24.1. STRONG AND WEAK MEASURABILITY 653

Then since∥∥x∗k − y∗

∥∥< ε, this implies

∥y∗∥ ≥ |y∗ (x)|= |(y∗− x∗k)(x)+ x∗k (x)| ≥ |x∗k (x)|− ε > ∥y∗∥−3ε

It follows that∥y∗∥−3ε ≤ sup

x∈D|y∗ (x)| ≤ ∥y∗∥

This proves the lemma because ε is arbitrary. ■The next theorem is another version of the Pettis theorem. First here is a definition.

Definition 24.1.14 A function y having values in X ′ is weak ∗ measurable, whenfor each x ∈ X, y(·)(x) is a measurable scalar valued function.

Theorem 24.1.15 If X ′ is separable and y : Ω→ X ′ is weak ∗ measurable meaningω → y(ω)(x) is a F valued measurable function, then y is strongly measurable.

Proof: It is necessary to show y−1 (B(a∗,r)) is measurable for a∗ ∈ X ′. This will sufficebecause the separability of X ′ implies every open set is the countable union of such ballsof the form B(a∗,r). It also suffices to verify inverse images of closed balls are measurablebecause every open ball is the countable union of closed balls. From Lemma 24.1.13,

y−1(

B(a∗,r))

= {ω : ∥y(ω)−a∗∥ ≤ r}

=

{ω : sup

x∈D|(y(ω)−a∗)(x)| ≤ r

}=

{ω : sup

x∈D|y(ω)(x)−a∗ (x)| ≤ r

}= ∩x∈Dy(·)(x)−1

(B(a∗ (x) ,r)

)which is a countable intersection of measurable sets by hypothesis. ■

The following are interesting consequences of the theory developed so far and are ofinterest independent of the theory of integration of vector valued functions.

Theorem 24.1.16 If X ′ is separable, then so is X.

Proof: Let D = {xm} ⊆ B, the unit ball of X , be the sequence promised by Lemma24.1.13. Let V be all finite linear combinations of elements of {xm} with rational scalars.Thus V is a separable subspace of X . The claim is that V = X . If not, then it follows thatthere exists x0 ∈ X \V . But by the Hahn Banach theorem there exists x∗0 ∈ X ′ satisfyingx∗0 (x0) ̸= 0, but x∗0 (v) = 0 for every v ∈V . Hence

∥∥x∗0∥∥= supx∈D

∣∣x∗0 (x)∣∣= 0, a contradic-tion. ■

Corollary 24.1.17 If X is reflexive, then X is separable if and only if X ′ is separable.

Proof: From the above theorem, if X ′ is separable, then so is X . Now suppose X isseparable with a dense subset equal to D. Then since X is reflexive, J (D) is dense in X ′′

where J is the James map satisfying Jx(x∗) ≡ x∗ (x) . Recall how this J preserves normsand maps onto X ′′ for X reflexive. Then since X ′′ is separable, it follows from the abovetheorem that X ′ is also separable. ■

Note how this shows that L1 (Rp,mp) is not reflexive because this is a separable space,but L∞ (Rp,mp) is clearly not. For example, you could consider X[0,r] for r a positiveirrational number. There are uncountably many of these functions in L∞ ([0,1]) and it isclear that

∥∥X[0,r]−X[0,r̂]∥∥

∞= 1.