A simplified construction of the Lebesgue integral
aa r X i v : . [ m a t h . C A ] M a y A SIMPLIFIED CONSTRUCTION OF THE LEBESGUEINTEGRAL
VILMOS KOMORNIK
Abstract.
We present a modification of Riesz’s construction ofthe Lebesgue integral, leading directly to finite or infinite integrals,at the same time simplifying the proofs. Introduction
Among the many approaches to the Lebesgue integral that of Riesz[8, 9, 11] is probably the shortest and most elementary. As Daniell’sabstract method [3], it is motivated by the research of weak sufficientconditions ensuring the relation R f n dx → R f dx for pointwise conver-gent sequences f n → f . Based on two elementary lemmas concerningmonotone sequences of step functions, the functions having a finiteLebesgue integral are constructed in two steps. It is completed byintroducing measurable functions having infinite integrals.This theory has regained popularity recently; see, e.g., Bear [1], Chae[2], Johnston [6], Roselli [12].We modify this method by defining directly the functions havinga finite or infinite integral. Technically we postpone the use of thecrucial “Lemma B” of Riesz to the second stage of the construction.This leads to even shorter and more transparent proofs, and yieldsthe Lebesgue measure at once as a byproduct. In our approach thefundamental lemmas have a symmetric use: just as Lemma A justifiesthe correctness of the first step of the extension of the integral, LemmaB plays the same role for the second step.For clarity the theory is presented in Sections 2–6 for functions de-fined on the real line. In Sections 7–8 we extend the results to arbi-trary measure spaces and we investigate product measures. In the finalSection 9 we explain that various difficulties and counter-examples dis-appear if we return to Riesz’s natural definition of measurability andto Fréchet’s σ -ring framework. We omit in the main text some proofsthat remain the same as in the classical setting; for the convenience ofthe reader they are recalled in an Appendix.The author is grateful to László Czách for stimulating discussionsand for indicating him the last examples of the paper. Date : Version 2018-05-18.
Key words and phrases.
Lebesgue integral, Riesz-Daniell approach, generalizedBeppo Levi theorem, Fubini–Tonelli theorem. Integral of step functions
As usual, N ⊂ R is called a null set if for each ε > it has a countablecover by intervals of total length < ε . We say that a property holds almost everywhere (a.e.) if it holds outside a null set.We identify two functions if they are equal almost everywhere, i.e.,we are working with equivalence classes of functions. Accordingly, wewrite f = g , f ≤ g , f ≥ g , f n → f , f n ր f or f n ց f if these relationshold a.e. (Sometimes we keep the words “a.e.” for clarity.)We denote by χ A the characteristic function of the set A . By a stepfunction we mean a function having a representation of the form ϕ = m X i =1 c i χ I i with bounded intervals I i and real numbers c i . We define its integral by the formula Z ϕ dx := m X i =1 c i | I i | where | I i | denotes the length of I i . Proposition 2.1. (i)
The step functions form a vector lattice R . (ii) R ϕ dx does not depend on the particular representation of ϕ . (iii) The integral is a positive linear functional on R .Proof. The proof is elementary and well-known if we consider individualfunctions and not equivalence classes. In the latter case (ii) follows fromBorel’s theorem stating that the non-degenerate intervals are not nullsets. (cid:3)
In the next two sections we extend the integral in two steps to largerfunction classes R and R . We recall without proof two elementaryresults of Riesz: they will ensure the correctness of these extensions to R and R , respectively. Lemma A. If ( ϕ n ) ⊂ R and ϕ n ( x ) ց , then R ϕ n dx ց . Lemma B. If ( ϕ n ) ⊂ R , ϕ n ( x ) ր f and sup R ϕ n dx < ∞ , then f isfinite a.e. The R function class A function f : R → R is said to belong to R if there exists a sequence ( ϕ n ) ⊂ R satisfying ϕ n ր f , and we set Z f dx := lim Z ϕ n dx. For the convenience of the reader we present the missing classical proofs ofSections 2–8 in an appendix at the end of the paper.
