684 CHAPTER 24. THE BOCHNER INTEGRAL

Also∫

Ωg(ω)( f (ω)) =

1

∥g∥p′−1Lp′ (Ω;X ′)

∫Ω

m

∑i=1

c∗i (di)∥c∗i ∥p′−1 XEi (ω)dµ

≥ 1

∥g∥p′−1Lp′ (Ω;X ′)

∫Ω

m

∑i=1

(∥c∗i ∥− ε)∥c∗i ∥p′−1 XEi (ω)dµ

=1

∥g∥p′−1Lp′ (Ω;X ′)

m

∑i=1∥c∗i ∥

p′µ (Ei)−

1

∥g∥p′−1Lp′ (Ω;X ′)

ε

m

∑i=1∥c∗i ∥

p′−1µ (Ei)

= ∥g∥Lp′ (Ω;X ′)− ε

Therefore, ∥g∥Lp′ (Ω;X ′) ≥ ∥θ (g)∥ ≥ |∫

Ωg(ω)( f (ω))| ≥ ∥g∥Lp′ (Ω;X ′)− ε and since ε is ar-

bitrary, it follows that ∥g∥Lp′ (Ω;X ′) = ∥θ (g)∥ whenever g is a simple function.

In general, let g∈ Lp′ (Ω;X ′) and let gn be a sequence of simple functions converging tog in Lp′ (Ω;X ′). Such a sequence exists by Lemma 24.1.2. Let gn (ω)→ g(ω) ,∥gn (ω)∥ ≤2∥g(ω)∥ . Then each gn is in Lp′ (Ω;X ′) and by the dominated convergence theorem theyconverge to g in Lp′ (Ω;X ′). Then for ∥·∥ the norm in (Lp (Ω;X))′ ,

∥θg∥= limn→∞∥θgn∥= lim

n→∞∥gn∥= ∥g∥.

This proves the theorem in case p = 1 and shows θ is the desired isometry.Next suppose p = 1 and g ∈ L∞ (Ω;X ′). It is still the case that ∥θg∥ ≤ ∥g∥L∞(Ω;X ′). As

above, one must choose f appropriately. In this case, assume µ is a finite measure. Beginwith g a simple function g(ω) = ∑

mi=1 c∗i XEi (ω) . Suppose ∥c∗1∥ is at least as large as all

other ∥c∗i ∥ modify if the largest of these occurs at k ̸= 1. Thus ∥g∥∞= ∥c∗i ∥X ′ . Now let

c∗1 (d1) ≥ ∥c∗1∥− εµ (Ei) ,∥d1∥X = 1, and let f (ω) ≡ d1µ(Ei)

XEi (ω) . Then∫

Ω∥ f∥dµ = 1.

Also

g(ω)( f (ω))c∗1 (d1)

µ (Ei)XEi (ω)≥ 1

µ (Ei)XEi (ω)(∥c∗1∥− εµ (Ei))

and so

|θg( f )| =

∣∣∣∣∫Ω

g(ω)( f (ω))dµ

∣∣∣∣≥ ∣∣∣∣∫Ω

(1

µ (Ei)XEi (ω)(∥c∗1∥− εµ (Ei))

)dµ

∣∣∣∣≥ ∥c∗1∥− εµ (Ω) = ∥g∥

∞− εµ (Ω)

Thus∥g∥

∞≥ ∥θg∥ ≥ ∥g∥

∞− εµ (Ω)

and so, since ε is arbitrary, it follows that ∥θg∥ = ∥g∥L∞(Ω;X ′). Extending from simplefunctions to functions in L∞ (Ω;X ′) goes as before. Approximate with simple functionsand pass to a limit. ■

Theorem 24.8.3 If X is a Banach space and X ′ has the Radon Nikodym property,then if (Ω,S ,µ) is a finite measure space,(Lp (Ω;X))′ ∼= Lp′ (Ω;X ′) and in fact the map-ping θ of Theorem 24.8.2 is onto.

Proof: Let l ∈ (Lp (Ω;X))′ and define F (E) ∈ X ′ by F (E)(x)≡ l (XE (·)x) .