2126 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

and Pn (t) is a martingale. Then from 63.2.9, Qn (t) is a submartingale and converges foreach t to something, denoted as [M] (t) in L1 (Ω) uniformly in t ∈ [0,T ]. This is becausePn (t) converges uniformly on [0,T ] to N (t) in L2 (Ω) and ||M (t)||2 does not depend on n.Then also [M] is a submartingale which equals 0 at 0 because this is true of Qn and becauseif A ∈Fs where s < t,∫

AE ([M] (t) |Fs)dP≡

∫A[M] (t)dP = lim

n→∞

∫A

(||M (t)||2−Pn (t)

)dP

= limn→∞

∫A

E(||M (t)||2−Pn (t) |Fs

)dP≥ lim inf

n→∞

∫A||M (s)||2−Pn (s)dP

= lim infn→∞

∫A

Qn (s)dP =∫

A[M] (s)dP.

Note that Qn (t) is increasing because as t increases, the definition allows for the pos-sibility of more nonzero terms in the sum. Therefore, [M] (t) is also increasing in t. Thefunction t→ [M] (t) is continuous because ||M (t)||2 = [M] (t)+N (t) and t→ N (t) is con-tinuous as is t → ||M (t)||2 . That is, off a set of measure zero, these are both continuousfunctions of t and so the same is true of [M] .

Now put back in Mα p∧δ in place of M. From the above, this has shown∣∣∣∣∣∣Mα p∧δ (t)∣∣∣∣∣∣2 = [Mα p∧δ

](t)+Np (t)

where Np is a martingale and[Mα p∧δ

](t) = lim

n→∞∑k≥0

∣∣∣∣∣∣Mα p∧δ(t ∧ τ

nk+1)−Mα p∧δ (t ∧ τ

nk)∣∣∣∣∣∣2

= limn→∞

∑k≥0

∣∣∣∣M (t ∧ τnk+1∧α p∧δ

)−M (t ∧ τ

nk ∧α p∧δ )

∣∣∣∣2 in L1 (Ω) , (63.2.10)

the convergence being uniform on [0,T ] . The above formula shows that[Mα p∧δ

](t) is a

Ft∧δ∧α p measurable random variable which depends on t ∧ δ ∧α p.(Note that t ∧ δ is areal valued stopping time even if δ = ∞.) Therefore, by Proposition 63.2.2, there exists arandom variable, denoted as

[Mδ](t) which is the pointwise limit as p→∞ of these random

variables which is Ft∧δ measurable because, for a given ω, when α p becomes larger than t,the sum in 63.2.10 loses its dependence on p. Thus from pointwise convergence in 63.2.10,[

Mδ

](t)≡ lim

n→∞∑k≥0

∣∣∣∣M (t ∧δ ∧ τnk+1)−M (t ∧δ ∧ τ

nk)∣∣∣∣2

In case δ = ∞, the above gives an Ft measurable random variable denoted by [M] (t) suchthat

[M] (t)≡ limn→∞

∑k≥0

∣∣∣∣M (t ∧ τnk+1)−M (t ∧ τ

nk)∣∣∣∣2

Now stopping with the stopping time δ , this shows that[Mδ

](t)≡ lim

n→∞∑k≥0

∣∣∣∣M (t ∧δ ∧ τnk+1)−M (t ∧δ ∧ τ

nk)∣∣∣∣2 = [M]δ (t)