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Lemma 66.0.13 Let Y ∈ G . Then for any stopping time τ,∫ t∧τ
0(Y,dMτ p) =
∫ t
0
(X[0,τ]Y,dMτ p
)=∫ t
0
(Y,dMτ∧τ p
)for ω off some set of measure zero.
Proof: From Lemma 66.0.9, there exists a sequence of elementary functions Y n suchthat t →
∫ t0 (Y
n,dM) converges uniformly to t →∫ t
0 (Y,dMτ p) on [0,T ] for each ω /∈ N, aset of measure zero. Then∫ t∧τ
0(Y,dMτ p) = lim
n→∞
∫ t∧τ
0(Y n,dMτ p)
= limn→∞
∫ t
0
(X[0,τ]Y
n,dMτ p)=∫ t
0
(X[0,τ]Y,dMτ p
)The last claim needs a little clarification. As shown in the above discussion proving Lemma66.0.12, while X[0,τ]Y n is no longer obviously an elementary function due to the fact thatτ has values which are not partition points, it is still the limit of a sequence of elementaryfunctions X[0,τk]Y
n and so the integral makes sense. Then from the inequality of Lemma66.0.9,
E
(∣∣∣∣∫ t
0
(X[0,τ]Y
n,dMτ p)−∫ t
0
(X[0,τ]Y,dMτ p
)∣∣∣∣2)≤ E
(∫ T
0∥Y n−Y∥2
H d [M]τ p
)and so by the same Borel Canteli argument of that lemma, there is a further subsequencefor which the convergence is uniform off a set of measure zero as n→ ∞. (Actually, thesame subsequence as in the first part of the argument works.) Therefore, the conclusionfollows.
What of the second equation? Let {Y n} be as above where uniform convergence takesplace for the stochastic integrals. Then from Lemma 66.0.12∫ t
0
(X[0,τ]Y
n,dMτ p)=∫ t
0
(Y n,dMτ∧τ p
)Hence
E
(∣∣∣∣∫ t
0
(Y n,dMτ∧τ p
)−∫ t
0
(X[0,τ]Y,dMτ p
)∣∣∣∣2)≤ E
(∫ T
0∥Y n−Y∥2
H d [M]τ p
)Now by the usual application of the Borel Canelli lemma, there is a subsequence and a setof measure zero off which
∫ t0 (Y
n,dMτ∧τ p) converges uniformly to∫ t
0 (Y,dMτ∧τ p) on [0,T ]and as n→ ∞, and also ∫ t
0
(Y n,dMτ∧τ p
)→∫ t
0
(X[0,τ]Y,dMτ p
)uniformly on t ∈ [0,T ]. Then from the above,∫ t
0
(Y n,dMτ∧τ p
)→∫ t
0
(X[0,τ]Y,dMτ p
)=∫ t∧τ
0(Y,dMτ p)
uniformly. Thus∫ t
0 (Y,dMτ∧τ p) =∫ t∧τ
0 (Y,dMτ p) .