24.1. STRONG AND WEAK MEASURABILITY 649

then x−1n (W ) = ∪{Ek : ck ∈W} . Thus, x−1

n (W ) is measurable for every Borel set W . Thisfollows from the observation that

{W : x−1

n (W ) is measurable}

is a σ algebra containingthe open sets. Since X is a metric space, it follows that if U is an open set in X , there existsa sequence of open sets, {Vn} which satisfies

V n ⊆U, V n ⊆Vn+1, U = ∪∞n=1Vn.

Then x−1 (Vm)⊆⋃

n<∞

⋂k≥n

x−1k (Vm)⊆ x−1

(V m).This implies

x−1 (U) =⋃

m<∞

x−1 (Vm)⊆⋃

m<∞

⋃n<∞

⋂k≥n

x−1k (Vm)⊆

⋃m<∞

x−1 (V m)⊆ x−1 (U).

Since x−1 (U) =⋃

m<∞

⋃n<∞

⋂k≥n

x−1k (Vm),it follows that x−1 (U) is measurable for every open

U . It remains to show x(Ω) is separable. Let D ≡ all values of the xn. Then x(Ω) ⊆ D,which has a countable dense subset. By Lemma 24.1.3, x(Ω) is separable. ■

Lemma 24.1.6 Let x ∈ X a normed linear space. Then there exists f ∈ X ′ such that∥ f∥= 1 and f (x) = ∥x∥.

Proof: Consider the one dimensional subspace M ≡{

αx∥x∥ : α ∈ F

}and define a con-

tinuous linear functional on M by g(

αx∥x∥

)≡ α. Then the operator norm of g is obtained

as ∥g∥ ≡ sup|α|≤1 |α|= 1. Extend g to all of X using the Hahn Banach theorem, calling the

extended function f . Then ∥ f∥= 1 and f (x) = f(∥x∥ x

∥x∥

)≡ ∥x∥. ■

The next lemma is interesting for its own sake. Roughly it says that if a Banach spaceis separable, then the unit ball in the dual space is weak ∗ separable. This will be usedto prove Pettis’s theorem, one of the major theorems in this subject which relates weakmeasurability to strong measurability. First here is a standard application which comesfrom earlier material on the Hahn Banach theorem.

Lemma 24.1.7 If X is a separable Banach space with B′ the closed unit ball in X ′, thenthere exists a sequence { fn}∞

n=1 ≡ D′ ⊆ B′ with the property that for every x ∈ X ,∥x∥ isobtained as ∥x∥= sup f∈D′ | f (x)| . If H is a dense subset of X ′ then D′ may be chosen to becontained in H.

Proof: Let {ak}∞k=1 be a countable dense set in X . Consider the mapping φ n : B′→ Fn

given by φ n ( f )≡ ( f (a1) , · · · , f (an)) .Then φ n (B

′) is contained in a compact subset of Fn because | f (ak)| ≤ ∥ak∥ . There-fore, there exists a countable dense subset of φ n (B

′) ,{φ n ( fk)}∞

k=1 . Pick hkj ∈ H ∩B′ such

that lim j→∞

∥∥∥ fk−hkj

∥∥∥ = 0. Then{

φ n

(hk

j

)}k, j

must also be dense in φ n (B′) . Let D′n ={

hkj

}k, j

. Thus D′n is a countable collection of f ∈ B′ which can be used to approximate

each ∥ak∥ ,k ≤ n. Indeed, if x is arbitrary, there exists fx ∈ B′ with fx (x) = ∥x∥ and so ifx = ak, then ∥ak∥ will be close to g(ak) for some g ∈ D′n. Define D′ ≡ ∪∞

n=1D′n.From the construction, D′ is countable and can be used to approximate each ∥am∥ . That

is, ∥am∥= sup{| f (am)| : f ∈ D′} Then, for x arbitrary, | f (x)| ≤ ∥x∥ and so

∥x∥ ≤ ∥x−am∥+∥am∥= ∥x−am∥+ sup{| f (am)| : f ∈ D′

}≤ ∥x−am∥+ sup

{| f (am− x)+ f (x)| : f ∈ D′

}≤ sup

{| f (x)| : f ∈ D′

}+2∥x−am∥ ≤ ∥x∥+2∥x−am∥ .