2716 CHAPTER 79. INCLUDING STOCHASTIC INTEGRALS

Proof: In the following argument, N will be a set of measure zero containing the one ofLemma 79.4.6 and the sequence will always be a subsequence of the subsequence of thatlemma. Recall that from Lemma 79.4.6, that for ω /∈ N,∥∥∥∥ un (t,ω)−u0 (ω)

+∫ t

0 (zn (s,ω)− f (s,ω))ds−∫ t

0 ΦdW

∥∥∥∥U ′→ 0 (79.4.58)

for this sequence which does not depend on ω or t. From now on, this or a further subse-quence will be meant.

From the hypothesis,

un→ u weakly in V , zn→ z weakly in V ′ (79.4.59)

Thusu(t)−u0 +

∫ t

0zds =

∫ t

0f ds+

∫ t

0ΦdW for a.e. ω (79.4.60)

and so ∥∥∥∥u(t)−u0 +∫ t

0zds−

(∫ t

0f ds+

∫ t

0ΦdW

)∥∥∥∥U ′

= 0 (79.4.61)

Note that in these weak convergences 79.4.59, we can assume the σ algebra is justB ([0,T ])×FT because the progressive measurability will be preserved in the limit dueto the Pettis theorem and the progressive measurability of each un,zn. However, we couldalso let the σ algebra be P the progressively measurable sets just as well.

Claim: There is a set of measure zero N, including the one obtained so far such that forω /∈ N

lim infn→∞⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩ ≥ 0

for a.e. t. The exceptional set, denoted as Mω includes those t for which some zn (t,ω) /∈A(un (t,ω) , t,ω).

Let ω /∈ N and t /∈Mω . First take a subsequence such that liminf = lim . Then supposethat

lim infn→∞⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩< 0

Then from the estimates, one can obtain that for a suitable subsequence, un (t,ω)→ψ (t,ω) weakly in V . Note, n = n(t,ω) . Here is why: From the above inequality, thereexists a subsequence {nk}, which may depend on t,ω , such that

limk→∞⟨znk (t,ω) ,unk (t,ω)−u(t,ω)⟩ (79.4.62)

= lim infn→∞⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩< 0. (79.4.63)

Now, condition 3 implies that for all k large enough,

b3∣∣∣∣unk (t,ω)

∣∣∣∣pV −b4 (t,ω)−λ

∣∣unk (t,ω)∣∣2H

<∣∣∣∣znk (t,ω)

∣∣∣∣V ′ ||u(t,ω)||V

≤(

b1∣∣∣∣unk (t,ω)

∣∣∣∣p−1V +b2 (t,ω)

)||u(t,ω)||V , (79.4.64)