Let $M_t$ be a $\mathcal F_t$ martingale. That is, for $t>s$ we have $$\mathbb{E}[M_t\mid\mathcal F_s]=M_s,\quad \text{a.s.}$$ Assume $\tau$ is a $\mathcal F_t$ stopping time. Is $(M_{\tau\wedge t})_{t\ge 0}$ a $\mathcal F_t$ martingale? In other words is it true that $$\mathbb{E}[M_{\tau\wedge t}\mid\mathcal F_s]=M_{\tau\wedge s},\quad \text{a.s.}$$ hold?
My thoughts: If the deterministic time $s$ can be considered as $\mathcal F_s$ stopping time, I guess this should follow from optional sampling theorem, but i'm not too sure about this, any help appreciated.