The optional stopping theorem (OST) gives that if $X$ is a martingale and $T$ is a bounded stopping time, then $\mathbb{E}[X_T]=\mathbb{E}[X_0]$.

Now take a Brownian motion $B=(B_t:t \geq 0)$, which is a martingale, and the stopping time $T=\inf\{t>0:B_t=1\}$. Obviously $B_T=1$ and OST says that $1=\mathbb{E}[B_T]=\mathbb{E}[B_0]=0$.

So what went wrong? Well quite simply, I did not check that $T$ was actually bounded. Why is it not bounded you say? No? Ok, well I’m going to tell you anyway; it is not bounded because there are many paths that trail below 0 which never hit 1. To properly show this, one must first calculate the supremum of the Brownian motion, then work out the probability that it is less than 1.

Rather nice reminder to always check that the stopping times are bounded before applying the OST. This was provided in a lecture by Nathanael Berestycki.