Kenneth Kuttler

21.1. STRONG AND WEAK MEASURABILITY 645

Thenx−1 (Vm)⊆

⋃n<∞

⋂k≥n

x−1k (Vm)⊆ x−1 (V m

This impliesx−1 (U) =

⋃m<∞

x−1 (Vm)

⊆⋃

m<∞

⋃n<∞

⋂k≥n

x−1k (Vm)⊆

⋃m<∞

x−1 (V m)⊆ x−1 (U).

Sincex−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 separa-ble. Let

D≡ all values of the simple functions xn

Then x(Ω)⊆D, which has a countable dense subset. By Lemma 21.1.3, x(Ω) is separable.

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

Proof: Consider the one dimensional subspace

M ≡{

αx∥x∥

: α ∈ F}

and define a continuous linear functional on M by g(

αx∥x∥

)≡ α. Then the norm of ∥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∥.

Lemma 21.1.6 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∥= supf∈D′| f (x)|

If H is a dense subset of X ′ then D′ may be chosen to be contained in H.

21.1. STRONG AND WEAK MEASURABILITY 645Thenx" (Vn) © U (xe! Vn) Cx! (Vin)n<ok>nThis impliesx l(U)= UJ x! (Vn)m<ooSU UM a1 Wn) © Ut Vm) Sat (W),m<on<ok>n m<ooSincex lU)= U U Ox! Vn);m<on<ok>nit follows that x! (U) is measurable for every open U. It remains to show x (Q) is separa-ble. LetD = all values of the simple functions x,,Then x (Q) CD, which has a countable dense subset. By Lemma 21.1.3, x (Q) is separable.|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 21.1.5 Let x € X anormed linear space. Then there exists f € X' such that || f|| = 1and f (x) = |lx||-Proof: Consider the one dimensional subspaceM= {ax [aE F|||and define a continuous linear functional on M by g (a;7) = a. Then the norm of ||g|| =SUP |g) <1 |o¢| = 1. Extend g to all of X using the Hahn Banach theorem calling the extendedfunction f. Then || f\| = 1 and f (x) =F (lx rn) = |x|). 0Lemma 21.1.6 /f X is a separable Banach space with B’ the closed unit ball in X', thenthere exists a sequence { f,}°_, = D! C B' with the property that for every x € X,\|x|| = sup |f (x)|feD'If H is a dense subset of X' then D' may be chosen to be contained in H.