2518 CHAPTER 74. A MORE ATTRACTIVE VERSION

where δ | fk|+Cδ |gkn|> | fkgnk| and limδ→0 Cδ =∞. Pick δ small enough that ε−2δa1/2 >ε/2. Then this is dominated by

≤ P(

2δ

∣∣∣ fk−a1/2∣∣∣> ε/2

)+P(2Cδ |gnk|> ε)

Fix n large enough that the second term is less than η for all k. Now taking k large enough,the above is less than η . It follows the expression in 74.6.26 and consequently in 74.6.25converges to 0 in probability.

Now consider the other term 74.6.24 using the n just determined. This term is of theform

qk−1

∑j=1

∫ t j+1

t j

⟨Y (s) ,X

(t j+1

)−X (t j)−Pn

(M(t j+1

)−M (t j)

)⟩ds =

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 < ∞])) (74.6.27)

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

])