I am trying to show that the last exit time of Brownian motion is a random variable, i.e. for $\tau$ defined as
$$\tau = \sup\{t > 0 : W_t = 0\}$$ it holds that $\{\tau < t\} \in \mathcal{F}$ for every $t > 0$. Here $W$ is BM on the filtered probability space $(\Omega,\mathcal{F}((\mathcal{F}_t))_t,P)$.
Here is my attempt: Fix $\omega \in \Omega'$. On $\Omega'$ BM is continuous and null at $0$. $\Omega'$ is measurable. $$\tau(\omega) < t \Leftrightarrow \lvert W_s(\omega) \rvert > 0 \quad \forall{s} \geq t$$
$$\lvert W_s(\omega) \rvert > 0 \quad \forall{s} \geq t \Leftrightarrow \forall{n} \in \mathbb{Z}_{>t}, \lvert W_s(\omega) \rvert > 0 \quad \forall{s}, t\leq s \leq n$$
Now fix $n$. By continuity of BM, $$\lvert W_s(\omega) \rvert > 0 \quad \forall{s}, t\leq s \leq n \Leftrightarrow \lvert W_{q^n_s}(\omega) \rvert > 0 \quad \forall{q^n_s} \in Q_n$$ where $Q_n = \{q(n-t)+t \vert q \in [0,1]\cap Q\}$
Something is wrong with the last step. BM can just bounce back at an irrational time point but I cannot fix it without using limits. I had a similar problem with the second hitting time at $0$ (Measurability of the zero-crossing time of Brownian motion). Is there a clever solution to this?
If I am not missing a point, hopefully this solves the problem in the last step.
$$\lvert W_s(\omega) \rvert > 0 \quad \forall{s}, t\leq s \leq n \Leftrightarrow \exists{q\in\mathbb{Q}}, q> 0, \lvert W_{q^n_s}(\omega) \rvert > q \quad \forall{q^n_s} \in Q_n$$ where $Q_n = \{q(n-t)+t \vert q \in [0,1]\cap Q\}$