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.

2254 CHAPTER 65. STOCHASTIC INTEGRATIONwhich converges to ®oJ~! in K where also the lengths of the sub intervals converge uni-formly to 0 as k + ©,Now let t be an arbitrary stopping time. The partition points corresponding to ®,; areMkk _ 4k = (4k 4k : ao ti{a a Let t% =f;,, on T (t7,1,;]. Then t, is a stopping time because[t% <t] EF,Here is why. If t € (¢,¢,,], then if ¢= 14, ,,the same as saying @ € E < tk, | = [t <t] € ¥;. On the other hand, if t < tha, thenit would follow that t;(@) <t would be[tT <t]= k < HK € Fx C F; because 7; can only take the values ti.. ; .Consider 27,,7,;Px. It is given that B, + ®oJ~! in K. Does it follow that Bat] Pk >Rig aP oJ—! in K? Consider first the indicator function. Let tT(@) € (tj iti 1]. Fixing ¢, ifKigq (t) =1, then also 2i, -,) (t) = 1 because t;, > t. Therefore, in this casepm. Kiar] (t)= Kia,t| (t).Next suppose 2, z(t) = 0 so that t(@) <t. Since the intervals defined by the partitionpoints have lengths which converge to 0, it follows that for all & large enough, t;(@) <talso and so 2jq,7,j (t) = 0. Therefore,jim Kia,ty(@)] (t)= Xa,t(o)] (t).It follows that 2jq7,;Px > Xa gqPos! in K. Now from 65.8.16, the function 2(q 7,)Pxis progressively measurable. Therefore, the same is true of 2jq,jPoJ aFrom the proof of Theorem 65.5.3, the part depending on maximal estimates and thefact that a 2at,|PrdW is a continuous martingale, there is a set of measure zero N, suchthat off this set, a suitable subsequence satisfiest t| Ka,tJPrdWw => | BRia,jPdWwa auniformly on [a, 7]. But also, since ®, + @oJ~! in K, a suitable subsequence satisfies,t t[ @,.dWw + / ddwJa auniformly on [a,7] a.e. @. In particular, [/\"* b,dW — ['" dW. Therefore,t t| Fantaw = lim Kia tJ PrdWa : ke Ja :TATElim ®, dwknw JatAT= @dwaThis 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 o off a set of measure zero.