62.2. KOLMOGOROV ČENTSOV CONTINUITY THEOREM 2051

I claim Y (t,ω) = X (t,ω) for a.e. ω. To see this, consider 62.1.2. From the construction ofXn, 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 .

Lemma 62.1.3 Let D be a dense subset of an interval, I = [0,T ] and suppose X : D→ Esatisfies ∣∣∣∣X (d)−X

(d′)∣∣∣∣≤C

∣∣d−d′∣∣γ

for all d′,d ∈ D. Then X extends uniquely to a continuous Y defined on [0,T ] such that∣∣∣∣Y (t)−Y(t ′)∣∣∣∣≤C

∣∣t− t ′∣∣γ .

Proof: Let t ∈ I and let dk → t where dk ∈ D. Then {X (dk)} is a Cauchy sequencebecause ||X (dk)−X (dm)|| ≤C |dk−dm|γ . Therefore, X (dk) converges. The thing it con-verges to will be called Y (t) . Note this is well defined, giving X (t) if t ∈D. Also, if dk→ tand d′k → t, then

∣∣∣∣X (dk)−X(d′k)∣∣∣∣ ≤ C

∣∣dk−d′k∣∣γ and so X (dk) and X

(d′k)

converge tothe same thing. Therefore, it makes sense to define Y (t)≡ limd→t X (d). It only remains toverify the estimate. But letting |d− t| and |d′− t ′| be small enough,∣∣∣∣Y (t)−Y

(t ′)∣∣∣∣ =

∣∣∣∣X (d)−X(d′)∣∣∣∣+ ε

≤ C∣∣d′−d

∣∣+ ε ≤C∣∣t− t ′

∣∣+2ε.

Since ε is arbitrary, this proves the existence part of the lemma. Uniqueness follows fromobserving that Y (t) must equal limd→t X (d). This proves the lemma.

62.2 Kolmogorov Čentsov Continuity Theorem

Lemma 62.2.1 Let rmj denote j

( T2m

)where j ∈ {0,1, · · · ,2m} . Also let Dm =

{rm

j

}2m

j=1and

D = ∪∞m=1Dm. Suppose X (t) satisfies∣∣∣∣∣∣X (rk

j+1

)−X

(rk

j

)∣∣∣∣∣∣≤ 2−γk (62.2.3)