2420 CHAPTER 70. MEASURABILITY WITHOUT UNIQUENESS
∫ t
0γ∗T F((un−g(ω))+
)µ(∣∣vT − U̇T
∣∣)ψ′ε
(vT − U̇T
)ds)=∫ t
0fds (70.5.54)
Thus t → v(t,ω) is continuous into V ′. This along with the estimate 70.5.44, impliesthat the conditions of Theorem 70.2.1 are satisfied. It follows that there is a function v̄which is product measurable into V ′ and weakly continuous in t and for each ω, a subse-quence vk(ω) such that vk(ω) (·,ω)⇀ v̄(·,ω) in V ′. Then by a repeat of the above argument,for each ω, there exists a further subsequence still denoted as vk(ω) which converges in V ′
to v(·,ω) which is a solution to the above integral equation which is continuous into V ′.Hence, v̄(·,ω) = v(·,ω) and since these are both weakly continuous into V ′ they must bethe same function. Hence, there is a product measurable solution v.
Next we pass to a limit as ε → 0. Denoting the product measurable solution to theabove integral equation as vk, where ε = 1/k. The estimate 70.5.42 is obtained as before.Then we get a subsequence, still denoted as vk which has the same convergences as in70.5.46 - 70.5.53. Thus we obtain these convergences along with the fact that vk is productmeasurable and for each ω, it is a solution of
vk (t)−v0 +∫ t
0Mvkds+
∫ t
0Aukds+
∫ t
0Pukds+
∫ t
0γ∗T F((ukn−g(ω))+
)µ(∣∣vkT − U̇T
∣∣)ψ′1/k
(vkT − U̇T
)ds =
∫ t
0fds (70.5.55)
Now in addition to these convergences, we can also obtain
ψ′1/k
(vkT − U̇T
)⇀ ξ in L∞
([0,T ] ;L∞ (ΓC)
3)
We have also
ψ′1/k
(vkT − U̇T
)·wT ≤ ψ1/k
(vkT − U̇T +wT
)−ψ1/k
(vkT − U̇T
)and so, passing to a limit, using the strong convergence of vkT to vT in L2 ([0,T ] ;W ) ,uniform convergence of ψ1/k to ∥·∥ , and pointwise convergence in W, we obtain using thedominated convergence theorem that for w ∈ V ,∫ t
0
∫ΓC
F((ukn−g(ω))+
)µ(∣∣vkT − U̇T
∣∣)ψ′1/k
(vkT − U̇T
)·wT dxds
→∫ t
0
∫ΓC
F((un−g(ω))+
)µ(∣∣vT − U̇T
∣∣)ξ ·wT dxds
where ∫ t
0
∫ΓC
ξ ·wT dαds≤∫ t
0
∫ΓC
∣∣vkT − U̇T +wT∣∣− ∣∣vkT − U̇T
∣∣dαds (70.5.56)
Then passing to the limit in the integral equation 70.5.55, we obtain that v is a solution foreach ω to the integral equation
v(t)−v0 +∫ t
0Mvds+
∫ t
0Auds+
∫ t
0Puds+