21.1. STRONG AND WEAK MEASURABILITY 649

a countable basis of open sets B for the weak topology on K. It follows that if U is anyweakly open set, covered by basic sets of the form BA (x,r) where A is a finite subset of 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)) = {s ∈Ω : ρA (x(s)− z)< r}

≡{

s ∈Ω : maxx∗∈A|x∗ (x(s)− z)|< r

}= ∪m

i=1 {s ∈Ω : |x∗i (x(s)− z)|< r}= ∪m

i=1 {s ∈Ω : |x∗i (x(s))− 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

{s ∈Ω : x(s) ∈ Bx∗ (z,r)} ∈F .

But this says the same as

{s ∈Ω : |x∗ (x(s))− 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 21.1.6.

Lemma 21.1.13 Let B be the closed unit ball in X. If X ′ is separable, there exists a se-quence {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

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.