Let $e\in \mathbb{R}^+$ and $B_t$ 1-dimensional Brownian motion. Consider $$X_t=\int^{t}_{0}\frac{B_s}{e}1_{B_s\in(-e,e)}dB_s.$$ How to show that $X_t \to 0$ in $L^2$ as $e\to0$?
Obviously the result should follow from Ito isometry. However I am stuck on evaluating the expectation $E[X_t^2]$ due to the indicator function.