674 CHAPTER 24. THE BOCHNER INTEGRAL

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 9.1.2.

By Egoroff’s theorem, Theorem 24.1.18, there exists a set of measure less than δ , Esuch that the convergence of { fn} to f is uniform off E. Therefore,∫

∥ f − fn∥dµ ≤∫

E(∥ f∥X +∥ fn∥X )dµ +

∫EC∥ f − fn∥X dµ

<2ε

3+∫

EC

ε

(µ (Ω)+1)3dµ < ε

if n is large enough. ■Note that a convenient way to achieve uniform integrability is to say { fn} is bounded

in Lp (Ω;X) for some p > 1. This follows from Holder’s inequality.∫E∥ fn∥dµ ≤

(∫E

)1/p′(∫Ω

∥ fn∥p dµ

)1/p

≤Cµ (E)1/p′ .

The following theorem is interesting.

Theorem 24.5.8 Let 1 ≤ p < ∞ and let p < r ≤ ∞. Then Lr ([0,T ] ,X) is a Borelsubset of Lp ([0,T ] ;X). Letting C ([0,T ] ;X) denote the continuous functions having valuesin X, C ([0,T ] ;X) is also a Borel subset of Lp ([0,T ] ;X) . Here the measure is ordinary onedimensional Lebesgue measure on [0,T ].

Proof: First consider the claim about Lr ([0,T ] ;X). Let

BM ≡{

x ∈ Lp ([0,T ] ;X) : ∥x∥Lr([0,T ];X) ≤M}.

Then BM is a closed subset of Lp ([0,T ] ;X) . Here is why. If {xn} is a sequence ofelements of BM and xn → x in Lp ([0,T ] ;X) , then passing to a subsequence, still denotedby xn, it can be assumed xn (s)→ x(s) a.e. Hence Fatou’s lemma can be applied to conclude∫ T

0∥x(s)∥r ds≤ lim inf

n→∞

∫ T

0∥xn (s)∥r ds≤Mr < ∞.

Now ∪∞M=1BM = Lr ([0,T ] ;X) . Note this did not depend on the measure space used. It

would have been equally valid on any measure space.Consider now C ([0,T ] ;X) . The norm on this space is the usual norm, ∥·∥

∞. The argu-

ment above shows ∥·∥∞

is a Borel measurable function on Lp ([0,T ] ;X) . This is becauseBM ≡{x ∈ Lp ([0,T ] ;X) : ∥x∥

∞≤M} is a closed, hence Borel subset of Lp ([0,T ] ;X). Now

let θ ∈ L (Lp ([0,T ] ;X) ,Lp (R;X)) such that θ (x(t)) = x(t) for all t ∈ [0,T ] and alsoθ ∈ L (C ([0,T ] ;X) ,BC (R;X)) where BC (R;X) denotes the bounded continuous func-tions with a norm given by ∥x∥ ≡ supt∈R ∥x(t)∥ , and θx has compact support.

For example, you could define

x̃(t)≡

x(t) if t ∈ [0,T ]x(2T − t) if t ∈ [T,2T ]x(−t) if t ∈ [−T,0]0 if t /∈ [−T,2T ]

674 CHAPTER 24. THE BOCHNER INTEGRALProof: Let € > 0 be given. Then by uniform integrability there exists 6 > 0 such thatif uw (E) < 6 then[Whilldu <e73.EBy Fatou’s lemma the same inequality holds for f. Fatou’s lemma shows f € L! (Q;X), fbeing measurable because of Theorem 9.1.2.By Egoroff’s theorem, Theorem 24.1.18, there exists a set of measure less than 6, Esuch that the convergence of {f,,} to f is uniform off E. Therefore,[lt fullawe <= [UUfll + Walled ae [oP Salled< ~4f —* ance3 | Jec (u(Q +3"if n is large enough.Note that a convenient way to achieve uniform integrability is to say {f,,} is boundedin L? (Q;X) for some p > 1. This follows from Holder’s inequality.1/p! \/p ;[Amlans (fae) (fpimirae) <cwcey”.E E QThe following theorem is interesting.Theorem 24.5.8 Let 1 < p< and let p <r <0. Then L" ((0,T],X) is a Borelsubset of L? ({0,T|;X). Letting C ({0,T];X) denote the continuous functions having valuesin X, C((0,7];X) is also a Borel subset of L? (|0,T]|;X). Here the measure is ordinary onedimensional Lebesgue measure on |0,T}.Proof: First consider the claim about L’ ((0,7];X). LetBy = {x EL? ((0,T]X):[lellieqosrxy) <M}Then By is a closed subset of L? ([0,T];X). Here is why. If {x,} is a sequence ofelements of By and x, + x in L? ([0,7];X), then passing to a subsequence, still denotedby x;, it can be assumed x, (s) — x(s) a.e. Hence Fatou’s lemma can be applied to concludeT T[ \x(o)ll'as < tim int f lIxn(s)||" ds <M" <0,0 neo JNow Uy,_;Bu = L’ ([0,T];X). Note this did not depend on the measure space used. Itwould have been equally valid on any measure space.Consider now C ((0,7];X). The norm on this space is the usual norm, ||-||,,. The argu-ment above shows ||-||,, is a Borel measurable function on L? ((0,7];X). This is becauseBu = {x € L? ((0,T];X) : ||x||,, <M} is aclosed, hence Borel subset of L? ({0,T];X). Nowlet 6 € L(L? ((0,T];X),L? (R;X)) such that 6 (x(t)) = x(t) for all ¢ € [0,7] and also6 € £(C((0,T];X),BC(IR;X)) where BC (IR;X) denotes the bounded continuous func-tions with a norm given by ||x|| = sup,<p ||x (¢)||, and @x has compact support.For example, you could definex(t) ift € [0,7]x(2T —t) ift € [T,2T]x(—t) ift € [-T,0}0 ifr ¢ [—T,2T]x(t) =