70.2. A MEASURABLE SELECTION 2395

Lemma 70.2.3 ω → Γn (ω) is a F measurable set valued map with values in X. If σ isa measurable selection, (σ (ω) ∈ Γn (ω) so σ = σ (·,ω) a continuous function. To havethis measurable would mean that σ

−1k (open) ∈F where the open set is in C ([0,T ]).) then

for each t, ω → σ (t,ω) is F measurable and (t,ω)→ σ (t,ω) is B ([0,T ])×F ≡Pmeasurable.

Proof: Let O be a basic open set in X . Thus O = ∏∞k=1 Ok where Ok is a proper open

set of C ([0,T ]) only for k ∈ {k1, · · · ,kr}. We need to consider whether

Γn− (O)≡ {ω : Γ

n (ω)∩O ̸= /0} ∈F .

Now Γn− (O) equals∩r

i=1

{ω : Γ

n (ω)ki∩Oki ̸= /0

}Thus we consider whether {

ω : Γn (ω)ki

∩Oki ̸= /0}∈F (70.2.1)

From the definition of Γn (ω) , this is equivalent to the condition that for some j ≥ n,

fki (u j (·,ω)) = (f(u j (·,ω)))ki∈ Oki

and so the above set in 70.2.1 is of the form

∪∞j=n

{ω : (f(u j (·,ω)))ki

∈ Oki

}Now ω → (f(u j (·,ω)))ki

is F measurable into C ([0,T ]) and so the above set is in F . Tosee this, let g ∈C ([0,T ]) and consider the inverse image of the ball B(g,r) ,{

ω :∥∥∥(f(u j (·,ω)))ki

−g∥∥∥

C([0,T ])< r}.

By continuity considerations,∥∥∥(f(u j (·,ω)))ki−g∥∥∥

C([0,T ])= sup

t∈Q∩[0,T ]

∣∣∣(f(u j (t,ω)))ki−g(t)

∣∣∣which is the sup of countably many F measurable functions. Thus it is F measurable.Since every open set is the countable union of such balls, it follows that the claim about Fmeasurability is valid. Thus Γn− (O) is F measurable whenever O is a basic open set.

Now X is a separable metric space and so every open set is a countable union of thesebasic sets. Let U be an open set in X and let U = ∪∞

l=1Ol where Ol is a basic open set asabove. Then

Γn− (U) = ∪∞

l=1Γn−(

Ol)∈F .

That there exists a measurable selection follows from the standard theory of measurablemulti-functions [10], [70]. This is proved in Theorem 70.1.2 above. For σ one of thesemeasurable selections, the evaluation at t is F measurable. Thus ω → σ (t,ω) is F mea-surable with values in R∞. Also t→ σ (t,ω) is continuous, and so it follows that in fact σ

is product measurable as claimed.