2270 CHAPTER 65. STOCHASTIC INTEGRATION
This is then no larger than 4η provided n is large enough. Pick such an n. Then for allk > K, this has shown that
P
([sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗∆k (s)
)◦ JdW (s)
∣∣∣∣≥ ε
])
≤ P
([sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗Pn∆k (s))◦ JdW (s)
∣∣∣∣≥ ε/2
])+
P
([sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗ (I−Pn)∆k (s))◦ JdW (s)
∣∣∣∣≥ ε/2
])
≤ P
([sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗Pn∆k (s))◦ JdW (s)
∣∣∣∣≥ ε/2
])+4η
By 65.14.24 this whole thing is less than 5η provided k is large enough. This has provedthat under the assumption that X is bounded uniformly off a set of measure zero,
limk→∞
P
([sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗∆k (s)
)◦ JdW (s)
∣∣∣∣≥ ε
])= 0
This is what was desired to show. It remains to remove the extra assumption that X isbounded.
Now to finish the argument, define the stopping time
τm ≡ inf{t > 0 : |X (t)|H > m} .
As observed in Lemma 65.14.2, this is a valid stopping time. Also define ∆τmk ≡ Xτm −(
X lk
)τm . Using this stopping time on X and X lk does not affect the pointwise convergence to
0 as k→ ∞ of ∆τmk on which the above argument depends.
Consider
Akε ≡
[sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗∆k (s)
)◦ JdW (s)
∣∣∣∣≥ ε
]
Then
P(Akε ∩ [τm = ∞])≤ P
([sup
t∈[0,T ]
∣∣∣∣∫ t
0R((
Z (s)◦ J−1)∗∆
τmk (s)
)◦ JdW (s)
∣∣∣∣≥ ε
])
which converges to 0 as k→ ∞ by the first part of the argument. This is because |Xτm | and∣∣∣(X lk
)τm∣∣∣ are both bounded by m and the same pointwise convergence condition still holds.
NowAkε = ∪∞
m=1Akε ∩ ([τm = ∞]\ [τm−1 < ∞])