72.6. THE ITO FORMULA 2457

≤ 2∣∣∣∣∫ t(k)

0R((

Z (s)◦ J−1)∗(X (s)−X lm (s)

))◦ JdW (s)

∣∣∣∣+2∣∣∣∣∫ t(m)

0R((

Z (s)◦ J−1)∗(X (s)−X lm (s)

))◦ JdW (s)

∣∣∣∣≤ 4 sup

t∈[0,T ]

∣∣∣∣∫ t

0R((

Z (s)◦ J−1)∗(X (s)−X lm (s)

))◦ JdW (s)

∣∣∣∣In Lemma 72.5.1 the above expression was shown to converge to 0 in probability. There-fore, by the usual appeal to the Borel Canteli lemma, there is a subsequence still referredto as {m} , such that it converges to 0 pointwise in ω for all ω off some set of measure 0as m→ ∞. It follows there is a set of measure 0 such that for ω not in that set, 72.6.20converges to 0 in R. Note that t > 0 is arbitrary. Similar reasoning shows the first term inthe non stochastic integral of 72.6.19 is dominated by an expression of the form

4∫ T

0

∣∣∣⟨Y (s) , X̄ (s)−X lm (s)

⟩∣∣∣ds

which clearly converges to 0 for ω not in some set of measure zero because X lm converges

in K to X̄ . Finally, it is obvious that

limm→∞

∫ t(k)

t(m)||Z (s)||2 ds = 0 for a.e. ω

due to the assumptions on Z.This shows that for ω off a set of measure 0

limm,k→∞

|X (t (k))−X (t (m))|2 = 0

and so {X (t (k))}∞

k=1 is a convergent sequence in H. Does it converge to X (t)? Let ξ (t) ∈H be what it converges to. Let v ∈V then

(ξ (t) ,v) = limk→∞

(X (t (k)) ,v) = limk→∞

⟨X (t (k)) ,v⟩= ⟨X (t) ,v⟩= (X (t) ,v)

and now, since V is dense in H, this implies ξ (t) = X (t).Now for every t ∈ D,

|X (t)|2 = |X0|2 +∫ t

0

(2⟨Y (s) , X̄ (s)⟩+ ||Z (s)||2

L2(Q1/2U,H)

)ds

+2∫ t

0R((

Z (s)◦ J−1)∗ X̄ (s))◦ JdW (s)

and so, using what was just shown along with the obvious continuity of the functions of ton the right of the equal sign, it follows the above holds for all t ∈ [0,T ] off a set of measurezero.

It only remains to verify t→ X (t) is continuous with values in H. However, the aboveshows t→|X (t)|2 is continuous and it was shown in Lemma 72.4.1 that t→X (t) is weakly