1594 CHAPTER 48. MULTIFUNCTIONS AND THEIR MEASURABILITY

argument for each γ . Then partition [0,T ]×Ω \Σ as follows. For γ = 1,2, · · · , considerSγ \ Sγ−1,γ = 1,2, · · · for S0 defined as /0. Then letting zγ be the selection for (t,ω) ∈ Sγ ,

let z(t,ω) = ∑∞γ=1 zγ (t,ω)XSγ\Sγ−1 (t,ω). The estimates imply z ∈ V ′ and so z ∈ Â(u) .

From the estimates, there exists h ∈ L1 ([0,T ]×Ω) such that

⟨z(t,ω) ,u(t,ω)− y(t,ω)⟩ ≥ −|h(t,ω)|

Thus, from the above inequality,

∥h∥L1 + ⟨z,u− y⟩V ′,V

≤∫

Ω

∫ T

0lim inf

k→∞⟨znk (t,ω) ,unk (t,ω)− y(t,ω)⟩+ |h(t,ω)|dtdP

≤ lim infk→∞

⟨znk ,unk − y

⟩V ′,V +∥h∥L1

= limn→∞⟨zn,un− y⟩V ′,V +∥h∥L1

which contradicts 48.7.42.The difficulty with this, is that it is hard to get the hypotheses holding. If you have

un → u weakly in V , then how do you get un (t,ω)→ u(t,ω) weakly in U ′, for a subse-quence, this for each ω not in a set of measure zero? The weak convergence does not seemto give pointwise weak convergence of the sort you need. More precisely, the pointwiseconvergence you get, might not be the right thing because in un it will be un(ω). This is theproblem with this theorem. It seems correct but not very useful.