2490 CHAPTER 73. THE HARD ITO FORMULA, IMPLICIT CASE

qk−1

∑j=1

∫ t j+1

t j

⟨Y (s) ,X r

k (s)−X lk (s)−Pn

(Mr

k (s)−Mlk (s)

)⟩ds

=∫ t

t1

⟨Y (s) ,X r

k (s)−X lk (s)−Pn

(Mr

k (s)−Mlk (s)

)⟩ds

where Mrk denotes the step function

Mrk (t) =

mk−1

∑i=0

M (ti+1)X(ti,ti+1] (t)

and Mlk is defined similarly. The term∫ t

t1

⟨Y (s) ,Pn

(Mr

k (s)−Mlk (s)

)⟩ds

converges to 0 for a.e. ω as k→ ∞ thanks to continuity of t → M (t). However, more isneeded than this. Define the stopping time

τ p = inf{t > 0 : ∥M (t)∥W > p} .

Then τ p = ∞ for all p large enough, this for a.e. ω. Let

Ak =

[∣∣∣∣∫ t

t1

⟨Y (s) ,Pn

(Mr

k (s)−Mlk (s)

)⟩ds∣∣∣∣> ε

]

P(Ak) =∞

∑p=0

P(Ak ∩ ([τ p = ∞]\ [τ p−1 < ∞])) (73.7.32)

NowP(Ak ∩ ([τ p = ∞]\ [τ p−1 < ∞]))≤ P(Ak ∩ ([τ p = ∞]))

≤ P([∣∣∣∣∫ t

t1

⟨Y (s) ,Pn

((Mτ p)r

k (s)− (Mτ p)lk (s)

)⟩ds∣∣∣∣> ε

])This is so because if τ p = ∞, then it has no effect but also it could happen that the defin-ing inequality may hold even if τ p < ∞ hence the inequality. This is no larger than anexpression of the form

Cn

ε

∫Ω

∫ T

0∥Y (s)∥V ′

∥∥∥(Mτ p)rk (s)− (Mτ p)l

k (s)∥∥∥

WdsdP (73.7.33)

The inside integral converges to 0 by continuity of M. Also, thanks to the stopping time,the inside integral is dominated by an expression of the form∫ T

0∥Y (s)∥V ′ 2pds

and this is a function in L1 (Ω) by assumption on Y . It follows that the integral in 73.7.33converges to 0 as k→ ∞ by the dominated convergence theorem. Hence

limk→∞

P(Ak ∩ ([τ p = ∞])) = 0.