65.14. A TECHNICAL INTEGRATION BY PARTS RESULT 2263

Then let t be given and pick n such that τn (ω)≥ t. Then from the first part, for that ω,

L∫ t

aΦdW ≡ L

∫ t

aX[a,τn]ΦdW

=∫ t

aLX[a,τn]ΦdW

=∫ t

aX[a,τn]LΦdW ≡

∫ t

aLΦdW

65.14 A Technical Integration By Parts ResultLet Z ∈ L2

([0,T ]×Ω,L2

(Q1/2U,H

))where this has reference to the usual diagram

U↓ Q1/2

U1 ⊇ JQ1/2U J←1−1

Q1/2U

Φn ↘ ↓ Φ

H

Also suppose X ∈ L2 ([0,T ]×Ω,H) , both X and Z being progressively measurable. Let{tn

j

}mn

j=1denote a sequence of partitions of the sort discussed earlier where

Xn (t)≡mn−1

∑j=0

X(tn

j)X[tn

j ,tnj+1)

(t)

converges to X in L2 ([0,T ]×Ω,H) . Thus Xn (t) is right continuous. Let

τnp = inf{t : |Xn (t)|H > p} .

This is the first hitting time of a right continuous adapted process so it is a stopping time.Also there exists a set of measure zero N such that for ω /∈ N, then given t,

τnp ≥ t

if p is large enough because of the assumption on X . Here is why. There exists a set ofmeasure 0 N such that if ω /∈ N, then∫ T

0|Xn (t)|2H dt =

mn−1

∑j=0|X (tn

k )|2H

(tn

j+1− tnj)< ∞.

It follows that there exists an upper bound, depending on ω which dominates each of thevalues

∣∣X (tnk

)∣∣2H . Then if p is larger than this upper bound, τn

p = ∞ > t.Next consider the expression

m−1

∑j=0

(∫ tnj+1∧t

tnj∧t

Z (u)dW,X(tn

j))

H

. (65.14.20)