2722 CHAPTER 79. INCLUDING STOCHASTIC INTEGRALS

These wn (t,ω) are bounded for each (t,ω) off a set of measure zero and so by Lemma79.4.5, there is a P measurable function (t,ω)→ ẑ(t,ω) and a subsequence defined bywn(t,ω) (t,ω)→ ẑ(t,ω) weakly as n(t,ω)→∞. Now A(u(t,ω) , t,ω) is closed and convex,and wn(t,ω) (t,ω) is in A(u(t,ω) , t,ω) , and so ẑ(t,ω) ∈ A(u(t,ω) , t,ω) and

⟨ẑ(t,ω) ,u(t,ω)− y(t,ω)⟩ ≤ α (t,ω) = lim infk→∞⟨znk (t,ω) ,unk (t,ω)− y(t,ω)⟩ (**)

Therefore, t → F (t,ω) has a measurable selection on Sγ excluding a set of measure zero,namely ẑ(t,ω) which will be called ẑγ (t,ω) in what follows.

Then F (t,ω) has a measurable selection on [0,T ]×Ω other than a set of measurezero. To see this, enlarge Σ to include the exceptional sets of measure zero in the aboveargument for each γ . Then partition [0,T ]×Ω \Σ as follows. For γ = 1,2, · · · , considerSγ \ Sγ−1,γ = 1,2, · · · for S0 defined as /0. Then letting ẑγ be the selection for (t,ω) ∈ Sγ ,

let z̃(t,ω) = ∑∞γ=1 ẑγ (t,ω)XSγ\Sγ−1 (t,ω). The estimates imply z̃ ∈ V ′ and so z̃ ∈ Â(u) .

From the estimates, there exists h ∈ L1 ([0,T ]×Ω) such that

⟨z̃(t,ω) ,u(t,ω)− y(t,ω)⟩ ≥ −|h(t,ω)|

Thus, from the above inequality,

∥h∥L1 + ⟨z̃,u− y⟩V ′,V

≤∫

Ω

∫ T

0lim inf

k→∞⟨znk (t,ω) ,unk (t,ω)− y(t,ω)⟩+ |h(t,ω)|dtdP

≤ lim infk→∞

⟨znk ,unk − y

⟩V ′,V +∥h∥L1

= limn→∞⟨zn,un− y⟩V ′,V +∥h∥L1

which contradicts 79.4.69.Finally, why is z ∈ Â(u)? We have shown that (t,ω)→ F (t,ω) has a measurable se-

lection. Thus, there exists for any y ∈ V , z̃(y) ∈ Â(u) such that for a.e. (t,ω) ,

⟨z̃(y)(t,ω) ,u(t,ω)− y(t,ω)⟩ ≤ lim infk→∞⟨znk (t,ω) ,u(t,ω)− y(t,ω)⟩

Using the same trick involving the estimates and Fatou’s lemma, there exists z̃ dependingon y such that z̃(y) ∈ Â(u) and

⟨z̃(y) ,u− y⟩V ′,V ≤ lim infk→∞

⟨znk ,u− y

⟩V ′,V = ⟨z,u− y⟩

It follows that since y is arbitrary, it must be the case that z ∈ Â(u) thanks to separationtheorems and the fact that Â(u) is convex and closed.

An examination of the above proof yields the following corollary.

Corollary 79.4.10 Replace 1n Fun with en where en is progressively measurable and

∥en∥U ′ → 0, ∥en (·,ω)∥U ′ω → 0

where the second convergence happens for each ω off a set of measure zero and keep theother assumptions the same. Then the same conclusion is obtained.