65.11. THE QUADRATIC VARIATION OF THE STOCHASTIC INTEGRAL 2259
Lemma 65.11.2 Let Φ,Φn all be in L2([0,T ] ,L2
(Q1/2U,H
))off some set of measure
zero. These are all progressively measurable. Thus there are all stochastically squareintegrable.
P(∫ T
0∥Φ∥2 ds
)= 1
Suppose also that for each ω /∈ N, the exceptional set,∫ T
0∥Φn−Φ∥2
L2dt→ 0
Then there exists a set of measure zero, still denoted as N and a subsequence, still denotedas n such that for each ω /∈ N,
limn→∞
∫ T
0ΦndW =
∫ T
0ΦdW
Proof: Define stopping times
τnp ≡ inf{
t ∈ [0,T ] :∫ t
0∥Φn∥2 ds > p
}Let τ p be similar but defined with reference to Φ. Then by Ito isometry,
E
(∣∣∣∣∫ T
0X[0,τnp]ΦndW −
∫ T
0X[0,τ p]ΦdW
∣∣∣∣2)
= E(∫ T
0
∥∥∥X[0,τnp]Φn−X[0,τ p]Φ∥∥∥2
L2dt)
(65.11.19)
The integrand in the right side is bounded by 2p2. Also this integrand converges to 0 foreach ω as n→ ∞. This is shown next.∫ T
0
∥∥∥X[0,τnp]Φn−X[0,τ p]Φ∥∥∥2
L2dt
≤ 2∫ T
0
(∥∥∥X[0,τnp]Φn−X[0,τnp]Φ∥∥∥2
L2+∥∥∥X[0,τnp]Φ−X[0,τ p]Φ
∥∥∥2
L2
)dt
≤ 2∫ T
0∥Φn−Φ∥2
L2dt +2
∫ T
0
∣∣∣X[0,τnp] (t)−X[0,τ p] (t)∣∣∣2 ∥Φ∥2
L2dt
The first converges to 0 by assumption. Problem is, it does not look like this second integralconverges to 0. We do know that
∫ t0 ∥Φn∥2 ds→
∫ t0 ∥Φ∥
2 ds uniformly so τnp→ τ p is likely.However, this does not imply X[0,τnp]→X[0,τ p]. However, it would converge in L2 (0,T )and so there is a subsequence such that convergence takes place a.e. t. Then restrictingto this subsequence, the second integral converges to 0. Actually, it may be easier thanthis. X[0,τ p] has a single point of discontinuity and convergence takes place at every otherpoint. Thus it appears that the integrand in 65.11.19 converges to 0 for each ω . Thus, bydominated convergence theorem the whole expectation converges to 0.