EBESGUE INTEGRAL 3
Proposition 3.1 (i), (ii) below will show that the definition of the inte-gral is correct.Riesz used instead of R the smaller class C := (cid:26) f ∈ R : Z f dx is finite (cid:27) . The elements of R may have infinite integrals, and they may eventake the value ∞ on non-null sets. Nevertheless, we show that theusual properties of C continue to hold in R by essentially the sameproofs. Proposition 3.1. (i) R f dx does not depend on the particular choice of ( ϕ n ) . (ii) The integral on R is an extension of the integral on R . (iii) If f, g ∈ R and f ≤ g , then R f dx ≤ R g dx . (iv) If f, g ∈ R and c is a nonnegative real number, then (3.1) cf, f + g, min { f, g } and max { f, g } also belong to R , and Z cf dx = c Z f dx, Z f + g dx = Z f dx + Z g dx. Proof. (i) and (iii) follow from Lemma 3.2 below. (ii) follows from (i)by using constant sequences of step functions. For (iv) we observe thatif ( ϕ n ) , ( ψ n ) ⊂ R , ϕ n ր f and ψ n ր g , then the sequences ( cϕ n ) , ( ϕ n + ψ n ) , (min { ϕ n , ψ n } ) and (max { ϕ n , ψ n } ) are non-decreasing, belong to R , and tend to the sequences in (3.1).The equalities follow from the linearity of the integral on R . (cid:3) Lemma 3.2.
Let ( ϕ n ) , ( ψ n ) ⊂ R and ϕ n ր f , ψ n ր g . If f ≤ g ,then lim Z ϕ n dx ≤ lim Z ψ n dx. Proof.
It suffices to show for each fixed m the inequality Z ϕ m dx ≤ lim n →∞ Z ψ n dx. Since the left side is finite, this is equivalent to the inequality lim n →∞ Z ϕ m − ψ n dx ≤ . Applying Lemma A to the sequence of functions ( ϕ m − ψ n ) + := max { ϕ m − ψ n , } ց as n → ∞ We use the conventions · ( ±∞ ) := 0 . We did not need this observation in the classical framework.
VILMOS KOMORNIK we obtain the still stronger relation lim n →∞ Z ( ϕ m − ψ n ) + dx ≤ . (cid:3) The class R is stable for the process used in its definition: Proposition 3.3. If ( f n ) ⊂ R and f n ր f , then f ∈ R and Z f n dx ր Z f dx. Proof.
Fix for each n a sequence ( ϕ n,k ) ⊂ R satisfying ϕ n,k ր f n .Then the formula ϕ k := sup n,i ≤ k ϕ n,i defines a non-decreasing sequence of step functions.For each fixed n , we have ϕ n,k ≤ ϕ k ≤ f for all k ≥ n ; letting k → ∞ we conclude that f n ≤ lim ϕ k ≤ f for each n . Since f n → f a.e., weconclude that ϕ k ր f a.e. Therefore f ∈ R and R ϕ k dx → R f dx .Since ϕ n ≤ f n ≤ f for all n , integrating and then letting n → ∞ weobtain that lim Z f n dx = lim Z ϕ n dx = Z f dx. (cid:3) The R function class and the Lebesgue measure It is natural to define R − f dx := − R f dx if f ∈ R . More generally,we write f − f ∈ R if f , f ∈ R and R f dx − R f dx is well defined,and for f = f − f ∈ R we set Z f dx := Z f dx − Z f dx. The function class R is well defined: if f , f ∈ R , then R f dx − R f dx is defined if and only if at least one of the two integrals is finite.Then at least one of the functions f , f is finite a.e. by Lemma B, sothat the function f − f is defined a.e.Proposition 4.1 (i), (ii) below will show that the definition of theintegral is also correct. Remark.
