2624 CHAPTER 77. STOCHASTIC INCLUSIONS

each cnk being in V . We can assume that ∥un (ω)∥ ≤ 2∥u(ω)∥ for all ω . Then by as-

sumption, there is a measurable selection for ω → A(cn

k ,ω)

denoted as ω → ynk (ω) . Thus

ω → ynk (ω) is measurable into V ′ and yn

k (ω) ∈ A(cn

k ,ω)

for all ω ∈Ω. Then consider

yn (ω) =mn

∑k=1

ynk (ω)XEn

k(ω)

It is measurable and for ω ∈ Enk it equals yn

k (ω) ∈ A(cn

k ,ω)= A(un (ω) ,ω) . Thus yn is a

measurable selection of ω → A(un (ω) ,ω) . By the estimates, for each ω these yn (ω) liein a bounded subset of V ′. The bound might depend on ω of course.

Now let {zi} be a countable dense subset of V. Then let X ≡∏∞i=1R. It is a Polish space.

Let

f(y j)(ω)≡

∞

∏i=1

⟨y j (ω) ,zi

⟩Γn (ω)≡ ∪k≥nf(yk)(ω),

the closure taken in X . Now yk (ω)∈ A(uk (ω) ,ω) and so by assumption, since ∥uk (ω)∥≤2∥u(ω)∥ it is bounded in V ′, this for each ω .

Thus the components of f(y j)(ω) lie in a compact subset of R, this for each ω . It

follows from Tychanoff’s theorem that Γn (ω) is a compact subset of the Polish space X .Claim: Γn is measurable into X .Proof of claim: It is necessary to show that Γ−n (U)≡ {ω : Γn (U)∩U ̸= /0} is measur-

able whenever U is open. It suffices to verify this for U a basic open set in the topology ofX . Thus let U = ∏

∞i=1 Oi where Oi is a proper open subset of R only for i ∈ { j1, · · · , jn} .

ThenΓ−n (U) = ∪ j≥n∩n

r=1{

ω :⟨y j (ω) ,z jr

⟩∈ O jr

}which is a measurable set thanks to y j being measurable.

In addition to this, Γn (ω) is compact, as explained above. Therefore, Γn is also stronglymeasurable meaning Γ−n (F) is measurable for all F closed. Now let Γ(ω)≡ ∩∞

n=1Γn (ω) .It is a nonempty closed subset of X and if F is closed in X ,

Γ− (F) = ∩∞

n=1Γ−n (F)

a measurable set since each Γ−n (F) is measurable. Thus Γ is a measurable multifunctionand so it has a measurable selection ω → z(ω) . Thus by definition, for each i, zi (ω) =

limn(ω)→∞

⟨yn(ω) (ω) ,zi

⟩for some subsequence indexed by n(ω). The sequence given

as{

yn(ω) (ω)}

is bounded in V ′ and so there is a subsequence still denoted as{

yn(ω)}

which converges weakly to y(ω) . Thus zi (ω) = ⟨y(ω) ,zi⟩ for each i. Since ω → zi (ω) ismeasurable, it follows from density of the {zi} that y is weakly, hence strongly measurable,this by the Pettis theorem. Now y(ω) = limn(ω)→∞ yn(ω) (ω) . But

yn(ω) (ω) ∈ A(un(ω) (ω) ,ω

)which is a convex closed set for which u→ A(u,ω) is upper semicontinuous and un(ω)→ uso in fact, y(ω) ∈ A(u(ω) ,ω). This is the claimed measurable selection.