2050 CHAPTER 62. STOCHASTIC PROCESSES

Proof: If this is not so, there exists ε,δ > 0 and points of I,sn, tn such that even though

|tn− sn|<1n,

P([||X (sn)−X (tn)|| ≥ ε])> δ . (62.1.1)

Taking a subsequence, still denoted by sn and tn there exists t ∈ I such that the above holdand

limn→∞

sn = limn→∞

tn = t.

Then

P([||X (sn)−X (tn)|| ≥ ε])

≤ P([||X (sn)−X (t)|| ≥ ε/2])+P([||X (t)−X (tn)|| ≥ ε/2]) .

But the sum of the last two terms converges to 0 as n→ ∞ by stochastic continuity of X att, violating 62.1.1 for all n large enough. This proves the lemma.

For a stochastically continuous process defined on a closed and bounded interval, therealways exists a measurable version. This is significant because then you can do things withproduct measure and iterated integrals.

Proposition 62.1.2 Let X be a stochastically continuous process defined on a closed inter-val, I ≡ [a,b]. Then there exists a measurable version of X.

Proof: By Lemma 62.1.1 X is uniformly stochastically continuous and so there existsa sequence of positive numbers, {ρn} such that if |s− t|< ρn, then

P([||X (t)−X (s)|| ≥ 1

2n

])≤ 1

2n . (62.1.2)

Then let{

tn0 , t

n1 , · · · , tn

mn

}be a partition of [a,b] in which

∣∣tni − tn

i−1

∣∣< ρn. Now define Xn asfollows:

Xn (t) ≡mn

∑i=1

X(tni−1)X[tn

i−1,tni )(t)

Xn (b) ≡ X (b) .

Then Xn is obviously B(I)×F measurable because it is the sum of functions which are.Consider the set, A on which {Xn (t,ω)} is a Cauchy sequence. This set is of the form

A = ∩∞n=1∪∞

m=1∩p,q≥m

[∣∣∣∣Xp−Xq∣∣∣∣< 1

n

]and so it is a B(I)×F measurable set. Now define

Y (t,ω)≡{

limn→∞ Xn (t,ω) if (t,ω) ∈ A0 if (t,ω) /∈ A