We may assume in the definition of R that f , f ≥ : choosefor i = 1 , a step function ϕ i satisfying ϕ i ≤ f i , and change f i to f i − max { ϕ , ϕ } .Riesz used instead of R the smaller class C := (cid:26) f ∈ R : Z f dx is finite (cid:27) . The elements of R may have infinite integrals, and they are not nec-essarily finite a.e. Nevertheless, most properties of the integral on C remain valid on R . EBESGUE INTEGRAL 5
Our first result readily follows from Proposition 3.1 for the originalclass C ; for the extended class R some new arguments are needed: Proposition 4.1. (i) R f dx does not depend on the particular choice of f and f . (ii) The integral on R is an extension of the integral on R . (iii) If f, g ∈ R and f ≤ g , then R f dx ≤ R g dx . (iv) If f ∈ R and c ∈ R , then cf ∈ R and R cf dx = c R f dx . (v) If f, g ∈ R and R f dx + R g dx is well defined, then f + g ∈ R and Z f + g dx = Z f dx + Z g dx. (vi) If f, g ∈ R , then max { f, g } and min { f, g } also belong to R .Proof. For (i) and (iii) we have to show that if f = f − f and g = g − g as in the definition of f, g ∈ R , and f ≤ g , then(4.1) Z f dx − Z f dx ≤ Z g dx − Z g dx. The inequality is obvious if R f dx = ∞ . Henceforth we assume that R f dx is finite.If R g dx is also finite, then our assumption f − f ≤ g − g impliesthat f + g ≤ g + f because f and g are finite a.e. by Lemma B,and then R f dx + R g dx ≤ R g dx + R f dx by Proposition 3.1 (iv).Since R g dx and R g dx are finite, hence (4.1) follows.If R g dx = ∞ , then choose a sequence ( ϕ n ) ⊂ R satisfying ϕ n ր g . We have f − f ≤ g − ϕ n for each n . Applying the precedingarguments with ϕ n in place of g we get Z f dx − Z f dx ≤ Z g dx − Z ϕ n dx, and (4.1) follows by letting n → ∞ .(ii) follows from (i) by choosing f := f and f := 0 if f ∈ R .(iv) is obvious.(v) Write f = f − f and g = g − g with f , f , g , g ∈ R as inthe definition of f, g ∈ R . If both integrals R f dx and R g dx arefinite, then Proposition 3.1 (iv) shows that f + g , f + g ∈ R , and R f + g dx = R f dx + R g dx is finite. Therefore f + g = ( f + g ) − ( f + g ) VILMOS KOMORNIK belongs to R , and using Proposition 3.1 (iv) again we obtain that Z f + g dx = Z f + g dx − Z f + g dx = (cid:18)Z f dx + Z g dx (cid:19) − (cid:18)Z f dx + Z g dx (cid:19) = (cid:18)Z f dx − Z f dx (cid:19) + (cid:18)Z g dx − Z g dx (cid:19) = Z f dx + Z g dx. If one of the integrals R f dx and R g dx is infinite, then bothintegrals R f dx and R g dx are finite by the definition of f ∈ R andby our assumption that R f dx + R g dx is well defined, and we mayrepeat the above proof.(vi) By symmetry we consider only the case of maximum. We haveto show that if f = f − f and g = g − g as in the definition of f, g ∈ R , then max { f − f , g − g } ∈ R . If f , g ∈ R , then max { f − f , g − g } = ( f + g ) + max {− g − f , − f − g } = ( f + g ) − min { g + f , f + g } is the difference of two elements of R . Furthermore, R f + g dx isfinite because f + g ∈ R , so that the difference belongs to R .In the general case we choose two sequences ( ϕ n ) , ( ψ n ) ⊂ R satis-fying ϕ n ր f and ψ n ր g . We have max { ϕ n − f , ψ n − g } ∈ R bythe preceding arguments, and max { ϕ n − f , ψ n − g } ր max { f − f , g − g } . Since Z max { ϕ n − f , ψ n − g } dx > −∞ for all n , because at least one of the integrals R f dx and R g dx arefinite, we may conclude by applying Theorem 4.2 below. (cid:3) The class R has a similar invariance property as R : Theorem 4.2 (Generalized Beppo Levi theorem) . Let ( f n ) ⊂ R and f n ր f . If R f n dx > −∞ for at least one n , then f ∈ R and (4.2) Z f n dx ր Z f dx. Examples.
The functions f n := − χ ( −∞ , + χ (0 ,n ) and f n := − χ ( n, ∞ ) show that the assumption lim R f n dx > −∞ cannot be omitted. In-deed, in the first case f = sign / ∈ R ; in the second case f = 0 ∈ R ,but (4.2) fails. EBESGUE INTEGRAL 7
The proof below is a simple adaptation (even a simplification) of theusual one for the smaller class C . Since this theorem is fundamentalin the present construction, we give the proof here for the convenienceof the reader. Proof of Theorem 4.2.
