21.8. THE RIESZ REPRESENTATION THEOREM 685

Continuing with the proof of Theorem 21.8.3, note that F≪ µ. Since X ′ has the RadonNikodym property, there exists g ∈ L1 (Ω;X ′) such that

F (E) =∫

Eg(s)dµ.

Also, from the definition 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. (21.8.37)

It follows from 21.8.37 that whenever h is a simple function,

l (h) =∫

Ω

g(s)(h(s))dµ. (21.8.38)

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.

Letting h be a simple function in Lp (Gn;X),

j∗l (h) = l ( jh) =∫

Gn

g(s)(h(s))dµ. (21.8.39)

Since the simple functions are dense in Lp (Gn;X), and g ∈ Lp′ (Gn;X ′), it follows 21.8.39holds for all h ∈ Lp (Gn;X). By Theorem 21.8.2,

∥g∥Lp′ (Gn;X ′) = ∥ j∗l∥(Lp(Gn;X))′ ≤ ∥l∥(Lp(Ω;X))′ .

By the monotone convergence theorem,

∥g∥Lp′ (Ω;X ′) = limn→∞∥g∥Lp′ (Gn;X ′) ≤ ∥l∥(Lp(Ω;X))′ .

Therefore g ∈ Lp′ (Ω;X ′) and since simple functions are dense in Lp (Ω;X), 21.8.38 holdsfor all h ∈ Lp (Ω;X) . Thus l = θg and the theorem is proved because, by Theorem 21.8.2,∥l∥= ∥g∥ and the mapping θ is onto because l was arbitrary.

As in the scalar case, everything generalizes to the case of σ finite measure spaces. Theproof is almost identical.

Lemma 21.8.5 Let (Ω,S ,µ) be a σ finite measure space and let X be a Banach spacesuch that X ′ has the Radon Nikodym property. Then there exists a measurable function, rsuch that r (x)> 0 for all x, such that |r (x)|< M for all x, and

∫rdµ < ∞. For

Λ ∈ (Lp(Ω;X))′, p≥ 1,