2234 CHAPTER 65. STOCHASTIC INTEGRATION

Note that it would make no difference in terms of the conclusion of this lemma if youdefined

Φlk (t)≡

mk

∑j=1

Φ

(tk

j−1

)X(tk

j−1,tkj ](t)

because the modified function equals the one given above off a countable subset of [0,T ] ,the union of the mesh points. One could change Φr

k similarly with no change in the conclu-sion.

Proof: For t ∈ R let γn (t) ≡ k/2n,δ n (t) ≡ (k+1)/2n, where t ∈ (k/2n,(k+1)/2n],

and 2−n < T/4. Also suppose Φ is defined to equal 0 on [0,T ]C×Ω. There exists a set ofmeasure zero N such that for ω /∈ N, t → ∥Φ(t,ω)∥ is in Lp (R). Therefore by continuityof translation, as n→ ∞ it follows that for ω /∈ N, and t ∈ [0,T ] ,∫

R||Φ(γn (t)+ s)−Φ(t + s)||pE ds→ 0

The above is dominated by∫R

2p−1 (||Φ(s)||p + ||Φ(s)||p)X[−2T,2T ] (s)ds

=∫ 2T

−2T2p−1 (||Φ(s)||p + ||Φ(s)||p)ds < ∞

Consider ∫Ω

∫ 2T

−2T

(∫R||Φ(γn (t)+ s)−Φ(t + s)||pE ds

)dtdP

By the dominated convergence theorem, this converges to 0 as n→ ∞. This is because theintegrand with respect to ω is dominated by∫ 2T

−2T

(∫R

2p−1 (||Φ(s)||p + ||Φ(s)||p)X[−2T,2T ] (s)ds)

dt

and this is in L1 (Ω) by assumption that Φ ∈ K. Now Fubini. This yields∫Ω

∫R

∫ 2T

−2T||Φ(γn (t)+ s)−Φ(t + s)||pE dtdsdP

Change the variables on the inside.∫Ω

∫R

∫ 2T+s

−2T+s||Φ(γn (t− s)+ s)−Φ(t)||pE dtdsdP

Now by definition, Φ(t) vanishes if t /∈ [0,T ] , thus the above reduces to∫Ω

∫R

∫ T

0||Φ(γn (t− s)+ s)−Φ(t)||pE dtdsdP

+∫

Ω

∫R

∫ 2T+s

−2T+sX

[0,T ]C ||Φ(γn (t− s)+ s)||pE dtdsdP