = 1 for all n ∈ N , we see that

limn→∞

∞∑k=1

ank = 1 �= 0 =

∞∑k=1

ak.

Therefore, (5.11) fails to hold.



b) Let (Ω,A, P ) be a probability space and X a real-valued random variable hav-ing finite expectation, that is, X ∈ L1(P ). Define f on R by f(t) = E(eitX).Note that the definition of f makes sense because |eitX | ≤ 1. We claimthat f ′(0) = iE(X). To prove this result, let {tn}∞n=1 be an arbitrary se-quence of nonzero real numbers that converges to 0. For each n ∈ N , defineYn =

(eitnX − 1

)/tn. Then (see Exercise 5.75)

f(tn) − f(0)

tn − 0=

∫Ω

eitnX − 1

tndP =

∫Ω

Yn dP. (5.12)

Now, for x ∈ R, we have |eix − 1| ≤ |x| and, therefore, |Yn| ≤ |X| for eachn ∈ N . As Yn → iX pointwise on Ω, we can apply the DCT to conclude that

limn→∞

f(tn) − f(0)

tn − 0= lim

n→∞

∫Ω

Yn dP =

∫Ω

iX dP = iE(X).

Because {tn}∞n=1 is an arbitrary sequence of nonzero real numbers convergingto 0, it follows that f ′(0) exists and equals iE(X).


5.64 Let (Ω,A, μ) = (N ,P(N ), μ), where μ is counting measure on P(N ). Define f :N → Rby f(n) = (−1)n/n for n ∈ N . Is

∫N f dμ defined? Explain your answer.



5.67 Let f(x) = x−1/2. Show that f ∈ L1((0, 1),M(0,1), λ(0,1)

).

5.68 Prove Theorem 5.8 on page 168.

5.69 Prove that Definition 5.14 on page 169 is well-posed. In other words, assume that fis defined μ-ae on Ω and that g and h are A-measurable functions that equal f on itsdomain. Show that for E ∈ A, either

∫Eg dμ =

∫Eh dμ or neither integral exists.

5.70 Show that for a fixed E ∈ A, the conclusion of the DCT remains valid if the hypothesesare satisfied only on E.

5.71 State and prove a version of the DCT for extended real-valued A-measurable functions.

★5.72 Suppose that f ∈ L1(Ω,A, μ). Further suppose that {En}∞n=1 is a sequence of pairwisedisjoint A-measurable sets. Prove that∫⋃∞

n=1En

f dμ =

∞∑n=1

∫En

f dμ.

★5.73 Let f ∈ L1(Ω,A, μ) and C = {xn}n a countable subset of Ω such that {xn} ∈ A foreach n. Prove that ∫

C

f dμ =∑n

f(xn)μ({xn}).

Deduce that if {an}∞n=1 ∈ �1, then∫Nf dμ =

∞∑n=1

an,

where f(n) = an.



5.74 Let∑∞

k=1ak be a convergent series of nonnegative numbers and, for n, k ∈ N , let bnk be

complex numbers with |bnk| ≤ M < ∞. Assume that limn→∞ bnk = bk for each k ∈ N .Prove that

limn→∞

∞∑k=1

akbnk =

∞∑k=1

akbk.

5.75 Provide a detailed justification of (5.12).

5.76 Let (Ω,A, P ) be a probability space and Y a real-valued random variable taking ononly finitely many values, say y1, y2, . . . , yn. Verify that

E(Y ) =

n∑k=1

ykP (Y = yk) (5.13)

where, by convention, {Y = y } = {x ∈ Ω : Y (x) = y }. Note: Equation (5.13) showsthat the mean of a random variable Y taking on only finitely many values is a weightedaverage of the values of Y , weighted according to their probabilities.

5.77 Let (Ω,A, μ) be a measure space, f ∈ L1(μ), and {En}∞n=1 a sequence of A-measurablesets with E1 ⊃ E2 ⊃ · · · . Prove that∫⋂∞

n=1En

f dμ = limn→∞

∫En

f dμ.

