Optional Stopping Theorem (continuous case) where stopping time is unbounded

Question

I am referring to the version of OST that states that: when M is a continuous Martingale and $\rho \leq \tau $ are bounded stopping times then $M_\infty := lim_{t \rightarrow \infty} M_t $ exists almost surely and $ M_{\infty} \in L^1 $.

Now I am looking for an example when we have an unbounded stopping time and the result of the theorem fails. I've tried the exponential martingale $ M_t := exp(B_t -\frac{t}{2})$ where $(B_t)_{t \geq 0}$ is a standard Brownian motion then considered the stopping time $ \tau := inf ${t $\geq$ 0 : $M_t = 2 $} which is unbounded but I couldn't see a contradiction to the theorem. Am I missing something or is there another example (also I would be interested to see other examples even if this does work out)

Answer 1

We have $$ M_t\xrightarrow[t\to+\infty]{\mathbb{P}\rm -a.s.} 0 $$ which makes Doob's theorem to work for the stopped martingale $\left(M_t^\tau\right)_{t\ge0}$.