74.2. PRELIMINARY RESULTS 2501

measure zero N̂ ⊆ [0,T ] such that BX (t) = B(X (t)) in Lq′ (Ω;V ′) for all t /∈ N̂. Thus forsuch t,BX (t)(ω) = B(X (t,ω)) for a.e.ω . In particular, for such t /∈ N̂,

⟨BX (t)(ω) ,X (t,ω)⟩= ∑i⟨B(X (t)) ,ei⟩2 a.e.ω.

D has empty intersection with N̂. There is also a set of Lebesgue measure zero Nω for eachω /∈ N defined by BX (t,ω) = B(X (t,ω)) for all t /∈ Nω .

Now define a stopping time.

σnq ≡ inf

{t :⟨

BX ln (t) ,X

ln (t)

⟩> q}, (74.2.3)

Thus this pertains to the nth partition. Since X ln is right continuous, this will be a well

defined stopping time. Thus, for t one of the partition points,⟨BXσn

q (t,ω) ,Xσnq (t,ω)

⟩≤ q (74.2.4)

From the definition of X ln and the observation that these partitions are nested,

limn→∞

σnq ≡ σq

exists because this is a decreasing sequence. There are more available times to consider asn gets larger and so when the inf is taken, it can only get smaller. Thus

[σq ≤ t] = ∩∞m=1∪∞

k=1∩n≥k

[σ

nq ≤ t +

1m

]∈ ∩∞

m=1Ft+(1/m) = Ft

since it is assumed that the filtration is normal. Thus this appears to be a stopping time.However, I don’t know how to use this.

Theorem 74.2.2 Let{

tnj

}mn

j=0be the above sequence of partitions of the sort in Lemma

74.0.2 such that if

X ln (t)≡

mn−1

∑j=0

X(tn

j)X[tn

j ,tnj+1)

(t)

then Xn→ X in Lp ([0,T ]×Ω,V ) with the other conditions holding which were discussedabove. In particular, BX (t) = B(X (t)) for t one of these mesh points. Then the expression

mn−1

∑j=0

⟨B(M(tn

j+1∧ t)−M

(tn

j ∧ t))

,X(tn

j)⟩

=mn−1

∑j=0

⟨BX(tn

j),(M(tn

j+1∧ t)−M

(tn

j ∧ t))⟩

(74.2.5)

is a local martingale ∫ t

0

⟨BX l

k ,dM⟩

with{

σnq}∞

q=1 being a localizing sequence.