646 CHAPTER 21. THE BOCHNER INTEGRAL

Proof: Let {ak}∞k=1 be a countable dense set in X and consider the mapping

φ n : B′→ Fn

given byφ n ( f )≡ ( f (a1) , · · · , f (an)) .

Then φ n (B′) is contained in a compact subset of Fn because | f (ak)| ≤ ∥ak∥ . Therefore,

there exists a countable dense subset of φ n (B′) ,{φ n ( fk)}∞

k=1 . Then 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 ={hk

j,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∥. Thus ∥ak∥is contained in φ n (B

′). DefineD′ ≡ ∪∞

n=1D′n.

From the construction, D′ is countable and can be used to approximate each ∥am∥ . Thatis,

∥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′

}≤ sup

{| f (am− x)+ f (x)| : f ∈ D′

}+∥x−am∥

≤ sup{| f (x)| : f ∈ D′

}+2∥x−am∥ ≤ ∥x∥+2∥x−am∥ .

Since am is arbitrary and the {am}∞

m=1 are dense, this establishes the claim of the lemma.

Note that the proof would work the same if H were only given to be weak ∗ dense.The next theorem is one of the most important results in the subject. It is due to Pettis

and appeared in 1938 [107].

Theorem 21.1.7 If x has values in a separable Banach space X, then x is weakly measur-able if and only if x is strongly measurable.

Proof: ⇒It is necessary to show x−1 (U) is measurable whenever U is open. Sinceevery open set is a countable union of balls, it suffices to show x−1 (B(a,r)) is measurablefor any ball, B(a,r) . Since every open ball is the countable union of closed balls, it sufficesto verify x−1

(B(a,r)

)is measurable. For D′ described in Lemma 21.1.6,

x−1(

B(a,r))

= {s : ∥x(s)−a∥ ≤ r}=

{s : sup

f∈D′| f (x(s)−a)| ≤ r

}= ∩ f∈D′ {s : | f (x(s)−a)| ≤ r}= ∩ f∈D′ {s : | f (x(s))− f (a)| ≤ r}

= ∩ f∈D′ ( f ◦ x)−1 B( f (a) ,r)

which equals a countable union of measurable sets because it is assumed that f ◦ x is mea-surable for all f ∈ X ′.

646 CHAPTER 21. THE BOCHNER INTEGRALProof: Let {a;}7_, be a countable dense set in X and consider the mapping@, :B! > F"given by$n (f) = (f (a1) .°°° ,f (an))-Then @,, (B’) is contained in a compact subset of F” because | f (a;)| < ||a,||. Therefore,there exists a countable dense subset of 9, (B’) ,{, (fe) }-_1 - Then pick hi € HMB’ suchthat lim j—4.0 | fe - n\| = 0. Then {6, (n') sk, j} must also be dense in @,, (B’). Let Di, ={ni k, i}. Thus D/, is a countable collection of f € B’ which can be used to approximateeach ||a||,k <n. Indeed, if x is arbitrary, there exists f, € B’ with f, (x) = ||x||. Thus |a,||is contained in @,, (B’). DefineD! =U?_,D),.From the construction, D’ is countable and can be used to approximate each ||a,,|| . Thatis,l!@m|| = sup {]f (4m)| + f € DYThen, for x arbitrary, | f (x)| < ||x|| and so|x = aml] + [lam] = ll = ml] + sup {|f (@m)| +f € D’Esup {If (am —x) + f(x]: f €D'} + |e amlsup {|f (x): f €D'} +2 ||x—aml] < lla] +2 lx— alIIA IA IASince ay is arbitrary and the {a,,};,_, are dense, this establishes the claim of the lemma.|Note that the proof would work the same if H were only given to be weak x dense.The next theorem is one of the most important results in the subject. It is due to Pettisand appeared in 1938 [107].Theorem 21.1.7 [fx has values in a separable Banach space X, then x is weakly measur-able if and only if x is strongly measurable.Proof: =It is necessary to show x! (U) is measurable whenever U is open. Sinceevery open set is a countable union of balls, it suffices to show x! (B(a,r)) is measurablefor any ball, B (a,r). Since every open ball is the countable union of closed balls, it sufficesto verify x7! (3 (a, r)) is measurable. For D’ described in Lemma 21.1.6,fEeD'= Neo {s:|f (x(s)—a)| <r} =O pen 9: |f (x(s)) -— F(a <r}= NMrep (fox) | B(f(a),r)which equals a countable union of measurable sets because it is assumed that f ox is mea-surable for all f € X’.x! (Blan) = (elo) a co} = {ss ap ro(9 -a <r}