5.78 Suppose that f : [0, 1] × (0, 1) → R is such that, for each fixed y ∈ (0, 1), the func-tion f [y] defined by f [y](x) = f(x, y) is M[0,1]-measurable. Further suppose that ∂f/∂yexists and is bounded on [0, 1] × (0, 1). Show that

d

dy

∫ 1

0

f(x, y) dx =

∫ 1

0

∂f

∂y(x, y) dx.

5.79 Let f ∈ L1(Ω,A, μ). Show that for each ε > 0, there is an A ∈ A such that μ(A) < ∞and

∫Ac |f | dμ < ε.

★5.80 Suppose that f ∈ L1(Ω,A, μ). Show that for each ε > 0, there is a δ > 0 such thatμ(E) < δ ⇒ ∫

E|f | dμ < ε.

★5.81 Let f ∈ L1(R,M, λ). Then we define the Fourier transform of f , denoted f , by

f(t) =

∫Re−itxf(x) dλ(x), t ∈ R.

a) Prove that f is continuous on R.

b) Prove that if∫R |xf(x)| dλ(x) < ∞, then f is differentiable on R and

f ′(t) =

∫R

(−ix)e−itxf(x) dλ(x), t ∈ R.

★5.82 Suppose that f ∈ L1(Ω,A, μ).a) Prove that, for each ε > 0, there exists a bounded A-measurable function g such

that∫Ω|f − g| dμ < ε.

b) Prove that, for each ε > 0, there exists an A-measurable simple function s suchthat

∫Ω|f − s| dμ < ε.



5.5 CONVERGENCE IN MEASURE

Until now, we have discussed three types of convergence for functions: pointwiseconvergence, uniform convergence, and almost-everywhere convergence. Anotherkind of convergence, important especially in probability theory, is convergencein measure.† Here is the definition.

DEFINITION 5.15 Convergence in Measure

Let (Ω,A, μ) be a measure space and {fn}∞n=1 a sequence of complex-valuedA-measurable functions on Ω. Then {fn}∞n=1 is said to converge in measureto the A-measurable function f , if for each ε > 0,

limn→∞μ ({x : |f(x) − fn(x)| ≥ ε }) = 0.

We often write fnμ→ f to indicate convergence in measure. Thus, fn

μ→ f if themeasure of the set where fn differs from f by more than any prescribed positivenumber tends to zero as n → ∞.

Is there a relationship between almost-everywhere convergence and conver-gence in measure? The following example shows that, generally speaking, thereis no relationship.


a) Let (Ω,A, μ) = (R,M, λ). Set f(x) ≡ 0 and fn(x) = x/n for x ∈ R andn ∈ N . Then fn → f pointwise and, hence, λ-ae. But fn �→ f in measure.Indeed, for ε > 0,

{x : |f(x) − fn(x)| ≥ ε } = (−∞,−nε) ∪ (nε,∞),

which has infinite Lebesgue measure for every n ∈ N . Thus, we see thatλ({x : |f(x) − fn(x)| ≥ ε }) �→ 0. Consequently, almost-everywhere conver-gence does not imply convergence in measure.

b) Let (Ω,A, μ) =([0, 1],M[0,1], λ[0,1]

). Consider the sequence of functions de-

fined by f1 = χ[0,1], f2 = χ[0,1/2], f3 = χ[1/2,1] and, in general, if n = k + 2j ,where 0 ≤ k < 2j , fn = χ[k2−j ,(k+1)2−j ]. Then, for ε > 0,

μ({x : |fn(x)| ≥ ε }) < 2

n→ 0

as n → ∞. So, fnμ→ 0. But, for each x ∈ [0, 1], the sequence {fn(x)}∞n=1

contains infinitely many 1s and infinitely many 0s. Thus, {fn}∞n=1 convergesfor no x ∈ [0, 1] and, in particular, fn �→ 0 μ-ae. Consequently, convergencein measure does not imply almost-everywhere convergence.

Example 5.11(a) shows that, in general, convergence almost everywhere doesnot imply convergence in measure. For finite measure spaces, however, the im-plication is correct.

