65.9. LOCALIZATION IN GENERAL 2253

Now observe

X[a,τ] (t)Φ(t) =n−1

∑k=0

X[a,τ] (t)Φ(tk)X(tk,tk+1] (t)

=n−1

∑k=0

X[τ≥t]Φ(tk)X(tk,tk+1] (t)

=n−1

∑k=0

X[τ>tk]Φ(tk)X(tk,tk+1] (t) (65.8.16)

The last step occurs because of the following reasoning. The kth term of the sum in themiddle expression above equals Φ(tk) if and only if t > tk and τ ≥ t. If the two conditionsdo not hold, then the kth term equals 0. As to the third line, if τ > tk and t ∈ (tk, tk+1], thenτ ≥ tk+1 ≥ t which is the same as the situation in the second line. The term equals Φ(tk).Note that X[τ>tk] (ω) is Ftk measurable, because [τ > tk] is the complement of [τ ≤ tk].Therefore, this is an elementary function. Thus, from 65.8.15 -65.8.16, X[a,τ] (t)Φ(t) is anelementary function and∫ t∧τ

aΦdW =

∫ t

a

n−1

∑k=0

X[τ>tk]Φ(tk)X(tk,tk+1] (t)dW =∫ t

aX[a,τ] (t)Φ(t)dW

From Proposition 65.1.5, if you have Φ,Ψ two of these elementary functions

E

(∥∥∥∥∫ t

aX[a,τ] (t)Φ(t)dW −

∫ t

aX[a,τ] (t)Ψ(t)dW

∥∥∥∥2

H

)=

∫ t

a

∫Ω

X[a,τ] (t)∥(Φ(s)−Ψ(s))◦ J∥2L2(Q1/2U,H) dPds

≤∫ t

a

∫Ω

∥(Φ(s)−Ψ(s))◦ J∥2L2(Q1/2U,H) dPds (65.8.17)

65.9 Localization In GeneralNext, what about the general case where Φ ∈ L2

([a,T ]×Ω;L2

(Q1/2U,H

))and is pro-

gressively measurable? Is it the case that for an arbitrary stopping time τ,∫ t∧τ

aΦdW =

∫ t

aX[a,τ]ΦdW?

This is the sort of thing which would be expected for an ordinary Stieltjes integral whichof course this isn’t. Let

L2([a,T ]×Ω;L2

(JQ1/2U,H

))= K

From Doob’s result Proposition 65.3.2 and Lemma 65.3.1, there exists a sequence ofelementary functions {Φk}

Φk (t) =mk−1

∑j=0

Φ

(tk

j

)X(tk

j ,tkj+1]

(t)