672 CHAPTER 21. THE BOCHNER INTEGRAL

Proof: The measurability of x follows from Theorem 21.1.10 if convergence happensfor each s. Otherwise, x is measurable by assumption. Then ∥xn (s)− x(s)∥ ≤ 2g(s) a.e.so, from Fatou’s lemma,∫

2g(s)dµ ≤ lim infn→∞

∫Ω

(2g(s)−∥xn (s)− x(s)∥)dµ

=∫

2g(s)dµ− lim supn→∞

∫Ω

∥xn (s)− x(s)∥dµ

and so,

lim supn→∞

∫Ω

∥xn (s)− x(s)∥dµ ≤ 0

Also, from Fatou’s lemma again,∫Ω

∥x(s)∥dµ ≤ lim infn→∞

∫Ω

∥xn (s)∥dµ <∫

g(s)dµ < ∞

so x ∈ L1. Then by the triangle inequality,

lim supn→∞

∥∥∥∥∫Ω

x(s)dµ−∫

xn (s)dµ

∥∥∥∥≤ lim supn→∞

∫Ω

∥xn (s)− x(s)∥dµ = 0

One can also give a version of the Vitali convergence theorem.

Definition 21.5.6 Let A ⊆ L1 (Ω;X). Then A is said to be uniformly integrable if forevery ε > 0 there exists δ > 0 such that whenever µ (E)< δ , it follows∫

E∥ f∥X dµ < ε

for all f ∈A . It is bounded if

supf∈A

∫Ω

∥ f∥X dµ < ∞.

Theorem 21.5.7 Let (Ω,F ,µ) be a finite measure space and let X be a separable Banachspace. Let { fn}⊆ L1 (Ω;X) be uniformly integrable and bounded such that fn (ω)→ f (ω)for each ω ∈Ω. Then f ∈ L1 (Ω;X) and

limn→∞

∫Ω

∥ fn− f∥X dµ = 0.

Proof: Let ε > 0 be given. Then by uniform integrability there exists δ > 0 such thatif µ (E)< δ then ∫

E∥ fn∥dµ < ε/3.

By Fatou’s lemma the same inequality holds for f . Fatou’s lemma shows f ∈ L1 (Ω;X), fbeing measurable because of Theorem 11.1.9.

672 CHAPTER 21. THE BOCHNER INTEGRALProof: The measurability of x follows from Theorem 21.1.10 if convergence happensfor each s. Otherwise, x is measurable by assumption. Then ||x, (s) —x(s)|| < 2g(s) ae.so, from Fatou’s lemma,[2e(s)au <_ tim int [2a (s)~ lvn(s)—x(9)|) duI 2¢(s)dy—lim sup f|lxs(s) —x(s)|| aaN—yoo»and so,lim sup | ||xn(s) —x(s)||du <0QnooAlso, from Fatou’s lemma again,J leoolldy stim int f sn (plan < f e(s)du <eQ no JO Qso x € L!. Then by the triangle inequality,lim sup [eran [s0(s)du| <tim sup | sn(s)—x(8)||qu=0N—yoo |] « NooOne can also give a version of the Vitali convergence theorem.Definition 21.5.6 Let <& C L'(Q;X). Then &f is said to be uniformly integrable if forevery € > 0 there exists 5 > 0 such that whenever U(E) < 6, it follows[\fledu<eEfor all f © &. It is bounded ifsup | Iifllydu <e.fea JQTheorem 21.5.7 Let (Q,.F, 1) be a finite measure space and let X be a separable Banachspace. Let { fy} CL! (Q;X) be uniformly integrable and bounded such that f, (@) > f (@)for each @ € Q. Then f € L' (Q;X) andlim [ fs—Sllx au =0.Proof: Let € > 0 be given. Then by uniform integrability there exists 6 > 0 such thatif u(E) < 6 thenJM fillan <e/3.EBy Fatou’s lemma the same inequality holds for f. Fatou’s lemma shows f € L! (Q;X), fbeing measurable because of Theorem 11.1.9.