2282 CHAPTER 66. THE INTEGRAL∫ t
0 (Y,dM)H
Let Y be one of those elementary functions which is in G , ∥Y (t)∥M∗ ∈ L2 (Ω).
Y (t) =mk−1
∑i=0
YiX(tki ,t
ki+1]
(t)
and consider X[0,τk]Y. Here Y will be always the same for the different partitions. It isjust that some of the Yi will be repeated on smaller and smaller intervals. Does it followthat X[0,τk]Y →X[0,τ]Y for each fixed ω? This depends only on the indicator function.Let τ (ω) ∈ (tk
j , tkj+1]. Fixing t, if X[0,τ] (t) = 1, then also X[0,τk] (t) = 1 because τk ≥ τ .
Therefore, in this case limk→∞ X[0,τk] (t) = X[0,τ] (t) . Next suppose X[0,τ] (t) = 0 so thatτ (ω) < t. Since the intervals defined by the partition points have lengths which convergeto 0, it follows that for all k large enough, τk (ω)< t also and so X[0,τk] (t) = 0. Therefore,
limk→∞
X[0,τk(ω)] (t) = X[0,τ(ω)] (t) .
It follows that X[0,τk]Y →X[0,τ]Y . Also it is clear from the dominated convergence theo-rem, ∥∥X[0,τk]Y −X[0,τ]Y
∥∥2H ≤ 4∥Y∥2
H ,
that
limk→∞
E(∫ T
0
∥∥X[0,τk]Y −X[0,τ]Y∥∥2
H d [Mτ p ]
)= 0
Thus X[0,τ]Y ∈ G . By Lemma 66.0.9, there is a subsequence, still denoted as X[0,τk]Y suchthat off a set of measure zero,∫ t
0
(X[0,τk]Y,dMτ p
)→∫ t
0
(X[0,τ]Y,dMτ p
)uniformly on [0,T ]. Therefore, from the localization for elementary functions and thisuniform convergence,∫ t
0
(X[0,τ]Y,dMτ p
)= lim
n→∞
∫ t
0
(X[0,τn]Y,dMτ p
)= lim
n→∞
∫ t∧τn
0(Y,dMτ p) =
∫ t∧τ
0(Y,dMτ p)
This proves most of the following lemma.
Lemma 66.0.12 Let Y be an elementary function. Then if τ is any stopping time, then offa set of measure zero,∫ t∧τ
0(Y,dMτ p) =
∫ t
0
(X[0,τ]Y,dMτ p
)=∫ t
0
(Y,dMτ∧τ p
)Proof: It remains to prove the second equation.∫ t∧τ
0(Y,dMτ p) ≡
m−1
∑i=0
(Yi,Mτ p (t ∧ ti+1∧ τ)−Mτ p (t ∧ ti∧ τ))
≡m−1
∑i=0
(Yi,Mτ∧τ p (t ∧ ti+1)−Mτ p (t ∧ ti)
)≡
∫ t
0
(Y,dMτ∧τ p
)