24.8. THE RIESZ REPRESENTATION THEOREM 685

Lemma 24.8.4 F defined above is a vector measure with values in X ′ and |F |(Ω)< ∞.

Proof of the lemma: Clearly F (E) is linear. Also

∥F (E)∥= sup∥x∥≤1

∥F (E)(x)∥ ≤ ∥l∥ sup∥x∥≤1

∥XE (·)x∥Lp(Ω;X) ≤ ∥l∥µ (E)1/p.

Let {Ei}∞i=1 be a sequence of disjoint elements of S and let E = ∪n<∞En.∣∣∣∣∣F (E)(x)−

n

∑k=1

F (Ek)(x)

∣∣∣∣∣=∣∣∣∣∣l (XE (·)x)−

n

∑i=1

l (XEi (·)x)

∣∣∣∣∣ (24.36)

≤ ∥l∥

∥∥∥∥∥XE (·)x−n

∑i=1

XEi (·)x

∥∥∥∥∥Lp(Ω;X)

≤ ∥l∥µ

(⋃k>n

Ek

)1/p

∥x∥.

Since µ (Ω)< ∞, limn→∞ µ

( ⋃k>n

Ek

)1/p

= 0 and so inequality 24.36 shows that

limn→∞

∥∥∥∥∥F (E)−n

∑k=1

F (Ek)

∥∥∥∥∥X ′

= 0.

To show |F |(Ω) < ∞, let ε > 0 be given, let {H1, · · · ,Hn} be a partition of Ω, and let∥xi∥ ≤ 1 be chosen in such a way that F (Hi)(xi)> ∥F (Hi)∥− ε/n. Thus

−ε +n

∑i=1∥F (Hi)∥<

n

∑i=1|l (XHi (·)xi)| ≤ ∥l∥

∥∥∥∥∥ n

∑i=1

XHi (·)xi

∥∥∥∥∥Lp(Ω;X)

≤ ∥l∥(∫

Ω

n

∑i=1

XHi (s)dµ

)1/p

= ∥l∥µ (Ω)1/p.

Since ε > 0 was arbitrary, ∑ni=1 ∥F (Hi)∥ < ∥l∥µ (Ω)1/p.Since the partition was arbitrary,

this shows |F |(Ω)≤ ∥l∥µ (Ω)1/p and this proves the lemma. ■Continuing with the proof of Theorem 24.8.3, note that F≪ µ. Since X ′ has the Radon

Nikodym property, there exists g ∈ L1 (Ω;X ′) such that F (E) =∫

E g(s)dµ. Also, from thedefinition of F (E) ,

l

(n

∑i=1

xiXEi (·)

)=

n

∑i=1

l (XEi (·)xi)

=n

∑i=1

F (Ei)(xi) =n

∑i=1

∫Ei

g(s)(xi)dµ. (24.37)

It follows from 24.37 that whenever h is a simple function,

l (h) =∫

Ω

g(s)(h(s))dµ. (24.38)

Let Gn ≡ {s : ∥g(s)∥X ′ ≤ n} and let j : Lp (Gn;X)→ Lp (Ω;X) be given by

jh(s) ={

h(s) if s ∈ Gn,0 if s /∈ Gn.