34.3. AN IMPORTANT FORMULA 1173

Consider ∫ 2T

−2T

(∫R||Φ(γn (t)+ s)−Φ(t + s)||pE ds

)dt

By the dominated convergence theorem, this converges to 0 as n→ ∞. Now Fubini. Thisyields ∫

R

∫ 2T

−2T||Φ(γn (t)+ s)−Φ(t + s)||pE dtds

Change the variables on the inside.∫R

∫ 2T+s

−2T+s||Φ(γn (t− s)+ s)−Φ(t)||pE dtds

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

∫ T

0||Φ(γn (t− s)+ s)−Φ(t)||pE dtds

+∫R

∫ 2T+s

−2T+sX

[0,T ]C ||Φ(γn (t− s)+ s)||pE dtds

=∫R

∫ T

0||Φ(γn (t− s)+ s)−Φ(t)||pE dtds

+∫R

∫ 2T+s

−2T+sX

[0,T ]C ||Φ(γn (t− s)+ s)−Φ(t)||pE dtds

Also by definition, γn (t− s)+ s is within 2−n of t and so the integrand in the integral onthe right equals 0 unless t ∈ [−2−n−T,T +2−n]⊆ [−2T,2T ]. Thus the above reduces to∫

R

∫ 2T

−2T||Φ(γn (t− s)+ s)−Φ(t)||pE dtds.

This converges to 0 as n→ ∞ as was shown above. Therefore,∫ T

0

∫ T

0||Φ(γn (t− s)+ s)−Φ(t)||pE dtds

also converges to 0 as n→ ∞. The only problem is that γn (t− s) + s ≥ t − 2−n and soγn (t− s)+ s could be less than 0 for t ∈ [0,2−n]. Since this is an interval whose measureconverges to 0 it follows∫ T

0

∫ T

0

∣∣∣∣Φ((γn (t− s)+ s)+)−Φ(t)

∣∣∣∣pE dtds

converges to 0 as n→ ∞. Let

mn (s) =∫ T

0

∣∣∣∣Φ((γn (t− s)+ s)+)−Φ(t)

∣∣∣∣pE dt