73.6. CONVERGENCE 2481

and whether the above is also a local martingale. Maybe it is well to pause and considerthe integral and and what it means. Z ◦ J−1 maps JQ1/2U to W and so

(Z ◦ J−1

)∗ maps W ′

to(JQ1/2U

)′. Thus

(Z ◦ J−1)∗BX ∈

(JQ1/2U

)′, so

(Z ◦ J−1)∗BX ◦ J ∈ Q1/2 (U)′ = L2

(Q1/2U,R

)Thus it has the right values.

Does the stochastic integral just written even make sense? The integrand is HilbertSchmidt and has values in R so it seems like we ought to be able to define an integral. Theproblem is that the integrand is not in L2

([0,T ]×Ω;L2

(Q1/2U,R

)).

By assumption, t→ BX (t) is continuous into V ′ thanks to the integral equation solved,and also BX (t) = B(X (t)) for t /∈ Nω a set of measure zero. For such t, it follows fromLemma 69.4.1,

⟨BX (t) ,X (t)⟩= ∑i⟨BX (t) ,ei⟩2V ′,V a.e.ω

and so t → ∑i ⟨BX (t) ,ei⟩2 is lower semicontinuous and so it equals ⟨BX (t) ,X (t)⟩ fora.e. t, this for each ω /∈ N, a single set of measure zero. Also, t → ∑i ⟨BX (t) ,ei⟩2V ′,V isprogressively measurable and lower semicontinuous in t so by Proposition 62.7.6, one candefine a stopping time

τ p ≡ inf

{t : ∑

i⟨BX (t) ,ei⟩2V ′,V > p

},τ0 ≡ 0 (73.6.20)

Instead of referring to this Proposition, you could consider

τmp ≡ inf

{t :

m

∑i=1⟨BX (t) ,ei⟩2V ′,V > p

}

which is clearly a stopping time because t→ ∑mi=1 ⟨BX (t) ,ei⟩2V ′,V is a continuous process.

Then observe that τ p = supm τmp . Then

[τ p ≤ t] = ∪m[τ

mp ≤ t

]∈Ft .

Is it the case that τ p = ∞ for all p large enough? Yes, this follows from Lemma 73.4.2.

Lemma 73.6.1 Suppose τ p = ∞ for all p large enough off a set of measure zero, then

P(∫ T

0

∣∣∣(Z ◦ J−1)∗BX ◦ J∣∣∣2 dt < ∞

)= 1

Also∫ t

0(Z ◦ J−1

)∗BX ◦ JdW can be defined as a local martingale.

Proof: Let

A≡{

ω :∫ T

0

∣∣∣(Z ◦ J−1)∗BX ◦ J∣∣∣2 dt = ∞

}