2281

Then you can obtain from this and the usual appeal to the Borel Cantelli lemma a set ofmeasure zero Nt and a subsequence still denoted with n satisfying that for all ω /∈ Nt and nlarge enough, ∣∣∣∣∣

[∫ (·)

0(Y,dMτ p)

](t)−

mn−1

∑k=0

∣∣∣∣∫ t∧tnk+1

t∧tnk

(Y,dMτ p)

∣∣∣∣2∣∣∣∣∣≤ 1

n

Hence [∫ (·)

0(Y,dMτ p)

](t)≤ 1

n+

mn−1

∑k=0

∫ t∧tnk+1

t∧tnk

∥Y∥2H d [M]τ p

=1n+∫ t

0∥Y∥2

H d [M]τ p

Then for that t, you have on taking a limit as n→ ∞,[∫ (·)

0(Y,dMτ p)

](t)≤

∫ t

0∥Y∥2

H d [M]τ p

Now take the union of Nt for t ∈ Q∩ [0,T ]. Denote this as N. Then if ω /∈ N, the aboveshows that for such t, [∫ (·)

0(Y,dMτ p)

](t)≤

∫ t

0∥Y∥2

H d [M]τ p

But both sides are continuous in t and so this inequality holds for all t ∈ [0,T ]. Thus thefollowing corollary is obtained.

Corollary 66.0.11 Let M be a continuous local martingale and τ p a localizing sequencewhich makes Mτ p an L2 martingale and assume that Y ∈ G . Then the quadratic variationof this martingale satisfies[∫ (·)

0(Y,dMτ p)

](t)≤

∫ t

0∥Y∥2

H d [M]τ p ≤∫ t

0∥Y∥2

H d [M]

for ω off a set of measure zero.

Does the localization stuff hold for an arbitrary stopping time? Let{

tki}

denote thekth partition of a sequence of nested partitions whose maximum length between successivepoints converges to 0. Let τ be a stopping time and 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 .