2618 CHAPTER 77. STOCHASTIC INCLUSIONS

and so limsupn→∞ ⟨Fun,un−w⟩ ≤ 0. Then as before, ξ = Fw and one obtains

(Bw)′+Fw = f , Bw(0) = Bu0, w ∈ X

contradicting uniqueness. Hence un→ u weakly as claimed.Now suppose (Ω,F ) is a measurable space and B = B(ω) and is measurable into

L (W,W ′) and f : [0,T ]×Ω→V ′ is product measurable, B ([0,T ])×F measurable whereB ([0,T ]) denotes the Borel sets. Also, it is assumed that for each ω, f (·,ω) ∈ V ′. Thefollowing lemma ties together these ideas. It is Lemma 77.2.18 proved above. It is statedhere for convenience.

Lemma 77.4.2 Let f (·,ω) ∈ V ′. Then if ω → f (·,ω) is measurable into V ′, it followsthat for each ω, there exists a representative f̂ (·,ω)∈V ′, f̂ (·,ω) = f (·,ω) in V ′ such that(t,ω)→ f̂ (t,ω) is product measurable. If f (·,ω) ∈ V ′ and (t,ω)→ f (t,ω) is productmeasurable, then ω → f (·,ω) is measurable into V ′. The same holds replacing V ′ withV .

Now consider the initial value problem

(B(ω)u(·,ω))′+Fu(·,ω) = f (·,ω) ,

B(ω)u(0,ω) = B(ω)u0 (ω) , u(·,ω) ∈ X (77.4.21)

where we also assume u0 is F measurable into W . From Lemma 77.4.2,

ω → ( f (·,ω) ,u0 (ω))

is measurable into V ′×W . That is,

( f ,u0)−1 (U) = {ω : ( f (·,ω) ,u0 (ω)) ∈U} ∈F

for U an open set in V ′×W . From Lemma 77.4.1, the map Φω which takes ( f ,u0) to thesolution u is demicontinuous. We desire to argue that u is measurable into V . In doing so,it is easiest to assume that B does not depend on ω . However, the dependence on ω can beincluded using the approximation assumption for B(ω) mentioned earlier.

Letting fn (·,ω)→ f (·,ω) where fn is a simple function and u0n (ω)→ u0 (ω) whereu0n is also a simple function, it follows that

Φω ( fn (·,ω) ,u0n (ω))→Φω ( f (·,ω) ,u0 (ω)) = u

weakly. Here Φω would be a continuous function of V ′×W .

Lemma 77.4.3 Suppose f (·,ω) ∈ V ′ for each ω and that (t,ω) → f (t,ω) is productmeasurable into V ′. Also u0 is F measurable into W and

B(ω) = k (ω)B, k (ω)≥ 0, k measurable

Then for each ω ∈Ω, there exists a unique solution u(·,ω) in V satisfying

(B(ω)u(·,ω))′+Fu(·,ω) = f (·,ω) ,

B(ω)u(0,ω) = B(ω)u0 (ω) , u(·,ω) ∈ X

This solution has a representative which satisfies (t,ω)→ u(t,ω) is product measurableinto V .