76.4. THE GENERAL CASE 2569
Proof: Let ε > 0. Then define
An =
{ω :∣∣∣∣∫ T
0
(Φn ◦ J−1)∗Bun ◦ JdW −
∫ T
0
(Φ◦ J−1)∗Bu◦ JdW
∣∣∣∣> ε
}Then
An = ∪∞p=1An∩ ([τ p = ∞]\ [τ p−1 < ∞]) , [τ0 < ∞]≡ /0
the sets in the union being disjoint. Then A∩ ([τ p = ∞])⊆{ω :
∣∣∣∣∣∫ T
0 X[0,τ p](Φn ◦ J−1
)∗Bun ◦ JdW
−∫ T
0 X[0,τ p](Φ◦ J−1
)∗Bu◦ JdW
∣∣∣∣∣> ε
}
Then as before,
E(∣∣∣∣∫ T
0X[0,τ p]
(Φn ◦ J−1)∗Bun ◦ JdW −
∫ T
0X[0,τ p]
(Φ◦ J−1)∗Bu◦ JdW
∣∣∣∣)
≤∫
Ω
((∫ T
0∥Φn−Φ∥2
L2X[0,τ p] ⟨Bun,un⟩
)1/2)
dP
+∫
Ω
(∫ T
0X[0,τ p] ∥Φ∥
2L2⟨Bun−Bu,un−u⟩dt
)1/2
dP (76.4.52)
Consider the second term. It is no larger than(∫Ω
∫ T
0X[0,τ p] ∥Φ∥
2L2⟨Bun−Bu,un−u⟩dtdP
)1/2
Now t → ⟨Bun,un⟩ is continuous and so X[0,τ p] ⟨Bun,un⟩(t) ≤ p. If not, then you wouldhave ⟨Bun,un⟩(t)> p for some t ≤ τ p and so, by continuity, there would be s < t ≤ τ p forwhich ⟨Bun,un⟩(s)> p contrary to the definition of τ p. Then X[0,τ p] ⟨Bun−Bu,un−u⟩ isbounded a.e. and also converges to 0 for a.e. t ≤ τ p as n→ ∞. Therefore, off a set of mea-sure zero, including the set where t→∥Φ∥2
L2is not in L1, the double integral converges to
0 by the dominated convergence theorem. As to the first integral in 76.4.52, it is dominatedby ∫
Ω
X[0,τ p] supt∈[0,τ p]
⟨Bun,un⟩1/2 (t)(∫ T
0∥Φn−Φ∥2
L2
)1/2
dP
≤
(∫Ω
supt∈[0,T ]
⟨Bun,un⟩dP
)1/2(∫Ω
∫ T
0∥Φn−Φ∥2
L2dtdP
)1/2
From the estimate 76.4.32,
≤C(∫
Ω
∫ T
0∥Φn−Φ∥2
L2dtdP
)1/2