73.6. CONVERGENCE 2485

Now ∑m P([τm = ∞]\ [τm−1 < ∞]) = 1 and so, one can apply the dominated convergencetheorem to conclude that

limk→∞

P(Ak) =∞

∑m=1

limk→∞

P(Ak ∩ ([τm = ∞]\ [τm−1 < ∞])) = 0

Lemma 73.6.4 Let X be as in Situation 73.2.1 and let X lk be as in Lemma 73.1.1 corre-

sponding to X above. Let X lk and X r

k both converge to X in K and also

BX lk ,BX r

k → BX in L2 ([0,T ]×Ω,W ′)

Say

X lk (t) =

mk

∑j=0

X (t j)X[t j ,t j+1)(t) , (73.6.22)

BX lk (t) =

mk

∑j=0

BX (t j)X[t j ,t j+1)(t) (73.6.23)

Then the sum in 73.6.23 is progressively measurable into W ′. As mentioned earlier, we cantake X (0)≡ 0 in the definition of the “left step function”.

Proof: This follows right away from the definition of progressively measurable.One can take a further subsequence such that uniform convergence of the stochastic

integral is obtained.

Lemma 73.6.5 Let X (s)−X lk (s)≡ ∆k (s) . Then the following limit occurs.

limk→∞

P

([sup

t∈[0,T ]

∣∣∣∣∫ t

0

(Z ◦ J−1)∗B∆k ◦ JdW

∣∣∣∣≥ ε

])= 0

The stochastic integral ∫ t

0

(Z ◦ J−1)∗BX ◦ JdW

makes sense because BX is W ′ progressively measurable and is in L2 ([0,T ]×Ω;W ′). Also,there exists a further subsequence, still denoted as k such that∫ t

0

(Z ◦ J−1)∗BX l

k ◦ JdW →∫ t

0

(Z ◦ J−1)∗BX ◦ JdW

uniformly on [0,T ] for a.e. ω .

Proof: This follows from Lemma 73.6.3. The last conclusion follows from the usualuse of the Borel Cantelli lemma. There exists a further subsequence, still denoted withsubscript k such that

P

([sup

t∈[0,T ]

∣∣∣∣∫ t

0

(Z ◦ J−1)∗B∆k ◦ JdW

∣∣∣∣≥ 1k

])< 2−k