672 CHAPTER 24. THE BOCHNER INTEGRAL

Thus, if N > M, and s is a point where g(s)< ∞,

∥xN+1 (s)− xM+1 (s)∥X ≤N

∑n=M+1

∥xn+1 (s)− xn (s)∥X

≤∞

∑n=M+1

∥xn+1 (s)− xn (s)∥X

which shows that {xN+1 (s)}∞

N=1 is a Cauchy sequence for each s /∈ E. Now let

x(s)≡{

limN→∞ xN (s) if s /∈ E,0 if s ∈ E.

Theorem 24.1.10 shows that x is strongly measurable. By Fatou’s lemma,∫Ω

∥x(s)− xN (s)∥p dµ ≤ lim infM→∞

∫Ω

∥xM (s)− xN (s)∥p dµ.

But if N and M are large enough with M > N,(∫Ω

∥xM (s)− xN (s)∥p dµ

)1/p

≤M

∑n=N∥xn+1− xn∥p ≤

∞

∑n=N∥xn+1− xn∥p < ε

and this shows, since ε is arbitrary, that

limN→∞

∫Ω

∥x(s)− xN (s)∥p dµ = 0.

It remains to show x ∈ Lp (Ω;X). This follows from the above and the triangle inequality.Thus, for N large enough,(∫

Ω

∥x(s)∥p dµ

)1/p

≤(∫

Ω

∥xN (s)∥p dµ

)1/p

+

(∫Ω

∥x(s)− xN (s)∥p dµ

)1/p

≤(∫

Ω

∥xN (s)∥p dµ

)1/p

+ ε < ∞. ■

Theorem 24.5.3 Lp (Ω;X) is complete. Also every Cauchy sequence has a subse-quence which converges pointwise.

Proof: If {xn} is Cauchy in Lp (Ω;X), extract a subsequence {xnk} satisfying∥∥xnk+1 − xnk

∥∥p ≤ 2−k

and apply Lemma 24.5.2. The pointwise convergence of this subsequence was establishedin the proof of this lemma. This proves the theorem because if a subsequence of a Cauchysequence converges, then the Cauchy sequence must also converge. ■

Observation 24.5.4 If the measure space is Lebesgue measure then you have continu-ity of translation in Lp (Rn;X) in the usual way. More generally, for µ a Radon measureon Ω a locally compact Hausdorff space, Cc (Ω;X) is dense in Lp (Ω;X) . Here Cc (Ω;X) isthe space of continuous X valued functions which have compact support in Ω. The proof ofthis little observation follows immediately from approximating with simple functions andthen applying the appropriate considerations to the simple functions.