1574 CHAPTER 48. MULTIFUNCTIONS AND THEIR MEASURABILITY

Then by the semicontinuity properties of a,b we obtain from routine considerations thatw(x) ∈ [a(γu(x) , t) ,b(γu(x) , t)] a.e. To see how you can do this, let

E =

{x : w(x)≥ b(γu(x) , t)+

1k

}.

Then∫Σ1

XE (x)(−b(γun (x) , t)) ≤∫

Σ1

XE (x)(−wn (x))→∫

Σ1

XE (x)(−w(x))

≤∫

Σ1

XE (x)(−b(γu(x) , t)− 1

k

)By lower semicontinuity of −b(·, t) and the boundedness assumption, we can use Fatou’slemma to take liminf of both sides and conclude that∫

Σ1

XE (x)(−b(γu(x) , t))≤∫

Σ1

XE (x)(−b(γu(x) , t)− 1

k

)an obvious contradiction unless α (E) = 0. Then taking the union of the exceptional setsfor all k, it follows that w(x) ≤ b(γu(x) , t) a.e. The other side of the inequality can beshown similarly. Letting zn ∈ B(un, t) and v ∈V, is it true that

lim infn→∞⟨zn,un− v⟩ ≥ ⟨z(v) ,un− v⟩

for some z(v) ∈ B(u, t)? Suppose not. Then from the above, there is a subsequence suchthat the limit equals the liminf but which has the inequality turned around for some v andall z ∈ B(u, t). Then from what was just shown, letting wn go with zn, there is a furthersubsequence such that wn→ w weakly in L2 (Σ1) and and γun→ γu strongly in L2 (Σ1) and

w(x) ∈ [a(γu(x) , t) ,b(γu(x) , t)] a.e. x

Then ∫Σ1

wn (x)(γun (x)− γv(x))→∫

Σ1

w(x)(γu(x)− γv(x)) = ⟨z,u− v⟩

where w ∈ B(u, t) and z = γ∗w so the liminf condition holds. Thus this second operator ispseudomonotone.

Do these have measurable selections? This is obvious. Letting u∈V, t→ γ∗a(γu(x) , t)is measurable into V ′ and is in B(u, t). Similarly t→A(u, t) is measurable into V ′. Note thaton the second operator, it was really only necessary to assume that there exists t → c(r, t)measurable with c(r, t) ∈ [a(r, t) ,b(r, t)] and totally eliminate the assumption that either aor b is measurable in t.

Now let t→ f (t) be measurable into V ′. Say

⟨ f (t) ,v⟩=∫

Ω

h(t)vdx+∫

Σ1

β (t)vdα