TITLE: Supermartingale vanishing at some stopping time
QUESTION [6 upvotes]: Let $\left\{X_t\right\}_{t\in[0, T]}$ be a continuous and non-negative supermartingale. We define the stopping time 
$$\tau_0:=\inf\{t\in[0,T]:X(t)=0\}\wedge T$$
and immediately obtain by continuity of $X$ that $\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)=0$ a.s.
I want to prove that $X(t)=0$ for almost every $\omega\in\{\tau_0<T\}$ where $t\in[\tau_0(\omega),T]$.
What I established so far (using optional stopping) is the following:
$$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)\mid F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$
Hence, as $X$ is non-negative, the desired result follows for the terminal time $t=T$. Can someone please help me to prove the whole result for all the other relevant times? Thanks a lot in advance!

REPLY [2 votes]: Let $\epsilon > 0$ and consider the stopping time
$$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$
Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$.  Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write
$$E[X_{\tau_\epsilon} ; \tau_0 < T] + E[X_{\tau_\epsilon} ; \tau_0 =T] \le E[X_{\tau_0} ; \tau_0 < T] + E[X_{\tau_0} ; \tau_0 =T].$$
Now $X_{\tau_0} = 0$ on the event $\{\tau_0 < T\}$.  On the event $\{\tau_0 = T\}$ we have $X_{\tau_0} = X_T$, and moreover, since $\tau_\epsilon \ge \tau_0$, we also have $\tau_\epsilon = T$ and $X_{\tau_\epsilon} = X_T$.  So our inequality becomes
$$E[X_{\tau_\epsilon} ; \tau_0 < T] + E[X_{T} ; \tau_0 =T] \le  0 + E[X_{T} ; \tau_0 =T]$$
showing that $E[X_{\tau_\epsilon} ; \tau_0 < T] \le 0$.  Since $X_{\tau_\epsilon}$ is nonnegative we have $X_{\tau_\epsilon} = 0$ almost everywhere on $\{\tau_0 < T\}$.  On the other hand, $\{\tau_\epsilon < T\} \subset \{\tau_0 < T\}$, and by continuity $X_{\tau_\epsilon} = \epsilon$ almost everywhere on $\{\tau_\epsilon < T\}$.  So we conclude $P(\tau_\epsilon < T) = 0$, i.e. almost surely on $\{\tau_0 < T\}$, there does not exist $t$ in $[\tau_0, T)$ with $X_t \ge \epsilon$.  Letting $\epsilon \downarrow 0$ along a sequence, we get that almost surely on $\{\tau_0 < T\}$, there does not exist $t \in [t_0, T)$ with $X_t > 0$.  This is the desired conclusion.