By omitting a finite number of initial terms wemay assume that R f n dx > −∞ for all n . Write f n = g n − h n with g n , h n ∈ R . Since R h n dx is finite, there exists ϕ n ∈ R satisfying ϕ n ≤ h n and R h n − ϕ n dx < − n . Changing g n and h n to g n − ϕ n and h n − ϕ n we may assume that f n = g n − h n , g n , h n ∈ R , h n ≥ and Z h n dx ≤ − n for n = 1 , , . . . . Finally, changing g n and h n by induction on n =2 , , . . . to h + · · · + h n − + g n and h + · · · + h n − + h n we may assume that f n = g n − h n , g n , h n ∈ R , h n ր and Z h n dx ≤ for n = 1 , , . . . . Applying Proposition 3.3 to the non-decreasing sequences ( h n ) and ( g n ) = ( f n + h n ) we obtain that h n ր h and g n ր g with suitablefunctions h, g ∈ R , and R h dx ≤ < ∞ . Hence f = g − h ∈ R and Z f n dx = Z g n dx − Z h n dx → Z g dx − Z h dx = Z f dx. (cid:3) Now we may greatly generalize the length of intervals. We write A ∈ M if χ A ∈ R , and we set µ ( A ) := R χ A dx in this case. Theorem4.2 yields at once the Theorem 4.3. µ is a σ -finite, complete measure on M . The following notions are used here. Given a family A of subsetsof a set X with ∅ ∈ A , by a measure on A we mean a nonnegativefunction µ : A → R such that µ ( ∅ ) = 0 , and satisfying the σ - additivity relations µ ( A ) = ∞ X n =1 µ ( A n ) whenever A ∈ A is the disjoint union of a sequence ( A n ) ⊂ A .This measure is called σ - finite if each A ∈ A has a countable coverby sets of finite measure, and complete if all subsets of a set of measurezero also belong to A .We call the elements of M measurable sets . Remark.
We will describe the structure of M in Proposition 6.2 (iv)below. VILMOS KOMORNIK The space L We write f ∈ L if f ∈ R and R f dx is finite, i.e., L := (cid:26) f ∈ R : Z f dx is finite (cid:27) . Each f ∈ L is finite a.e. by Lemma B.Observe that L coincides with the class C of Riesz. We are goingto show that L has a simple structure, and the integral on L has someremarkable properties. Proposition 5.1. (i) L is a vector lattice. (ii) The integral is a positive linear functional on L .Proof. (i) We show that if f, g ∈ L and c ∈ R then cf, f + g, min { f, g } and max { f, g } also belong to L . Write f = f − f and g = g − g with f , f , g , g ∈ R having finite integrals. Then the claim follows from the representa-tions cf = cf − cf if c ≥ ,cf = ( − c ) f − ( − c ) f if c < ,f + g = ( f + g ) − ( f + g ) , min { f, g } = min { f + g , g + f } − ( f + g ) , max { f, g } = max { f + g , g + f } − ( f + g ) , because by Proposition 3.1 (iv) the right side of each equality is thedifference of two functions from R having finite integrals.(ii) The integral is linear by Proposition 3.1 (iv) and by the definitionof the integral on R , and it is monotone by Proposition 4.1 (iii). (cid:3) Theorem 4.2 yields at once the fundamental
Theorem 5.2 (Beppo Levi) . If ( f n ) ⊂ L , f n ր f and sup R f n dx < ∞ , then f ∈ L and Z f n dx ր Z f dx. Corollary 5.3. (i)
The formula k f k := R | f | dx defines a norm on L . (ii) A set A is a null set ⇐⇒ A ∈ M and µ ( A ) = 0 .Proof. (i) The only non-trivial property is that if R | f | dx = 0 , then f = 0 . This follows by applying Theorem 5.2 with f n := n | f | .(ii) We have to show that χ A = 0 ⇐⇒ R χ A dx = 0 . The implication = ⇒ follows from the definition of the integral. The converse implicationfollows from (i). (cid:3) EBESGUE INTEGRAL 9
We recall that Theorem 5.2 also implies the following two importanttheorems:
Theorem 5.4 (Fatou) . Let ( f n ) ⊂ L and f n → f . If f n ≥ for all n , and lim inf R f n dx < ∞ , then f ∈ L and Z f dx ≤ lim inf Z f n dx. Theorem 5.5 (Lebesgue) . Let ( f n ) ⊂ L and f n → f . If there exists g ∈ L such that | f n | ≤ g for all n , then f ∈ L , and R f n dx → R f dx . Measurable functions
We may simplify the manipulation of integrable functions by intro-ducing the notion of measurability. This will also allow us to precisethe structure of the family M of measurable sets, and to generalizeFatou’s theorem for functions having infinite integrals.Following Riesz we call a function f : R → R measurable if thereexists a sequence ( ϕ n ) ⊂ R such that ϕ n → f a.e. Proposition 6.1. (i) If f ∈ R , then f is measurable. (ii) If f and g are measurable, then | f | , f g , max { f, g } and min { f, g } are also measurable. Furthermore, f /g and f ± g are also mea-surable whenever they are defined a.e.Proof. (i) Write f = f − f with f , f ∈ R . If ( ϕ n ) , ( ψ n ) ⊂ R and ϕ n ր f , ψ n ր f , then ( ϕ n − ψ n ) ⊂ R , and ϕ n − ψ n → f .