2678 CHAPTER 78. A DIFFERENT APPROACH

Theorem 78.4.3 Assume the above conditions, 78.3 - 78.3, and 78.4.2. Let u0 be F0measurable and (t,ω)→X[0,t] (t) f (t,ω) is B ([0, t])×Ft product measurable into V ′ foreach t. Also assume that for each ω, there is at most one solution (u,u∗) to the evolutionequation

Bu(ω)(t)−Bu0 (ω)+∫ t

0u∗ (·,ω)ds =

∫ t

0f (s,ω)ds, (78.4.32)

u∗ (·,ω) ∈ A(u(·,ω) ,ω)

for t ∈ [0,T ]. Then there exists a unique solution (u(·,ω) ,u∗ (·,ω)) in V[0,T ]×V ′[0,T ] to theabove integral equation for each ω, t ∈ (0,T ) . This solution satisfies

(t,ω)→ (u(t,ω) ,u∗ (t,ω))

is progressively measurable into V ×V ′.

Proof: First note that Theorem 78.3.2 there exists a solution on [0,T −σ ] for each smallσ > 0. Then by uniqueness, there exists a solution on (0,T ). Let T denote subsets of(0,T −σ ] which contain T −σ such that for S ∈T , there exists a solution uS for each ω tothe above integral equation on [0,T −σ ] such that (t,ω)→X[0,s] (t)uS (t,ω) is B ([0,s])×Fs measurable for each s ∈ S. Then {T −σ} ∈T . If S,S′ are in T , then S≤ S′ will meanthat S ⊆ S′ and also uS (t,ω) = uS′ (t,ω) in V for all t ∈ S, similar for u∗S and u∗S′ . Notehow we are considering a particular representative of a function in V[0,T−σ ] and V ′[0,T−σ ]

because of the pointwise condition. Now let C denote a maximal chain. Is ∪C ≡ S∞

all of (0,T −σ ]? What is uS∞? Define uS∞

(t,ω) the common value of uS (t,ω) for all S inC , which contain t ∈ S∞. If s∈ S∞, then it is in some S∈C and so the product measurabilitycondition holds for this s. Thus S∞ is a maximal element of the partially ordered set. Is S∞

all of (0,T −σ ]? Suppose ŝ /∈ S∞,T −σ > ŝ > 0.From Theorem 78.3.2 there exists a solution to the integral equation 78.4.32 on [0, ŝ]

called u1 such that (t,ω)→ u1 (t,ω) is B ([0, ŝ])×Fŝ measurable, similar for u∗1. Bythe same theorem, there is a solution on [0,T −σ ], u2 which is B ([0,T −σ ])×F[0,T−σ ]

measurable. Now by uniqueness, u2 (·,ω) = u1 (·,ω) in V[0,ŝ], similar for u∗i . Therefore,no harm is done in re-defining u2,u∗2 on [0, ŝ] so that u2 (t,ω) = u1 (t,ω) , for all t ∈ [0, ŝ] ,similar for u∗. Denote these functions as û, û∗. By uniqueness, uS∞

(·,ω) = û(·,ω) inLp ([0, ŝ] ,V ). Thus no harm is done by re-defining û(s,ω) to equal uS∞

(s,ω) for s < ŝ andu1 (ŝ,ω) at ŝ. As to s > ŝ also re define û(s,ω)≡ uS∞

(s,ω) for such s. By uniqueness, thetwo are equal in V[ŝ,T−σ ] and so no change occurs in the solution of the integral equation.Now S∞ was not maximal after all. S∞∪{ŝ} is larger. This contradiction shows that in fact,S∞ = (0,T−σ ]. Thus there exists a unique progressively measurable solution to 78.4.32 on[0,T −σ ] for each small σ . Thus we can simply use uniqueness to conclude the existenceof a unique progressively measurable solution on [0,T ).

Theorem 78.4.4 Assume the above conditions, 78.3 - 78.3, and 78.4.2. Let u0 be F0measurable and (t,ω)→X[0,t] (t) f (t,ω) is B ([0, t])×Ft product measurable into V ′ foreach t ∈ [0,T −σ ]. Also let t → q(t,ω) be continuous and q is progressively measurableinto V. Suppose there is at most one solution to

Bu(t,ω)+∫ t

0u∗ (s,ω)ds =

∫ t

0f (s,ω)ds+Bu0 (ω)+Bq(t,ω) , (78.4.33)