I got an example for a strict local martingale that I do not fully understand.
For the the one dimensional Brownian Motion $B_t$ and a stopping time $\tau=inf\{t \geq 0 |B_t=-1\}$ we know via Doob's optional sampling theorem that $B^{\tau}$ is a martingale. Now let
$M_t:=\begin{cases} 1+B_{t/(1-t)}^{\tau} &\text{ for t} \in [0,1) \\ 0 &\text{ for t} \in [1,\infty), \end{cases} $
with resprect to the generated filtration:
$ \mathcal{F}^`_t:= \begin{cases} \mathcal{F}_{t/(1-t)} &\text{ for t} \in [0,1) \\ \mathcal{F_{\infty}} &\text{ for t} \in [1,\infty), \end{cases}$
where $\mathcal{F_{\infty}}:=\sigma(\cup_{t\in\mathbb{R}^+} \mathcal{F}_t)$. Why is $M_t$ a strict local martingale? I guess the problem lies at $t=1$ and the fact that BM is not closable. But then i do not get the difference between $B_{t/(1-t)}^{\tau}$ at $t=1$ and $B_t^{\tau}$ (which is a martingale) at $t=\infty$.
I appreciate your time and thank you in advance!