77.2. SOME FUNDAMENTAL THEOREMS 2603

Such an index exists because there exists a convergent sequence XNC (ω)un(ω) that con-verge weakly to u. In fact,

{ω : k (ω) = l}={

ω : d (ul ,u)≤ 2−k}∩∩l−1

j=1

{ω : d (u j,u)> 2−k

}.

Since u is product measurable and each ul is also product measurable, these are all mea-surable sets with respect to F and so ω → k (ω) is F measurable. Now, we have thatXNC (ω)uk(ω)→ u weakly in Lp′ ([0,T ] ;V ′), for each ω , and each function is F measur-able because

uk(ω) (t,ω) =∞

∑j=1

X[k(ω)= j]u j (t,ω) ,

and every term in the sum is F measurable.Theorem 77.2.4 can be generalized in a very nice way. It is a better result because

you don’t need to assume anything so strong as to have the functions bounded. One doesnot need any assumption that the limit is weakly continuous into V ′. You can also havethe functions take values in either V or V ′. The following is not dependent on there beinga measure but in the applications there is typically a probability measure and often a setof measure zero which occurs in a natural way so an exceptional set of measure zero isincluded in the statement of the theorem.

Theorem 77.2.10 Let V be a reflexive separable Banach space with dual V ′, and let p, p′

be such that p > 1 and 1p + 1

p′ = 1. Let the functions t → un (t,ω), for n ∈ N, be inLp ([0,T ] ;V )≡ V and (t,ω)→ un (t,ω) be B ([0,T ])×F ≡P measurable into V . Sup-pose there is a set of measure zero N ⊆Ω such that if ω /∈ N, then

∥un (·,ω)∥V ≤C (ω) ,

for all n. (Thus, by weak compactness, for each ω, each subsequence of {un} has a furthersubsequence that converges weakly in V to v(·,ω) ∈ V . (v not known to be P measur-able))

Then, there exists a product measurable function u such that t→ u(t,ω)is in V and foreach ω /∈ N, a subsequence un(ω) such that un(ω) (·,ω)→ u(·,ω) weakly in V .

We prove the theorem in steps given below. Let X = ∏∞k=1 C ([0,T ]) and note that when

it is equipped with the product topology, then one can consider X as a metric space usingthe metric

d (f,g)≡∞

∑k=1

2−k ∥ fk−gk∥1+∥ fk−gk∥

,

where f = ( f1, f2, . . .),g = (g1,g2, . . .) ∈ X , and the norm is the maximum norm in thespace C ([0,T ]). With this metric, X is complete and separable.

Lemma 77.2.11 Let {fn} be a sequence in X and suppose that each one of the componentsfnk is bounded by C =C(k) in C0,(1/p′) ([0,T ]). Then, there exists a subsequence

{fn j

}that

converges to some f ∈ X as n j→ ∞. Thus, {fn} is pre-compact in X.