2162 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

Then since {X (t)} is a submartingale, it follows {A(t)} is also a submartingale.It remains to show several things. First, it is necessary to show A(t) is increasing in t

and A(0) = 0. To see this, let s < t,s, t ∈ ∪∞n=1∪

mnk=0 tn

k . Then letting n large enough both s, tare in ∪mn

k=0tnk . Only consider such n. Let t = tn

k(t),s = tnk(s) and let h ∈ L∞ (Ω) ,h≥ 0. Then∫

Ω

(A(t)−A(s))hdP =∫

Ω

(X (t)−M (t)− (X (s)−M (s)))hdP∫Ω

(X (t)−E (X (a)−A(a) |Ft)− (X (s)−E (X (a)−A(a) |Fs)))hdP. (63.7.35)

Now by Lemma 63.7.8, the following weak limit holds.

E (X (a)−A(a) |Ft) = limk→∞

E

 Mnk (a)︷ ︸︸ ︷X (a)−Ank (a)|Ft

= lim

k→∞Mnk (t)

A similar formula holds for s in place of t. Then the expression in 63.7.35 equals

= limk→∞

∫Ω

(X (t)−Mnk (t)− (X (s)−Mnk (s)))hdP

= limk→∞

∫Ω

(Ank (t)−Ank (s))hdP≥ 0

Since h ≥ 0 is arbitrary, this shows A(t)−A(s) ≥ 0 a.e. Not requiring h ≥ 0, the aboveargument also shows that for s, t ∈ ∪∞

n=1∪mnk=0 tn

k ,

A(t)−A(s) = weak limp→∞

Anp (t)−Anp (s) . (63.7.36)

Now consider the claim that A(0) = 0. Recall

A(0)≡ X (0)−E (X (a)−A(a) |F0) =−E (X (a)−A(a) |F0)

and so

A(0) = limk→∞−E (X (a)−Ank (a) |F0)

= limk→∞−E (Mnk (a) |F0) = lim

k→∞−Mnk (0) = 0.

This proves the theorem except for the claim that A(t) is natural. Let ξ (t) be a boundedright continuous martingale. I need to consider

E(∫

(0,t]ξ− (s)dA(s)

)and show it equals ξ (t)A(t). First consider the case t = a. By Lemma 63.7.5,

E(∫

(0,a]ξ− (s)dA(s)

)≡ E

(limk→∞

mnk

∑j=1

ξ

(tnk

j−1

)(A(

tnkj

)−A

(tnk

j−1

)))(63.7.37)