2160 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE
Thus the expected value of the first sum equals 0. Similarly, the expected value of the sec-ond sum equals 0. Hence this has shown that for any bounded right continuous martingale,{ξ (s)} and t > 0,
E(ξ (t)
(A(t)−A′ (t)
))= 0.
Now let ξ be a bounded random variable and let ξ (t) be a right continuous version of themartingale E (ξ |Ft) . Then
0 = E(E (ξ |Ft)
(A(t)−A′ (t)
))= E
(E(ξ(A(t)−A′ (t)
)|Ft))
= E(ξ(A(t)−A′ (t)
))and since ξ is arbitrary, it follows that A(t) = A′ (t) a.e. which proves uniqueness.
Because of the uniqueness assertion, it suffices to prove the theorem on an arbitraryinterval, [0,a].
Without loss of generality, it can be assumed X (0) = 0 since otherwise, you could sim-ply consider X (t)−X (0) in its place and then at the end, add X (0) to M (t) . Let
{tnk
}mnk=0
be a sequence of partitions of [0,a] such that these are equally spaced points, limn→∞ tnk+1−
tnk = 0, and
{tnk
}mnk=0 ⊆
{tn+1k
}mn+1k=0 . Then consider the submartingale,
{X(tnk
)}mnk=0 . Theo-
rem 63.7.1 implies there exists a unique martingale, and increasing submartingale,
{M (tnk )}
mnk=0 and {A(tn
k )}mnk=0
respectively such that M (0) = 0 = A(0) ,
X (tnk ) = Mn (tn
k )+An (tnk ) .
and An(tnk
)is Ftn
k−1measurable. Recall how these were defined.
An (tnk ) =
k
∑j=1
E(
X(tn
j)−X
(tn
j−1)|Ftn
j−1
), An (0) = 0
Mn (tnk ) = X (tn
k )−An (tnk ) .
I want to show that {An (a)} is equi integrable. From this there will be a weakly conver-gent subsequence and nice things will happen. Define T n (ω) to equal tn
j−1 where tnj is the
first time where An(
tnj ,ω)≥ λ or T n (ω) = a if this never happens. I want to say that T n is
a stopping time and so I need to verify that[T n ≤ tn
j
]∈Ftn
jfor each j. If ω ∈
[T n ≤ tn
j
],
then this means the first time, tnk , where An
(tnk ,ω
)≥ λ is such that tn
k ≤ tnj+1. Since An
k isincreasing in k, [
T n ≤ tnj]
= ∪ j+1k=0 [A
n (tnk )≥ λ ]
=[An (tn
j+1)≥ λ
]∈Ftn
j.
Note T n only has the values tnk . Thus for t ∈ [tn
j−1, tnj ),
[T n ≤ t] =[T n ≤ tn
j−1]∈Ftn
j−1⊆Ft .