Prove $\tau=\inf\{t\in[0,T]:M_t=0\}\wedge T$ a stopping time for a continuous martingale $(M_t)_{t \geq 0}$

Question

I have a question about a positive continuous martingale. Let $(M_t)_{t\in[0,T]}$ be a continuous martingale such that $P(M_t>0)=1$ for all $t\in[0,T]$. Set $\tau=\inf\{t\in[0,T]:M_t=0\}\wedge T$. I want to shaw what $\tau$ is a stopping time. It is clearly bounded, so if $\tau$ is indeed a stopping time, I can apply the optional sampling theorem.

Answer 1

Hints:

1. We are done if we can show that $$\{\tau \leq t\} = \{\omega; \inf_{r \in \mathbb{Q} \cap [0,t]} |M_r(\omega)|=0\} =: \Omega_t$$ for any $t \leq T$.
2. $\{\tau \leq t\} \subseteq \Omega_t$: Fix $\omega \in \{\tau \leq t\}$. Then, there exists $s \in [0,t]$ such that $M_s(\omega)=0$. Use the continuity of $s \mapsto M_s(\omega)$ to deduce that there exists a sequence $(s_j)_{j \in \mathbb{N}} \subseteq [0,t] \cap \mathbb{Q}$ such that $s_j \to s$ and $M_{s_j}(\omega) \to M_s(\omega)=0$.
3. $\Omega_t \subseteq \{\tau \leq t\}$: This is equivalent to $$\Omega_t^c \supseteq \{\tau \leq t\}^c = \{\tau>t\}.$$ So let $\omega \in \Omega_t^c$, that is $$\inf_{r \in \mathbb{Q} \cap [0,t]} |M_r(\omega)|>0.$$ Decude from the continuity of the sample paths that $$\inf_{r \in [0,t]} |M_r(\omega)|>0.$$ Conclude that $\tau(\omega)>t$.

Remark: The proof boils down to the following, very fundamental, result:

Let $f:[0,t] \to \mathbb{R}$ be a continuous function for some fixed $t>0$. Then the following statements are equivalent.

1. There exists $s \in [0,t]$ such that $f(s)=0$.
2. $\inf_{r \in [0,t] \cap \mathbb{Q}} |f(r)|=0$.

Exactly the same argumentation shows that $$\tau_F := \inf\{t \in [0,T]; M_t \in F\}$$ is a stopping time for any closed set $F$.