62.3. FILTRATIONS 2065

Proposition 62.3.10 Let X be a stochastically continuous adapted process for a normalfiltration defined on a closed interval, I ≡ [0,T ]. Then X has a progressively measurableadapted version.

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 . (62.3.16)

Then let{

tn0 , t

n1 , · · · , tn

mn

}be a partition of [0,T ] in which

∣∣tni − tn

i−1

∣∣< ρn. Now define Xn asfollows:

Xn (t)(ω) ≡mn

∑i=1

X(tni−1)(ω)X[tn

i−1,tni )(t)

Xn (T ) ≡ X (T ) .

Then (s,ω)→ Xn (s,ω) for (s,ω) ∈ [0, t]×Ω is obviously B([0, t])×Ft measurable. Con-sider 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 and A∩ [0, t]×Ω is B([0, t])×Ft measurable foreach t ≤ T because each Xq in the above has the property that its restriction to [0, t]×Ω isB([0, t])×Ft measurable. Now define

Y (t,ω)≡{

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

I claim that for each t,Y (t,ω) = X (t,ω) for a.e. ω. To see this, consider 62.3.16. Fromthe construction of Xn, it follows that for each t,

P([||Xn (t)−X (t)|| ≥ 1

2n

])≤ 1

2n

Also, for a fixed t, if Xn (t,ω) fails to converge to X (t,ω) , then ω must be in infinitelymany of the sets,

Bn ≡[||Xn (t)−X (t)|| ≥ 1

2n

]which is a set of measure zero by the Borel Cantelli lemma. Recall why this is so.

P(∩∞k=1∪∞

n=k Bn)≤∞

∑n=k

P(Bn)<1

2k−1

Therefore, for each t,(t,ω)∈A for a.e. ω. Hence X (t) =Y (t) a.e. and so Y is a measurableversion of X . Y is adapted because the filtration is normal and hence Ft contains all sets ofmeasure zero. Therefore, Y (t) differs from X (t) on a set which is Ft measurable.

There is a more specialized situation in which the measurability of a stochastic processautomatically implies it is adapted. Furthermore, this can be defined easily in terms of a π

system of sets.