Let $$M_t=\int_0^t X_sdB_s,$$ being a martingale. Does $$\mathbb E[\int_0^t X_s^2ds]<\infty \ \ ?$$
I'm asking this question because, I know that if $$\int_0^t X_s^2ds<\infty \ \ a.s.$$ then $(M_t)$ is a local martingale, i.e. there is a sequence $(tau_n)$ of stopping time s.t. $\tau_n\nearrow \infty $ a.s. s.t. $(M_{\tau_n\wedge t})_t$ is a martingale for all $n$. Now, does the $(M_{\tau_n\wedge t})_t$ satisfy $$\mathbb E\left[\int_0^{\tau_n\wedge t} X_s^2ds\right]<\infty ,$$ and thus, we can use itô isometry on $(M_{\tau_n\wedge t})$.