Let $B_t$ be a Brownian motion. Let $\tau$ be a stopping time with $\mathbb E(\tau)< \infty $
Show that $\mathbb E(B_\tau)=0$
I know that since $B_t$ is a martingale we have $\mathbb E(B_{\tau \wedge n})=B_0=0$ but I can't see how I could use any covergence theorem to take the limit into the expactation. Is this the right approach at all? Any ideas?
Since $(B_t^2-t)_{t \geq 0}$ is a martingale, it follows from the optional stopping theorem that
$$\mathbb{E}(B_{\tau \wedge n}^2) = \mathbb{E}(\tau \wedge n).$$
This implies
$$\mathbb{E}((B_{\tau \wedge n}-B_{\tau \wedge m})^2) = \mathbb{E}(\tau \wedge n)-\mathbb{E}(\tau \wedge m) \xrightarrow[]{m,n \to \infty} 0.$$
This shows that $(B_{\tau \wedge n})_{n \geq 1}$ is an $L^2$-Cauchy sequence and so $B_{\tau \wedge n} \to B_{\tau}$ in $L^2$. Hence, in particular, $B_{\tau \wedge n} \to B_{\tau}$ in $L^1$ and so $$\mathbb{E}(B_{\tau \wedge n}) \xrightarrow[]{n \to \infty} \mathbb{E}(B_{\tau}).$$