2254 CHAPTER 65. STOCHASTIC INTEGRATION
which converges to Φ◦ J−1 in K where also the lengths of the sub intervals converge uni-formly to 0 as k→ ∞.
Now let τ be an arbitrary stopping time. The partition points corresponding to Φk are{tk
j
}mk
j=0. Let τk = tk
j+1 on τ−1(tkj , t
kj+1]. Then τk is a stopping time because
[τk ≤ t] ∈Ft
Here is why. If t ∈ (tkj , t
kj+1], then if t = tk
j+1, it would follow that τk (ω) ≤ t would be
the same as saying ω ∈[τ ≤ tk
j+1
]= [τ ≤ t] ∈ Ft . On the other hand, if t < tk
j+1, then
[τk ≤ t] =[τ ≤ tk
j
]∈Ftk
j⊆Ft because τk can only take the values tk
j .
Consider X[a,τk]Φk. It is given that Φk→Φ◦J−1 in K. Does it follow that X[a,τk]Φk→X[a,τ]Φ◦ J−1 in K? Consider first the indicator function. Let τ (ω) ∈ (tk
j , tkj+1]. Fixing t, if
X[a,τ] (t) = 1, then also X[a,τk] (t) = 1 because τk ≥ τ . Therefore, in this case
limk→∞
X[a,τk] (t) = X[a,τ] (t) .
Next suppose X[a,τ] (t) = 0 so that τ (ω) < t. Since the intervals defined by the partitionpoints have lengths which converge to 0, it follows that for all k large enough, τk (ω) < talso and so X[a,τk] (t) = 0. Therefore,
limk→∞
X[a,τk(ω)] (t) = X[a,τ(ω)] (t) .
It follows that X[a,τk]Φk →X[a,τ]Φ◦ J−1 in K. Now from 65.8.16, the function X[a,τk]Φk
is progressively measurable. Therefore, the same is true of X[a,τ]Φ◦ J−1.From the proof of Theorem 65.5.3, the part depending on maximal estimates and the
fact that∫ t
a X[a,τk]ΦkdW is a continuous martingale, there is a set of measure zero N, suchthat off this set, a suitable subsequence satisfies∫ t
aX[a,τk]ΦkdW →
∫ t
aX[a,τ]ΦdW
uniformly on [a,T ] . But also, since Φk→Φ◦ J−1 in K, a suitable subsequence satisfies,∫ t
aΦkdW →
∫ t
aΦdW
uniformly on [a,T ] a.e. ω. In particular,∫ t∧τk
a ΦkdW →∫ t∧τ
a ΦdW. Therefore,∫ t
aX[a,τ]ΦdW = lim
k→∞
∫ t
aX[a,τk]ΦkdW
= limk→∞
∫ t∧τk
aΦkdW
=∫ t∧τ
aΦdW
This has proved the following major localization lemma. This is a marvelous result. Itsays that the stochastic integral acts algebraically like an ordinary Stieltjes integral, one foreach ω off a set of measure zero.