I have the following random time $\tau = \inf\{t > 0: W_t = 0\}$ where $(W_t)_{t\geq 0}$ is Brownian motion with almost surely continuous paths and $W_0 = 0$ a.s. I need to prove that $\tau$ is measurable (not necessarily a stopping time or an optional time). I first fix $\omega \in \Omega' \subset \Omega$. On $\Omega'$ $W$ is continuous and null at $0$. Then, I claim
$$\tau_0(\omega) > s \Leftrightarrow \lvert W_t(\omega) \rvert > 0 \quad \forall{t}, 0 < t \leq s$$ By continuity of $t \mapsto W_t(\omega)$, \begin{equation}\lvert W_t(\omega) \rvert > 0 \quad \forall{t}, 0 < t \leq s \Leftrightarrow \lvert W_{q_t}(\omega) \rvert > 0 \quad \forall{q_t} \in Q_s \qquad (\triangle)\end{equation} where $Q_s = \{qs \vert q \in (0,1] \cap Q\}$. Then $$\{\tau_0(\omega) > s\} = \underbrace{\cap_{q_t \in Q_s} \underbrace{\{\lvert W_{q_t} \rvert > 0\}}_{\in \mathcal{F}}}_{\in \mathcal{F}, \text{ by countable intersection}}$$
Now I don't think this is correct. In particular, $(\triangle)$ seems wrong. Take $s = 1$ for example. Then consider the function $t \mapsto -t/r + 1$ where $r \in [0,1]\setminus \mathbb{Q}$. Can someone help me fix this proof?
Recall the following elementary statement:
Applying this lemma, we find that
$$\{\tau \leq T\} = \{\omega; \exists t \in [0,T]: W_t(\omega)=0\} = \left\{\omega; \inf_{q \in \mathbb{Q} \cap [0,T]} |W_q(\omega)| = 0\right\}$$
Because $\mathbb{Q} \cap [0,T]$ is countable and each $W_q$ is measurable, we know that
$$\omega \mapsto \inf_{q \in \mathbb{Q} \cap [0,T]} |W_q(\omega)| $$
is measurable and this implies that $\{\tau \leq T\}$ is measurable for each $T$. Consequently, $\tau$ is measurable.
Remark: This reasoning works, more generally, if $\tau$ is of the form $$\tau := \inf\{t >0; W_t \in F\}$$ for some closed set $F$.