77.5. THE MAIN RESULT 2633

Since all terms but the first are in V ′, the equation holds in V ′. Also, the equation in 77.5.58shows that (B(ω)u(·,ω))′ ∈ V ′.

It only remains to show that there is a product measurable solution. The above argumenthas shown that there exists a solution for each ω . This is another application of Theorem77.2.10. For the sequence defined in the convergences 77.5.53 - 77.5.55, there is an estimate77.5.52. Therefore, the conditions of this theorem hold and there exists a subsequencedenoted with ε (ω) such that

uε(ω) (·,ω) → û(·,ω) weakly in V ,

u∗ε(ω) (·,ω) → û∗ (·,ω) weakly in V ′

where the û and û∗ are product measurable. Now the above argument shows that for eachω there exists a further subsequence, still denoted with ε (ω) such that convergence to asolution to the evolution inclusion is obtained (u(·,ω) ,u∗ (·,ω)). Then by uniqueness oflimits, û(·,ω) = u(·,ω) in V , similar for u∗ and û∗. Hence there is a solution to the aboveevolution problem which satisfies the claimed product measurability.

One can give a very interesting generalization of the above theorem.

Theorem 77.5.7 In the context of Theorem 77.5.6,let q(t,ω) be a product measurablefunction into V such that t→ q(t,ω) is continuous, q(0,ω) = 0.

Then, there exists a solution u of the integral equation

Bu(t,ω)+∫ t

0z(s,ω)ds =

∫ t

0f (s,ω)ds+Bu0 (ω)+Bq(t,ω) ,

where (t,ω) → u(t,ω) is product measurable. Moreover, for each ω , it follows thatBu(t,ω) = B(u(t,ω)) for a.e. t and z(·,ω) ∈ A(u(·,ω) ,ω) for a.e. t, z is product mea-surable into V ′. Also, for each a ∈ [0,T ] ,

Bu(t,ω)+∫ t

az(s,ω)ds =

∫ t

af (s,ω)ds+Bu(a,ω)+Bq(t,ω)−Bq(a,ω)

Proof: Define a stopping time

τr ≡ inf{t : |q(t,ω)|> r}

Then this is the first hitting time of an open set by a continuous random variable and so itis a valid stopping time. Then for each r, let

Ar (ω,w)≡ A(ω,w+qτr (·,ω)) ,

where the notation means qτr (t) ≡ q(t ∧ τr). Then, since qτr is uniformly bounded, all ofthe necessary estimates and measurability for the solution to the above corollary hold forAr replacing A. Therefore, there exists a solution wr to the inclusion

(Bwr)′ (·,ω)+Ar (wr (·,ω) ,ω) ∋ f (·,ω) , Bwr (0,ω) = Bu0 (ω)