2590 CHAPTER 76. IMPLICIT STOCHASTIC EQUATIONS
Passing to the limit and using the strong convergence described above along with the uni-form convergence of φ n to φ ,∫ T
0(h(s,ω)ζ (s) ,v(s)−u(s))H ds≤
∫ T
0h(s,ω)(φ (v(s))−φ (u(s)))ds
Hence,∫ T
0(h(s,ω)φ (v(s))−h(s,ω)φ (u(s)))− (h(s,ω)ζ (s) ,v(s)−u(s))H ds≥ 0
for any choice of v ∈Hω . It follows that for a.e. s,h(s,ω)ζ (s) ∈ ∂v (h(s,ω)φ (u(s))).This has shown that for each ω /∈ N, there exists a solution to the integral equation
Bu(t)−Bu0 +∫ t
0A(s,u)ds+
∫ t
0h(s,ω)ζ (s,ω)ds =
∫ t
0ΦdW (76.7.69)
where for a.e. s, h(s,ω)ζ (s,ω) ∈ ∂v (h(s,ω)φ (u(s))). Suppose you have two such solu-tions (u1,ζ 1) and (u2,ζ 2). Then
Bu1 (t)−Bu2 (t)+∫ t
0A(s,u1)−A(s,u2)ds+
∫ t
0h(s,ω)(ζ 1 (s,ω)−ζ 2 (s,ω))ds = 0
Then from monotonicity of the subgradient it follows that u1 = u2. Then the two integralequations yield that for a.e. t(
B(
u1−∫ (·)
0ΨdW
))′(t)+A(s,u1 (t))+h(t,ω)ζ 1 (t,ω) =
=
(B(
u2−∫ (·)
0ΨdW
))′(t)+A(s,u2 (t))+h(t,ω)ζ 2 (t,ω) = 0
Therefore, for a.e. t, h(t,ω)ζ 1 (t,ω) = h(t,ω)ζ 2 (t,ω). Thus the solution to the integralequation for each ω off a set of measure zero is unique.
At this point it is not clear that (t,ω)→ u(t,ω) is progressively measurable. We claimthat for ω /∈ N it is not necessary to take a subsequence in the above. This is becausethe above argument shows that if un fails to converge weakly, then there would exist twosubsequences converging weakly to two different solutions to the integral equation whichwould contradict uniqueness.
Therefore, for ω /∈ N, un (·,ω)→ u(·,ω) weakly in Vω for a single sequence. Usingthe estimate 76.3.24 it also follows that for a further subsequence still denoted as un,
un ⇀ ū in Lp ([0,T ]×Ω;V )
where the measurable sets are just the product measurable sets B ([0,T ])×FT . By Lemma76.3.4 for ω off a set of measure zero, u(·,ω) = ū(·,ω) in Vω where ū is progressivelymeasurable. It follows that in all of the above, we could substitute ū for u at least for ω offa single set of measure zero. Thus u can be assumed progressively measurable. The aboveargument along with technical details related to exponential shift considerations proves thefollowing theorem.