2280 CHAPTER 66. THE INTEGRAL∫ t
0 (Y,dM)H
which converges to 0. Thus there is a subsequence still denoted with index k such that
P
(sup
t∈[0,T ]
∣∣∣∣∫ t
0
(Y k,dMτ p
)−∫ t
0
(Y k+1,dMτ p
)∣∣∣∣2 > 2−k
)< 2−k
and so there exists a set of measure zero N such that for ω /∈ N,
supt∈[0,T ]
∣∣∣∣∫ t
0
(Y k,dMτ p
)−∫ t
0
(Y k+1,dMτ p
)∣∣∣∣2 ≤ 2−k
for all k large enough and so for this subsequence, the convergence is uniform. Hencet→
∫ t0 (Y
n,dMτ p) has a continuous version obtained from the uniform limit of these.Finally,
E
(∣∣∣∣∫ t
0(Y,dMτ p)
∣∣∣∣2)
= limn→∞
E
(∣∣∣∣∫ t
0(Y n,dMτ p)
∣∣∣∣2)
≤ limn→∞
E(∫ t
0∥Y n∥2
H d [M]τ p
)= E
(∫ t
0∥Y∥2
H d [M]τ p
)What is the quadratic variation of the martingale in the above lemma? I am not going
to give it exactly but it is easy to give an estimate for it. Recall the following result. It isTheorem 63.6.4.
Theorem 66.0.10 Let H be a Hilbert space and suppose (M,Ft) , t ∈ [0,T ] is a uniformlybounded continuous martingale with values in H. Also let
{tnk
}mnk=1 be a sequence of parti-
tions satisfying
limn→∞
max{∣∣tn
i − tni+1∣∣ , i = 0, · · · ,mn
}= 0, {tn
k }mnk=1 ⊆
{tn+1k
}mn+1k=1 .
Then
[M] (t) = limn→∞
mn−1
∑k=0
∣∣M (t ∧ tnk+1)−M (t ∧ tn
k )∣∣2H
the limit taking place in L2 (Ω). In case M is just a continuous local martingale, the abovelimit happens in probability.
In the above Lemma, you would find the quadratic variation according to this theoremas follows. [∫ (·)
0(Y,dMτ p)
](t) = lim
n→∞
mn−1
∑k=0
∣∣∣∣∫ t∧tnk+1
t∧tnk
(Y,dMτ p)
∣∣∣∣2H
where the limit is in probability. Thus
limn→∞
P
(∣∣∣∣∣[∫ (·)
0(Y,dMτ p)
](t)−
mn−1
∑k=0
∣∣∣∣∫ t∧tnk+1
t∧tnk
(Y,dMτ p)
∣∣∣∣2∣∣∣∣∣≥ ε
)= 0