(ii) If ( ϕ n ) ⊂ R and ϕ n → f , then ( | ϕ n | ) ⊂ R and | ϕ n | → | f | . Theproof of the other statements is analogous. (cid:3) The sign function shows that the not every measurable function be-longs to R . In our next result we collect several useful statements,including a partial converse of Proposition 6.1 (i), the description ofthe family M of measurable sets, and a generalization of Fatou’s the-orem for functions having infinite integrals. The proofs will rely onLebesgue’s theorem.We recall that by a σ - ring in X we mean a family M of subsetsof X containing ∅ , the difference A \ B of any two sets A, B ∈ M ,and the union ∪ A n of any disjoint sequence ( A n ) ⊂ M . Then we have ∪ A n ∈ M and ∩ A n ∈ M for any countable sequence ( A n ) ⊂ M , evenif it is not disjoint. Proposition 6.2. (i) If f is measurable, g ∈ L and | f | ≤ g , then f ∈ L . (ii) If f is measurable and nonnegative , then f ∈ R . (iii) The measurable sets form a σ -ring. (iv) If ( f n ) is a sequence of measurable functions and f n → f , then f is measurable. (v) Let ( f n ) be a sequence of nonnegative measurable functions. If f n → f , then f is also a nonnegative measurable function, and (6.1) Z f dx ≤ lim inf Z f n dx. Part (ii) and Proposition 6.1 (i) justify the terminology of Section 4:a set is measurable if and only if its characteristic function is measur-able.
Proof. (i) If ( ϕ n ) ⊂ R and ϕ n → f , then the functions f n := med {− g, ϕ n , g } belong to the lattice L , and f n → f . Since | f n | ≤ g for all n , we mayconclude by applying Lebesgue’s theorem.(ii) Choose a non-decreasing sequence A ⊂ A ⊂ · · · of sets of finitemeasure such that f = 0 outside their union. The functions f n ( x ) := min { f ( x ) , nχ A n } belong to L by (i), and f n ր f . We conclude by applying Theorem4.2.(iii) We have ∅ ∈ M because χ ∅ = 0 ∈ R ⊂ R . Theorem 4.2yields that ∪ A n ∈ M for any disjoint sequence ( A n ) ⊂ M .It remains to show that if A, B ∈ M , then A \ B ∈ M . By (ii) itsuffices to observe that if χ A , χ B are measurable, then χ A \ B = χ A − χ A χ B is also measurable.(iv) Fix a positive function g ∈ L , and set h n := gf n g + | f n | and h := gfg + | f | . Then h n is measurable and | h n | ≤ g , so that h n ∈ L by (i). Since h n → h , by Lebesgue’s theorem h ∈ L , and hence h is measurable by(i). Since f and h have the same sign, | f | h = f | h | , and hence f = ghg − | h | is also measurable by Proposition 6.1 (ii).(v) f is a nonnegative measurable function by (iv), hence all integralsin (6.1) are defined by (ii). The relation (6.1) is obvious if the rightside of (6.1) is finite; otherwise it follows from Fatou’s theorem. (cid:3) Here med { x, y, z } denotes the middle number among x , y and z . For x ≤ z itis equal to max { x, min { y, z }} . EBESGUE INTEGRAL 11 Generalization to arbitrary measure spaces
The above construction of the Lebesgue integral may be greatly gen-eralized a follows. Let µ : P → R be a finite measure on a semiring P in a set X , i.e., on a family of sets P ⊂ X having the followingproperties: • ∅ ∈ P ; • if P, Q ∈ P , then P ∩ Q ∈ P ; • if P, Q ∈ P , then there is a finite disjoint sequence P , . . . , P n in P such that P \ Q = P ∪ · · · ∪ P n .Two simple examples are the length of bounded intervals in R andthe counting measure on the finite subsets of any given set X : µ ( P ) isthe number of elements of P .We say that N ⊂ X is a null set if for each ε > there exists asequence ( P n ) ⊂ P satisfying N ⊂ ∪ P n and P µ ( P n ) < ε . We identifytwo functions if they are equal a.e., i.e., outside a null set.We denote by R the vector space of step functions spanned by thecharacteristic functions of the sets P ∈ P , and we define the integralof step functions by the formula Z n X i =1 c i χ P i dµ := n X i =1 c i µ ( P i ) . Now we may repeat the above construction of the integral and mea-sure, and all theorems of Sections 2–6 remain valid. Moreover, onlythe proof of Proposition 6.2 (iv) has to be modified because there aremeasure spaces containing no positive measurable functions. It sufficesto use a nonnegative function g ∈ L satisfying g ( x ) = 0 = ⇒ f n ( x ) = 0 for all n, and setting h n = h := 0 when g = 0 . See, e.g., [11] or [7] for details.8. The Fubini–Tonelli theorem
Given two finite measures µ : P → R and ν : Q → R where P is asemiring in X and Q is a semiring in Y , P × Q := { P × Q : P ∈ P and Q ∈ Q} is a semiring in the product space × Y , and the formula ( µ × ν )( P × Q ) := µ ( P ) ν ( Q ) defines a finite measure on P × Q .We are going to describe the relationship between the three corre-sponding integrals. The notation R i ( X ) , R i ( X ) , R i ( X × Y ) will referto the spaces R i for the measures µ , ν and µ × ν , respectively, and wewill write dx , dy and dx dy instead of dµ , dν and d ( µ × ν ) . Theorem 8.1 (Fubini–Tonelli) . We have (8.1) Z X × Y f ( x, y ) dx dy = Z X (cid:16)Z Y f ( x, y ) dy (cid:17) dx whenever the left side is defined, i.e., f ∈ R ( X × Y ) . We emphasize that the integrals in (8.1) may be infinite.