† In probability theory, the terminology convergence in probability is used in place ofconvergence in measure.


5.5 Convergence in Measure 175

PROPOSITION 5.11

Suppose that (Ω,A, μ) is a finite measure space and that {fn}∞n=1 is a sequence ofcomplex-valued A-measurable functions that converges μ-ae to the A-measurable

function f . Then fnμ→ f .

PROOF Let B = {x : fn(x) �→ f(x) }. Then, by assumption, μ(B) = 0. For ε > 0, defineEn = {x : |f(x) − fn(x)| ≥ ε } and E =

⋂∞n=1

(⋃∞k=n Ek

). We must show that

limn→∞ μ(En) = 0.Note that x ∈ E if and only if x ∈ En for infinitely many n. It follows easily

that E ⊂ B and, so, μ(E) = 0. Because μ(Ω) < ∞ and⋃∞

k=n Ek ⊃ ⋃∞k=n+1 Ek

for each n ∈ N , we conclude from Theorem 5.1(c) on page 147 that

lim supn→∞

μ(En) ≤ limn→∞μ

( ∞⋃k=n

Ek

)= μ(E) = 0.

Hence, limn→∞ μ(En) = 0, as required.

As we discovered in Example 5.11(b), convergence in measure does not implyalmost-everywhere convergence. Nonetheless, we do have the following usefulresult.

PROPOSITION 5.12

Suppose that {fn}∞n=1 is a sequence of complex-valued A-measurable functionsthat converges in measure to the A-measurable function f . Then there is asubsequence {fnk

}∞k=1 of {fn}∞n=1 such that fnk→ f μ-ae.

PROOF We can, for each k ∈ N , choose an nk ∈ N such that

μ

({x : |f(x) − fnk

(x)| ≥ 1

k

})≤ 2−k. (5.14)

Furthermore, {nk}∞k=1 can be selected so that n1 < n2 < · · · . Now let us defineEk = {x : |f(x) − fnk

(x)| ≥ k−1 } and E =⋂∞

k=1

(⋃∞j=k Ej

). Note that x ∈ E if

and only if |f(x) − fnk(x)| ≥ k−1 for infinitely many k.

From (5.14), we see that∑∞

k=1 μ(Ek) < ∞ and, consequently, by Exercise 5.14on page 150, μ(E) = 0. We claim that fnk

→ f on Ec. So, let x ∈ Ec andlet ε > 0 be given. Choose k1 ∈ N so that k−1

1 ≤ ε. Since x /∈ E, it follows thatthere is a k2 ∈ N such that x /∈ Ek for k ≥ k2. Let K = max{k1, k2}. Then wehave that |f(x) − fnk

(x)| < k−1 ≤ ε for all k ≥ K.

The DCT for Convergence in Measure

By employing Proposition 5.12, we can prove that the dominated convergencetheorem remains valid when almost-everywhere convergence is replaced by con-vergence in measure. That is, we have the following result.



THEOREM 5.10

Let (Ω,A, μ) be a measure space. Suppose that {fn}∞n=1 is a sequence of complex-valued A-measurable functions that converges in measure to the A-measurablefunction f . Further suppose that there is a nonnegative Lebesgue integrablefunction g such that |fn| ≤ g μ-ae for each n ∈ N . Then∫

E

f dμ = limn→∞

∫E

fn dμ (5.15)

for each E ∈ A.

PROOF Let E ∈ A. To prove (5.15) it suffices, by Exercise 2.32 on page 45, to show thatevery subsequence of

{∫Efn dμ

}∞n=1

has a subsequence that converges to∫Ef dμ.

So, let {nk}∞k=1 be a subsequence of N . As fnμ→ f , it is clear that fnk

μ→ f . Ap-plying Proposition 5.12, we deduce that {fnk

}∞k=1 has a subsequence {fnkj}∞j=1

with fnkj→ f μ-ae. Clearly, we have |fnkj

| ≤ g μ-ae for each j ∈ N and, hence,

by the DCT (Theorem 5.9 on page 169),∫E

f dμ = limj→∞

∫E

fnkjdμ.

