$\tau$ be a stopping time, $(M_t)_{t\ge 0}$ a martingale. Is $(M_{\tau\wedge t})_{t\ge 0}$ a martingale? 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.
 A: Assume $M$ is a supermartingale. First, for a fixed $n\ge 1$, let
$$
D_n = \left\{\frac{k}{2^n}, k= 0,1,2,\ldots\right\}\subset D_{n+1}\subset \cdots
$$
be the set of non-negative dyadic rationals of order no greater than $n$. It follows that
$$
M = (M_t,\mathcal{F}_t; t\in D_n)
$$
is a super martingale (discrete time).
Second, we construct a stopping time $T_n$ such that $T_n\ge T$ and $T_n$ only take values in $D_n$. Indeed, let
$$
T_n(\omega) = \inf\left\{t\in D_n; t\ge T(\omega)\right\}.
$$
Then $T_n\ge T_{n+1} \ge \cdots $ and $T_n$ is a stopping time. Fix $0\le s\le t$, we wish to show that
$$
\mathbb{E}\left[M_{t\wedge T}|\mathcal{F_s}\right]\le M_s
$$
almost surely. Similarly define
$$
t_n = \inf\left\{u\in D_n; u\ge t\right\} \ge t_{n+1}\ge\cdots
$$
and
$$
s_n = \inf\left\{u\in D_n; u\ge s\right\} \ge s_{n+1}\ge\cdots
$$
It follows from the discrete time optional sampling theorem that
$$
\mathbb{E}\left[M_{t_n\wedge T_n}|\mathcal{F}_{s_m}\right] \le M_{s_m \wedge T_n}
$$
for any integers $m\ge n$. Letting $m\to\infty$, we have $s_m \to s$ decreasing and $\mathcal{F}_{s_m}\to\mathcal{F}_s$. By Lévy's Downward theorem and the continuity of process $M$, we have
$$
\mathbb{E}\left[M_{t_n\wedge T_n}|\mathcal{F}_{s}\right] \le M_{s_m \wedge T_n}
$$
Observe that $(M_{t_n\wedge T_n},\mathcal{F}_{t_n\wedge T_n})$ is a backward martingale, whence it is uniformly integrable. Letting $n\to\infty$, we arrive at
$$
\mathbb{E}\left[M_{t\wedge T}|\mathcal{F}_{s}\right] \le M_{s\wedge T}.
$$
This completes the proof.
(from Prof. Hui Wang at Dam Brown University: Continuous time process and Brownian motion)