Remark.
Under the same assumption we also have Z X × Y f ( x, y ) dx dy = Z Y (cid:16)Z X f ( x, y ) dx (cid:17) dy by symmetry. Proof.
We may repeat the proof given in [11] for the special case f ∈ C ( X × Y ) = L ( X × Y ) ; it becomes even simpler because we do nothave to check the boundedness of the sequences of integrals. (cid:3) Example.
We recall that the theorem does not hold under the weakerassumption that both successive integrals exist and are equal. Forexample, let µ = ν be the counting measure on X = Y := Z , and f ( x, y ) := if x = y + 1 , − if x = y − , otherwise . Then Z X (cid:16)Z Y f ( x, y ) dy (cid:17) dx = Z Y (cid:16)Z X f ( x, y ) dx (cid:17) dy = 0 , but the integral R X × Y f ( x, y ) dx dy is undefined.9. σ -ring or σ -algebra? In this section we argue in favor of the measurability à la
Riesz,and the σ -rings à la Fréchet [4], instead of σ -algebras, i.e., σ -ringscontaining the fundamental set X .The measurability notion adopted here differs from the one usedin most modern texts. They coincide if X has a countable cover bysets of finite measure (or equivalently by sets belonging to P ), like theLebesgue measure in R n and the probability measures.Otherwise, like for the counting measure on an uncountable set X ,the present definition is more restrictive: for example the constantfunctions are not measurable.In the latter case it is tempting to adopt a weaker definition, bycalling a function f locally measurable if f χ P is measurable for all P ∈P . Indeed, we may extend the integral to locally measurable functions f by setting R f dx := ∞ if f is nonnegative and non-measurable, andthen setting R f dx := R f + dx − R f − dx whenever the right side iswell defined. This extended integral is still monotone. EBESGUE INTEGRAL 13
However, the Fubini–Tonelli theorem may fail for locally measurablefunctions:
Example.
Let µ be the zero measure and ν the counting measure onthe finite subsets of an uncountable set X . Then the characteristicfunction f of the set D := { ( x, x ) : x ∈ X } is locally measurable for the product measure, Z X × X f ( x, y ) dx dy = ∞ and Z X (cid:16)Z X f ( x, y ) dx (cid:17) dy = 0 . Next we take a closer look of Theorem 4.3. It may be shown (see,e.g., [5], Chapter 13) that the measure µ : M → R is the only possibleextension of its restriction to the initial semiring P .If X is measurable, then M is not only a σ -ring, but also a σ - algebra . Otherwise we may extend µ further to a σ -algebra M bysetting µ ( A ) := R χ ( A ) dx whenever the characteristic function of A is locally measurable. However, this extension is not unique in general: Example.
Consider the zero measure µ on the semiring P of finite sub-sets of an uncountable set X . Then M = 2 X , and µ ( A ) = ( if A is countable, ∞ if A is uncountable.But the zero measure on X is also an extension of µ .Moreover, the two measures already differ on the smallest σ -algebra A containing P , i.e., on the family of countable subsets and their com-plements. In fact, there are infinitely many other extensions of µ to A :the formula µ α ( A ) = ( if A is countable, α if X \ A is countabledefines a different extension of µ for each ≤ α ≤ ∞ .10. Appendix
For the convenience of the reader we reproduce here some knownproofs that were admitted in the text.
