74.5. CONVERGENCE 2513
Proof: By assumption, d [M] = kdt for some k∈ L1 ([0,T ]×Ω) and so BXτ p ∈G whereG was the class of functions for which one can write
∫ t0 ⟨BX ,dM⟩. By the Burkholder Davis
Gundy inequality,
P(
supt
∣∣∣∣∫ t∧τ p
0
⟨B(
X lk
)−BX ,dM
⟩∣∣∣∣> ε
)= P
(sup
t
∣∣∣∣∫ t
0X[0,τ p]
⟨B(
X lk
)−BX ,dM
⟩∣∣∣∣> ε
)≤ C
ε
∫Ω
(∫ T∧τ p
0
∥∥∥B(
X lk
)−BX
∥∥∥2
Wkdt)1/2
dP
=Cε
∫Ω
(∫ T
0X[0,τ p]
∥∥∥B(
X lk
)−BX
∥∥∥2
Wkdt)1/2
dP (74.5.19)
Let
Ak =
[sup
t
∣∣∣∣∫ t
0
⟨BX l
k−BX ,dM⟩∣∣∣∣> ε
]Then, since τ p = ∞ for all p large enough,
Ak = ∪∞p=0Ak ∩ ([τ p = ∞]\ [τ p−1 ̸= ∞])
Consider BX lτ pk . If t > τ p, what of the values of BX lτ p
k ? It equals BX (s) where s is one ofthe mesh points s≤ τ p because this is a left step function. Therefore,⟨
BX lτ pk (s) ,X lτ p
k (s)⟩=⟨
B(
X lτ pk (s)
),X lτ p
k (s)⟩
= ∑i⟨BX (s) ,ei⟩2 ≤ p
As to X[0,τ p]BX , it follows that for all t ≤ τ p you have ∑i ⟨BX (t) ,ei⟩2≤ p and so, since this
equals ⟨B(X (t)) ,X (t)⟩ a.e. t, it follows that∥∥∥X[0,τ p]BX (t)
∥∥∥W ′
is bounded by a constant
depending on p for a.e.t. It follows that BX and BX lk are bounded. Now by Lemma 74.5.1,
BX lk (t)→ BX (t) a.e. t and the term
∥∥B(X l
k
)−BX
∥∥2W is essentially bounded. Therefore, in
74.5.19, the integral converges to 0. From this formula,
P(Ak ∩ ([τ p = ∞]\ [τ p−1 ̸= ∞]))≤ P(Ak ∩ ([τ p = ∞]))
≤ Cε
∫Ω
(∫ T
0X[0,τ p]
∥∥∥B(
X lk
)−BX
∥∥∥2
Wkdt)1/2
dP
Thuslimk→∞
P(Ak ∩ ([τ p = ∞]\ [τ p−1 ̸= ∞])) = 0
Then
P(Ak) =∞
∑p=1
P(Ak ∩ ([τ p = ∞]\ [τ p−1 ̸= ∞]))