652 CHAPTER 24. THE BOCHNER INTEGRAL

Proof: By Corollary 21.5.11 on Page 559, there exists a metric d, such that the met-ric space topology with respect to d coincides with the weak topology on K. Since K iscompact, it follows that K is also separable. Hence it is completely separable and so thereexists a countable basis of open sets B for the weak topology on K. It follows that if U isany weakly open set, covered by basic sets of the form BA (x,r) where A is a finite subsetof X ′, there exists a countable collection of these sets of the form BA (x,r) which covers U .

Suppose now that x is weakly measurable. To show x−1 (U)∈F whenever U is weaklyopen, it suffices to verify x−1 (BA (z,r)) ∈F for any set, BA (z,r) . Let A = {x∗1, · · · ,x∗m} .Then

x−1 (BA (z,r)) = {ω ∈Ω : ρA (x(ω)− z)< r}

≡{

ω ∈Ω : maxx∗∈A|x∗ (x(ω)− z)|< r

}= ∪m

i=1 {ω ∈Ω : |x∗i (x(ω)− z)|< r}= ∪m

i=1 {ω ∈Ω : |x∗i (x(ω))− x∗i (z)|< r}

which is measurable because each x∗i ◦ x is given to be measurable.Next suppose x−1 (U) ∈F whenever U is weakly open. Then in particular this holds

when U = Bx∗ (z,r) for arbitrary x∗. Hence

{ω ∈Ω : x(ω) ∈ Bx∗ (z,r)} ∈F .

But this says the same as

{ω ∈Ω : |x∗ (x(ω))− x∗ (z)|< r} ∈F

Since x∗ (z) can be a completely arbitrary element of F, it follows x∗ ◦x is an F valued mea-surable function. In other words, x is weakly measurable according to the former definition.■

One can also define weak ∗ measurability and prove a theorem just like the Pettis theo-rem above. The next lemma is the analogue of Lemma 24.1.7.

Lemma 24.1.13 Let B be the closed unit ball in X. If X ′ is separable, there exists asequence {xm}∞

m=1 ≡ D⊆ B with the property that for all y∗ ∈ X ′,∥y∗∥= supx∈D |y∗ (x)| .

Proof: Let {x∗k}∞k=1be the dense subset of X ′. Define φ n : B → Fn by the formula

φ n (x)≡ (x∗1 (x) , · · · ,x∗n (x)) .Then

∣∣x∗k (x)∣∣ ≤ ∥∥x∗k∥∥ and so φ n (B) is contained in a compact subset of Fn. There-

fore, there exists a countable set, Dn ⊆ B such that φ n (Dn) is dense in φ n (B) . That is,{(x∗1 (x) , · · · ,x∗n (x)) : x ∈ Dn} is dense in φ n (B) . D≡ ∪∞

n=1Dn.It remains to verify this works. Let y∗ ∈ X ′. I want to show that ∥y∗∥= supx∈D |y∗ (x)|.

There exists y,∥y∥ ≤ 1, such that

|y∗ (y)|> ∥y∗∥− ε.

By density, there exists one of the x∗k from the countable dense subset of X ′ such that also

∥x∗k − y∗∥< ε, so |x∗k (y)|> ∥y∗∥−2ε

Now x∗k (y) ∈ φ k (B) and so there exists x ∈ Dk ⊆ D⊆ B such that also

|x∗k (x)|> ∥y∗∥−2ε.