Proof of Lemma A in R . We may fix a compact interval [ a, b ] and anumber M > such that ϕ , and hence all ϕ n are bounded by M andvanish outside [ a, b ] .For any fixed ε > there exists a countable sequence of open intervalsof total length < ε such that outside their union U all ϕ n are continuous. Then ϕ n → uniformly on the compact set [ a, b ] \ U by Dini’stheorem, so that ≤ ϕ n < ε on this set for all sufficiently large n .Then we also have ≤ Z ϕ n dx ≤ M ε + ε ( b − a ) , and the lemma follows by the arbitrariness of ε . (cid:3) In the proof of Lemma A for a general measure we may use the σ -additivity of the measure instead of the topological (compactness)arguments: see [11] or [7]. Proof of Lemma B.
Changing ϕ n to ϕ n − ϕ we may assume that thefunctions ϕ n are nonnegative. Fix an upper bound A > of the inte-grals R ϕ n dx , and set E ε := (cid:26) x ∈ R : f ( x ) > Aε (cid:27) , ε > . Since { x ∈ R : f ( x ) = ∞} ⊂ E ε , it suffices to show that each E ε has a countable cover by intervals of total length ≤ ε .Setting f := 0 and E ε,n := (cid:26) x ∈ R : f n ( x ) > Aε ≥ f n − ( x ) (cid:27) , n = 1 , , . . . ,E ε is the union of the disjoint sets E ε,n by the monotonicity of ( f n ) .Since each E ε,n is a finite union of disjoint intervals I n, , . . . , I n,k n , itremains to show that ∞ X n =1 k n X k =1 | I n,k | ≤ ε. By the definition of E ε,n we have for each m = 1 , , . . . the inequality Aε m X n =1 k n X k =1 | I n,k | ≤ m X n =1 Z E n ϕ n dx ≤ Z ϕ m dx ≤ A, and we conclude by letting m → ∞ . (cid:3) Proof of Fatou’s theorem 5.4.
Setting h n := inf { f k : k ≥ n } we have h n ր f . Since h n ≤ f n and hence sup R h n dx < ∞ by ourassumption lim inf R f n dx < ∞ , the relation f ∈ L will follow byapplying Theorem 5.2 if we show that ( h n ) ⊂ L .For each fixed n we set h n,m := min { f k : n ≤ k ≤ m } , m = n, n + 1 , . . . . Then ( h n,m ) ∞ m = n ⊂ L by Proposition 5.1 (i). Since h n,m ց h n , and R h n,m dx ≥ for all m , we may apply Theorem 5.2 to get − h n ∈ L ,and hence h n ∈ L . EBESGUE INTEGRAL 15
Finally, applying Theorem 4.2 and using the relations h n ≤ f n weget the required estimate: Z f dx = lim Z h n dx = lim inf Z h n dx ≤ lim inf Z f n dx. (cid:3) Proof of Lebesgue’s theorem 5.5.
Applying Fatou’s Theorem to the se-quences ( g − f n ) and ( g + f n ) we obtain that g − f, g + f ∈ L and Z g − f dx ≤ lim inf Z g − f n dx = Z g dx − lim sup Z f n dx, Z g + f dx ≤ lim inf Z g + f n dx = Z g dx + lim inf Z f n dx. Since L is a vector space, hence f ∈ L , and lim sup Z f n dx ≤ Z f dx ≤ lim inf Z f n dx. (cid:3) Let us give finally the complete proof of the Fubini–Tonelli theorem.First we recall from [11] an elementary lemma clarifying the relation-ship between one- and two-dimensional null sets:
Lemma C. If E is a null set in X × Y , then the “sections” { y ∈ Y : ( x, y ) ∈ E } of E are null sets in Y for almost every x ∈ X .Proof. Fix a sequence of “rectangles” R n = P n × Q n in P × Q , coveringeach point of E infinitely many times, and satisfying ∞ X n =1 ( µ × ν )( R n ) < ∞ . We may get such a sequence by applying the definition of null sets with ε = 2 − n for n = 1 , , . . . , and then combining the resulting covers intoone sequence.By the definition of the integral of step functions we have ( µ × ν )( R n ) = Z X × Y χ R n ( x, y ) dx dy = Z X (cid:16)Z Y χ R n ( x, y ) dy (cid:17) dx (their common value is µ ( P n ) ν ( Q n ) ), so that the series ∞ X n =1 Z X (cid:16)Z Y χ R n ( x, y ) dy (cid:17) dx is convergent. Applying the Beppo Levi theorem we obtain that theseries ∞ X n =1 Z Y χ R n ( x, y ) dy is convergent for a.e. x ∈ X . If x is such a point, then anotherapplication of the Beppo Levi theorem implies that the series ∞ X n =1 χ R n ( x , y ) is convergent for a.e. y ∈ Y . If y is such a point, then ( x , y ) / ∈ E ,because at the points of E we have P χ R n = ∞ . (cid:3) Proof of Theorem 8.1.
