Kenneth Kuttler

294 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL

Theorem 10.9.7 Let { fn} be a uniformly integrable set of complex valued func-tions, µ(Ω)< ∞, and fn(x)→ f (x) a.e. where f is a measurable complex valued function.Then f ∈ L1 (Ω) and

limn→∞

∫Ω

| fn− f |dµ = 0. (10.12)

Proof: First it will be shown that f ∈ L1 (Ω). By uniform integrability, there existsδ > 0 such that if µ (E) < δ , then

∫E | fn|dµ < 1 for all n. By Egoroff’s theorem, there

exists a set E of measure less than δ such that on EC, { fn} converges uniformly. There-fore, for p large enough, and n > p,

∫EC

∣∣ fp− fn∣∣dµ < 1 which implies

∫EC | fn|dµ <

1+∫

Ω

∣∣ fp∣∣dµ.Then since there are only finitely many functions, fn with n≤ p, there exists

a constant, M1 such that for all n,∫

EC | fn|dµ < M1. But also,∫Ω

| fm|dµ =∫

EC| fm|dµ +

∫E| fm| ≤M1 +1≡M.

Therefore, by Fatou’s lemma,∫

Ω| f |dµ ≤ liminfn→∞

∫| fn|dµ ≤ M, showing that f ∈ L1

as hoped.Now S∪{ f} is uniformly integrable so there exists δ 1 > 0 such that if µ (E) < δ 1,

then∫

E |g|dµ < ε/3 for all g ∈ S∪{ f}.By Egoroff’s theorem, there exists a set, F with µ (F) < δ 1 such that fn converges

uniformly to f on FC. Therefore, there exists m such that if n>m, then∫

FC | f − fn|dµ < ε

3 .It follows that for n > m,∫

Ω

| f − fn|dµ ≤∫

FC| f − fn|dµ +

∫F| f |dµ +

∫F| fn|dµ <

3= ε,

which verifies 10.12. ■

10.10 One Dimensional Lebesgue Stieltjes IntegralLet F be an increasing function defined on R. Let µ be the Lebesgue Stieltjes measuredefined in Theorems 9.9.1 and 9.7.4. The conclusions of these theorems are reviewed here.

Theorem 10.10.1 Let F be an increasing function defined on R, an integratorfunction. There exists a function µ : P (R)→ [0,∞] which satisfies the conditions of The-orem 9.7.4 in terms of measures of intervals and the inner and outer regularity properties.

The Lebesgue integral taken with respect to this measure, is called the Lebesgue Stielt-jes integral. Note that any real valued continuous function is measurable with respect to S .This is because if f is continuous, inverse images of open sets are open and open sets are inS . Thus f is measurable because f−1 ((a,b)) ∈S . Similarly if f has complex values thisargument applied to its real and imaginary parts yields the conclusion that f is measurable.This will be denoted here by

∫f dµ but it is often the case that it is denoted as

∫f dF.

In the case of most interest, where F (x) = x, how does the Lebesgue integral comparewith the Riemann integral? The short answer is that if f is Riemann integrable, then itis also Lebesgue interable and the two integrals coincide. It is customary to denote theLebesgue integral in this context as

∫ ba f dm.

Theorem 10.10.2 Suppose f is Riemann integrable on an interval [a,b]. Then f isalso Lebesgue integrable and the two integrals are the same.

294 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRALTheorem 10.9.7 Lez {fn} be a uniformly integrable set of complex valued func-tions, U(Q) <%, and f(x) > f(x) ae. where f is a measurable complex valued function.Then f € L' (Q) andlim | \fr—fldu =0. (10.12)no JOProof: First it will be shown that f € L'(Q). By uniform integrability, there exists6 > 0 such that if u(E) < 6, then f,|fn|du < 1 for all n. By Egoroff’s theorem, thereexists a set E of measure less than 6 such that on E©, { f,} converges uniformly. There-fore, for p large enough, and n > p, {pc | fp — fn|du < 1 which implies fic |fn|du <1+ fo | Fp|du.Then since there are only finitely many functions, f,, with n < p, there existsaconstant, M, such that for all n, fic | fn|du < M1. But also,[linia = | lnlde+ [fal <i +1 =m,Q EC ETherefore, by Fatou’s lemma, fo |f|du <liminf,. f|fnl|du <M, showing that f € L!as hoped.Now GU {f} is uniformly integrable so there exists 6; > 0 such that if u(E) < 61,then f,|g|du < €/3 for all ge GU{f}.By Egoroff’s theorem, there exists a set, F with u(F) < 6, such that f, convergesuniformly to f on F©. Therefore, there exists m such that if n > m, then frc |f — fnldp < §.It follows that for n > m,E €E— + —E[it-tlaws [op tldus [isldur [ ildu<5+5+5 =e,which verifies 10.12. Hi10.10 One Dimensional Lebesgue Stieltjes IntegralLet F be an increasing function defined on R. Let yt be the Lebesgue Stieltjes measuredefined in Theorems 9.9.1 and 9.7.4. The conclusions of these theorems are reviewed here.Theorem 10.10.1 Let F be an increasing function defined on R, an integratorfunction. There exists a function pb : Y (IR) — [0,°°] which satisfies the conditions of The-orem 9.7.4 in terms of measures of intervals and the inner and outer regularity properties.The Lebesgue integral taken with respect to this measure, is called the Lebesgue Stielt-jes integral. Note that any real valued continuous function is measurable with respect to 7.This is because if f is continuous, inverse images of open sets are open and open sets are inSY. Thus f is measurable because f—! ((a,b)) € -Y. Similarly if f has complex values thisargument applied to its real and imaginary parts yields the conclusion that f is measurable.This will be denoted here by { fd but it is often the case that it is denoted as [ fdF.In the case of most interest, where F (x) =x, how does the Lebesgue integral comparewith the Riemann integral? The short answer is that if f is Riemann integrable, then itis also Lebesgue interable and the two integrals coincide. It is customary to denote theLebesgue integral in this context as [ ° fdm.Theorem 10.10.2 Suppose f is Riemann integrable on an interval |a,b]. Then f isalso Lebesgue integrable and the two integrals are the same.