812 CHAPTER 30. CONTINUOUS STOCHASTIC PROCESSES

because if ω ∈ N both sides are 0 and if ω ∈ NC then the above limit in 30.10 holds andso if ||Y (t)(ω)−X (t)(ω)|| > ε, the same is true of ||X (d)(ω)−X (t)(ω)|| whenever dis close enough to t and so by Fatou’s lemma,

P([∥Y (t)−X (t)∥> ε]) =∫

X[∥Y (t)−X(t)∥>ε]∩NC (ω)dP

≤∫

lim infd→t

X[∥X(d)−X(t)∥>ε] (ω)dP

≤ lim infd→t

∫X[∥X(d)−X(t)∥>ε] (ω)dP

≤ lim infd→t

P([∥X (d)−X (t)∥α > ε

α])

≤ lim infd→t

ε−α

∫[∥X(d)−X(t)∥α>εα ]

∥X (d)−X (t)∥α dP

≤ lim infd→t

Cεα|d− t|1+β = 0.

Therefore,

P([∥Y (t)−X (t)∥> 0]) = P(∪∞

k=1

[∥Y (t)−X (t)∥> 1

k

])≤

∞

∑k=1

P([∥Y (t)−X (t)∥> 1

k

])= 0. ■

A few observations are interesting. In the proof, the following inequality was obtained.∥∥X(d′)(ω)−X (d)(ω)

∥∥ ≤ 2T γ (1−2−γ)

(T 2−(n+1)

)γ

≤ 2T γ (1−2−γ)

(∣∣d−d′∣∣)γ

which was so for any d′,d ∈ D with |d′−d| < T 2−(M(ω)+1). Thus the Holder continuousversion of X will satisfy

∥Y (t)(ω)−Y (s)(ω)∥ ≤ 2T γ (1−2−γ)

(|t− s|)γ

provided |t− s|< T 2−(M(ω)+1). Does this translate into an inequality of the form

∥Y (t)(ω)−Y (s)(ω)∥ ≤ 2T γ (1−2−γ)

(|t− s|)γ

for any pair of points t,s ∈ [0,T ]? It seems it does not for any γ < 1 although it does yield

∥Y (t)(ω)−Y (s)(ω)∥ ≤C (|t− s|)γ

where C depends on the number of intervals having length less than T 2−(M(ω)+1) whichit takes to cover [0,T ] . First note that if γ > 1, then the Holder continuity will imply t →