We have to show the following: f ( x, · ) ∈ R ( Y ) for almost each x ∈ X ; (10.1) Z Y f ( · , y ) dy ∈ R ( X ); (10.2) the two sides of (8.1) are equal.(10.3)These properties obviously hold if f is the characteristic function ofa “rectangle” P × Q . Taking linear combinations we see that they holdfor all step functions f ∈ R ( X × Y ) as well.Next let f ∈ R ( X × Y ) . Choose a sequence ( ϕ n ) ⊂ R ( X × Y ) anda null set E ⊂ X × Y such that ϕ n ( x, y ) ր f ( x, y ) for each ( x, y ) ∈ ( X × Y ) \ E ; then(10.4) Z X × Y ϕ n ( x, y ) dx dy ր Z X × Y f ( x, y ) dx dy by the definition of the integral.By Lemma C we may fix a set X ⊂ X such that X \ X is a nullset in X , and for each fixed x ∈ X , ( ϕ n ( x, · )) ⊂ R ( Y ) and ϕ n ( x, · ) ր f ( x, · ) in Y, implying f ( x, · ) ∈ R ( Y ) and the relation Z Y ϕ n ( x, y ) dy ր Z Y f ( x, y ) dy for a.e. x ∈ X. Since ( R Y ϕ n ( · , y ) dy ) ⊂ R ( X ) , this implies R Y f ( · , y ) dy ∈ R ( X ) andthe relation(10.5) Z X (cid:18)Z Y ϕ n ( x, y ) dy (cid:19) dx ր Z X (cid:18)Z Y f ( x, y ) dy (cid:19) dx. In particular, we have established (10.1) and (10.2). The equality (10.3)follows by observing that the left sides of (10.4) and (10.5) are equal,and that (10.3) is already known for step functions.If f ∈ R ( X × Y ) and R X × Y f ( x, y ) dx dy < ∞ , then applying LemmaB we also see that f is finite a.e. in X × Y , and R Y f ( x, y ) dy is finitefor every x ∈ X .Finally, if f ∈ R ( X × Y ) , then writing f = f − f with f , f ∈ R ( X × Y ) we have the required equality for f and f in place of f . EBESGUE INTEGRAL 17
Since one of the integrals R X × Y f ( x, y ) dx dy and R X × Y f ( x, y ) dx dy is finite, by the preceding paragraph we may take the difference of theseequalities to obtain the required identity for f . (cid:3) We end the appendix by recalling a not too well-known exampleof Weir [13, p. 43] (see also Johnston [6, pp. 54–55]). The strictinclusion R ( R follows from the lower boundedness of the elementsof R . The converse does not hold: a nonnegative element of R doesnot necessarily belong to R . To see this we enumerate the rationalnumbers in (0 , into a sequence ( r n ) and we set S := (0 , ∩ (cid:0) ∪ ∞ n =1 ( r n − − n − , r n + 2 − n − ) (cid:1) . Then S ⊂ (0 , and < µ ( S ) < .Now f := χ (0 , and g := χ S belong to R , so that f − g ∈ R , and f − g ≥ . However, f − g / ∈ R because R f − g dx > , and ϕ ≤ forevery step function satisfying ϕ ≤ f − g . References [1] H.S. Bear,
A Primer of Lebesgue Integration , Second Edition, Academic Press,2002.[2] S.B. Chae,
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Measure Theory , D. Van Nostrand Co., Inc., Princeton, N.J.,1950.[6] W. Johnston,
The Lebesgue Integral for Undergraduates , MAA Press, 2015.[7] V. Komornik,
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L’évolution de la notion d’intégrale depuis Lebesgue , Ann. Inst.Fourier 1 (1949), 29-42 ; see also [10] I, 327-340.[9] F. Riesz,
Les ensembles de mesure nulle et leur rôle dans l’analyse , Proceed-ings of the First Hungarian Mathematical Congress (1952), 214–224; see [10]I, 363–372.[10] F. Riesz,
Oeuvres complètes I-II (ed. Á. Császár), Akadémiai Kiadó, Bu-dapest, 1960.[11] F. Riesz, B. Sz.-Nagy,
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