2720 CHAPTER 79. INCLUDING STOCHASTIC INTEGRALS

converge to 0, thus,

limn→∞

∫Ω

∫ T

0|⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩|dtdP = 0 (79.4.71)

From the above, it follows that there exists a further subsequence {nk} not depending ont,ω such that ∣∣⟨znk (t,ω) ,unk (t,ω)−u(t,ω)⟩

∣∣→ 0 a.e. (t,ω) . (79.4.72)

By the pseudomonotone limit condition for A there exists wt,ω ∈ A(u(t,ω) , t,ω) suchthat for a.e.(t,ω)

α (t,ω) ≡ lim infk→∞⟨znk (t,ω) ,unk (t,ω)− y(t,ω)⟩

= lim infk→∞⟨znk (t,ω) ,u(t,ω)− y(t,ω)⟩ ≥ ⟨wt,ω ,u(t,ω)− y(t,ω)⟩.

Note that u is progressively measurable and if A(·, t,ω) were single valued, this would givea contradiction at this point. We continue with the case where A is set valued. This casewill make use of the measurable selection in Lemma 79.4.5.

On the exceptional set, let α (t,ω)≡ ∞, and consider the set

F (t,ω)≡ {w ∈ A(u(t,ω) , t,ω) : ⟨w,u(t,ω)− y(t,ω)⟩ ≤ α (t,ω)} ,

which then satisfies F (t,ω) ̸= /0. Now F (t,ω) is closed and convex in V ′.We will let Σ be a progressively measurable set of measure zero which includes

N× [0,T ]∪{(t,ω) : ω /∈ NC, t ∈Mω

}.

Claim ∗: (t,x)→ F (t,ω) has a P measurable selection off a set of measure zero.Proof of claim: Letting B(0,C (t,ω)) contain A(u(t,ω) , t,ω) , we can assume (t,ω)→

C (t,ω) is P measurable by using the estimates and the measurability of u. For γ ∈ N, letSγ ≡ {(t,ω) : C (t,ω)< γ} . If it is shown that F has a measurable selection on Sγ , then itfollows that it has a measurable selection. Thus in what follows, assume that (t,ω) ∈ Sγ .

Define for (t,ω) ∈ Sγ

G(t,ω)≡{

w : ⟨w,u(t,ω)− y(t,ω)⟩< α (t,ω)+1n

, (t,ω) ∈ ΣC ∩Sγ

}∩B(0,γ)

Thus, it was shown above that this G(t,ω) ̸= /0 at least for large enough γ that Sγ ̸= /0. ForU open, G− (U) is defined by

G− (U)≡{

(t,ω) ∈ Sγ : for some w ∈U ∩B(0,γ) ,⟨w,u(t,ω)− y(t,ω)⟩< α (t,ω)+ 1

n

}(*)

Let{

w j}

be a dense subset of U ∩B(0,γ). This is possible because V ′ is separable. Theexpression in ∗ equals

∪∞k=1

{(t,ω) ∈ Sγ : ⟨wk,u(t,ω)− y(t,ω)⟩< α (t,ω)+

1n

}