2154 CHAPTER 63. THE QUADRATIC VARIATION OF A MARTINGALE

and by 63.7.28

limp→∞

np

∑k=1

f(xp

k−1

)X(xk−1,xk] (x) = f− (x)

for each x ∈ (a,b]. Therefore, since f is bounded, 63.7.29 follows from the dominatedconvergence theorem. The last claim follows the same way. This proves the lemma.

Definition 63.7.6 An increasing stochastic process, {A(t)} which is right continuous issaid to be natural if A(0) = 0 and whenever {ξ (t)} is a bounded right continuous martin-gale,

E (A(t)ξ (t)) = E(∫

(0,t]ξ− (s)dA(s)

). (63.7.30)

Hereξ− (s,ω)≡ lim

r→s−,r∈Dξ (r,ω)

a.e. where D is a countable dense subset of [0, t] . By Corollary 62.8.2 the right side of63.7.30 is not dependent on the choice of D since if ξ− is computed using two differentdense subsets, the two random variables are equal a.e.

Some discussion is in order for this definition. Pick ω ∈ Ω. Then since A is rightcontinuous, the function t → A(t,ω) is increasing and right continuous. Therefore, onecan do the Lebesgue Stieltjes integral defined in Definition 63.7.4 for each ω whenever f isBorel measurable and bounded. Now it is assumed {ξ (t)} is bounded and right continuous.By Lemma 63.7.5 ξ− (t)≡ limr→t−,r∈D ξ (r) is measurable and by this lemma,

∫(0,t]

ξ− (s)dA(s) = limp→∞

np

∑k=1

ξ(t pk−1

)(A(t pk

)−A

(t pk−1

))where

{t pk

}npk=1 is a sequence of partitions of [0, t] such that

limp→∞

max{∣∣t p

k − t pk−1

∣∣ : k = 1,2, · · · ,np}= 0. (63.7.31)

and D≡ ∪∞p=1∪

npk=1

{t pk

}npk=1 .

Also, if t → A(t,ω) is right continuous, hence Borel measurable, then for ξ (t) theabove bounded right continuous martingale, it follows it makes sense to write∫

(0,t]ξ (s)dA(s) .

Consider the right sum,np

∑k=1

ξ(t pk

)(A(t pk

)−A

(t pk−1

))This equals ∫

(0,t]

np

∑k=1

ξ(t pk

)X(t p

k−1,tpk ](s)dA(s)