This completes the proof.


5.83 Show that if fnμ→ f and fn

μ→ g, then f = g μ-ae.

★5.84 Suppose that f , f1, f2, . . . are in L1(Ω,A, μ) and that∫Ω|f − fn| dμ → 0 as n → ∞.

Show that fn → f in measure.

5.85 Let (Ω,A, μ) be a measure space. We say that a sequence {fn}∞n=1 of complex-valuedA-measurable functions on Ω converges almost uniformly to the complex-valuedA-measurable function f if for each ε > 0, there is a set A ∈ A such that μ(A) < εand fn → f uniformly on Ac.a) Prove that almost-uniform convergence implies convergence in measure; that is,

if fn → f almost uniformly, then fn → f in measure.b) Prove that almost-uniform convergence implies almost-everywhere convergence; that

is, if fn → f almost uniformly, then fn → f μ-ae.c) Does almost-uniform convergence imply pointwise convergence? Why or why not?

5.86 Provide a detailed justification for all statements in Example 5.11(b) on page 174.

5.87 Suppose that f , g, f1, f2, . . . are complex-valued A-measurable functions. Furthersuppose that fn → g μ-ae and that fn → f in measure. Prove that f = g μ-ae and,hence, that fn → f μ-ae.

5.88 Fatou’s lemma for convergence in measure: Suppose that {fn}∞n=1 is a sequenceof nonnegative A-measurable functions that converges in measure to f . Prove that∫

E

f dμ ≤ lim infn→∞

∫E

fn dμ

for each E ∈ A. Hint: Select a subsequence of{∫

Efn dμ

}∞n=1

that converges to

lim infn→∞∫Efn dμ.


5.5 Convergence in Measure 177

5.89 Establish the following fact: If {fn}∞n=1 converges in measure, then it is also Cauchyin measure, that is, for each ε > 0, μ({x : |fn(x) − fm(x)| ≥ ε }) → 0 as m, n → ∞.

5.90 Prove the following strengthened version of Proposition 5.12. Suppose that {fn}∞n=1 isa sequence of complex-valued A-measurable functions that converges in measure to f .Then there is a subsequence {fnk}∞k=1 of {fn}∞n=1 such that fnk → f almost uniformly.Hint: Show that there is a subsequence {nk}∞k=1 of N such that

μ({

x : |fnk (x) − fnk+1(x)| ≥ 2−k})

≤ 2−k.

You will also need to apply the Weierstrass M -test.

5.91 Suppose that (Ω,A, μ) is a measure space and that f , f1, f2, . . . are complex-valuedA-measurable functions on Ω. Show that

{x : limn→∞

fn(x) = f(x) } =

∞⋂m=1

( ∞⋃n=1

( ∞⋂k=n

{x : |f(x) − fk(x)| < 1/m }))

.

★5.92 Suppose that (Ω,A, μ) is a finite measure space and that f , f1, f2, . . . are complex-valued A-measurable functions on Ω. Prove that fn → f μ-ae if and only if

limn→∞

μ

( ∞⋃k=n

{x : |f(x) − fk(x)| ≥ ε })

= 0

for each ε > 0. Compare this equation with the definition of convergence in measure.

5.93 Suppose that (Ω,A, μ) is a finite measure space and that {fn}∞n=1 is a sequence ofcomplex-valued A-measurable functions that converges in measure to f . Further sup-pose that g: C → C is continuous. Prove that g ◦ fn → g ◦ f in measure. Hint: For agiven ε > 0, let an = μ

({x :∣∣g(f(x)

)− g(fn(x)

)∣∣ ≥ ε})

. Show that each subsequenceof {an}∞n=1 has a subsequence that converges to 0.

5.94 Egorov’s theorem: Suppose that (Ω,A, μ) is a finite measure space and that f , f1,f2, . . . are complex-valued A-measurable functions on Ω. Prove that, if fn → f μ-ae,then fn → f almost uniformly.

Constantin Carath´eodory (1873–1950) - Elsevier.comA COURSE IN REAL ANALYIS 2/E John N. McDonald...

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