I am trying to prove the following:
$$\lim_{n\rightarrow \infty} E\left(\int_{0}^{t \wedge \tau_{n}} e^{-\delta s} v(X_{s})X_{s} dW_{s}\right)=0$$
where $\tau_{n}$ is a sequence of stopping times with the property that $P(\tau_{n}\rightarrow \infty)=1$, $v(x)$ is a function with a jump discontinuity at one point and linear growth, and $X_{s}$ is a continuous process that satisfy $E(\int_{0}^{\infty}e^{-\delta s}X_{s}^2ds)<\infty$
I think that proving the following inequality is enough
$$E\left(\int_{0}^{T} \lbrace e^{-\delta s} v(X_{s})X_{s}\rbrace^2 ds\right)<\infty$$
How can I prove this? Is the fact that $v$ is bounded in any interval and $X_{s}$ is continuous enough?