Show that this is a stopping time

Question

Show that $\sigma=\inf \{ t\ge 0 : |B_t|= \log t \}$ is a stopping time with respect to $(\mathcal F_t^B)_{t\ge0}$.

I've been trying to put the set $\{\sigma\le t\}$ equal to a countable union and then showing that this union belongs in $\mathcal F_t^B$. I am struggling to derive a countable union to this as I am only familiar with when $B_t$ is equal to a number and not a function of $t$.

definitions
$(\Omega,\mathcal F, \mathbb P)$ is a probability space. $(\cal F_t)_{t \ge 0}$ is a stochastic basis: that is, for each real $t \ge 0$, $\mathcal F_t$ is a sigma-algebra contained in $\mathcal F$, and if $s \le t$ then $\mathcal F_s \subseteq \mathcal F_t$. Perhaps we even assume the stochastic basis is right-continuous, namely $$ \mathcal F_t = \bigcap_{s>t}\mathcal F_s $$ for all $t$. $(B_t)$ is a stochastic process adapted to $(\mathcal F_t)$: that is, for each $t$, $B_t : \Omega \to \mathbb R$ is a random variable, measurable with respect to $\mathcal F_t$. Define random variable $\sigma$ as above, $\sigma : \Omega \to [0,+\infty]$. We want to show $\sigma$ is a stopping time for $(\mathcal F_t)$: that is, for every real $t$, $$ \{\sigma \le t\} := \{\omega \in \Omega : \sigma(\omega) \le t\} $$ belongs to $\mathcal F_t$.

Answer 1

Hint: $$[\sigma\leqslant t]=\bigcup_{r\in\mathbb Q,r\leqslant t}[|B_r|=\log r]$$ Of course, to see why this identity holds is part of the fun...

Answer 2

Recall that a random variable $\tau : \Omega \to [0, \infty]$ is a stopping time with respect to the filtration $(\mathscr{F}_t)_{t \geq0 }$ if $\{ \tau \leq t \} \in \mathscr{F}_t$ for all $t \geq 0$. Therefore, to see that $\tau = \inf \{ t \geq 0 : \left| B_t \right| = \log t \}$ is a stopping time, consider that \begin{eqnarray*} \{ \tau \leq t \} &=& \bigcap_{0 \leq s \leq t} \{ \omega : \left| B_s(\omega) \right| = \log s \} \\ &=& \bigcup_{s \in [0,t] \cap \mathbb{Q}} \{ \omega : \left| B_s(\omega) \right| = \log s \}, \end{eqnarray*} where the last equality follows from the right-continuity of the Brownian motion and the density of $\mathbb{Q}$ in $\mathbb{R}$. Now the sets $\{ \omega : \left| B_s(\omega) \right| = \log s \}$ are all $\mathscr{F}_s$-measurable, for $s \leq t$. Moreover, since the filtration is increasing, these sets are all $\mathscr{F}_t$-measurable. Since we have a countable union of $\mathscr{F}_t$-measurable sets, it follows that $\{ \tau \leq t \} \in \mathscr{F}_t$ and $\tau$ is a stopping time.