While I was reading the proof for uniqueness of the solution of SDE it a doubt came to my mind, namely why are we "allowed" to use Ito isometry (under our assumptions).
The theorem I am referring to could be found in "Introduction to stochastic integration" by Kuo.
Assumptions: $\sigma, f$ are measurable functions on $[a,b]\times \mathbb R$, and satisfy Lipschitz condition.
The starting point $\xi$ is an $\mathcal F_a$ measurable r.v. with $E(\xi^2)<\infty$.
He starts by assuming that $X_t$ and $Y_t$ are two solutions of an Stochastic Integral Equation. Put $Z_t=X_t-Y_t$.
Then $$Z_t=\int_a^t \big(\sigma(s,X_s)-\sigma(s,Y_s) \big)dB(s)+\int_a^t \big(f(s,X_s)-f(s,Y_s)\big)ds$$
Then he uses Cauchy-Schwarz inequality ($(a+b)^2\leq 2(a^2+b^2)$), then takes expectations on both sides.
Taking the first term on the right we have
$$\mathbb E\bigg[\bigg(\int_a^t \big(\sigma(s,X_s)-\sigma(s,Y_s) \big)dB(s)\bigg)^2\bigg]$$
At this point he applies Ito's isometry; my doubt is, why are we allowed to apply it? In order for $X_t$ to be a solution we need $\sigma(\cdot)$ to be adapted and $\int_a^b \sigma(\cdot)^2 ds<\infty $ almost surely, but this does not assure (at least, I am not able to see it) that the difference $\sigma(s,X_s)-\sigma(s,Y_s)$ is square integrable with respect to $P\otimes \lambda$.
Do you mind explaining me how this works?
Thanks!
EDIT:
In Karatzas book the author actually defines two stopping times
$$\tau_n^{(X)}:=\inf\{t\geq 0: X_t\geq n\}$$ $$\tau_n^{(Y)}:=\inf\{t\geq 0: X_t\geq n\}$$
Then if we work with the stopping times it's clear that
$$\mathbb E\bigg[\bigg(\int_a^{t\wedge \tau_n^{(Y)}\wedge \tau_n^{(X)}} \big(\sigma(s,X_s)-\sigma(s,Y_s) \big)dB(s)\bigg)^2\bigg]=\mathbb E\bigg(\int_a^{t\wedge \tau_n^{(Y)}\wedge \tau_n^{(X)}} \big(\sigma(s,X_s)-\sigma(s,Y_s) \big)^2ds\bigg)$$
because the right hand side is finite.