Let $\mathcal{F}$ be the complete filtration generated by the Brownian motion $B$, and let $a<0<b$. Define the stopping times $\tau_a=\inf\{t\ge 0|B_t=a\}$ and $\tau_b=\inf\{t\ge 0|B_t=b\}$. Then is the indicator $1_{\tau_b<\tau_a}$ measurable with respect to $\mathcal{F}_{\tau_b}$?
The claim holds if one of $\tau_a$ and $\tau_b$ is replaced by a deterministic time. Does the claim hold with the hitting times? Intuitively, if we know the history of $B$ up to the time once it first hits $b$, when should know whether it already hits $a$ or not, right?
We define $A_n$ as following and we see that $$\begin{aligned}A_n:&=\{\tau_b +1/n> \tau_a\}\cap\{\tau_b\leq t\}=\\ &= \bigcup_{q \in \mathbb{Q}}\{\tau_b+1/n> q\}\cap\{\tau_a\leq q\}\cap \{\tau_b\leq t\}=\\ &= \bigcup_{q \in \mathbb{Q},q< t+1/n}\{\tau_b+1/n\leq q\}^c\cap\{\tau_a\leq q\}\cap \{\tau_b\leq t\}\in \mathscr{F}_{t+1/n}\\ \end{aligned}$$ and that $A_n\supseteq A_{n+1},\forall n$ and $\cap_nA_n=\{\tau_b\geq \tau_a\}\cap\{\tau_b\leq t\}$. Furthermore, we note $\cap_nA_n=\cap_{k\geq \ell}A_k\in \mathscr{F}_{t+1/\ell}$ for any $\ell \in \mathbb{N}$ since $\mathscr{F}_{t+1/k}\subseteq \mathscr{F}_{t+1/\ell},\forall k \geq \ell$. That is, $\cap_nA_n \in \mathscr{F}_{t+1/\ell},\,\forall \ell \in \mathbb{N}$. But then $\cap_nA_n \in \mathscr{F}_{t^+}=\cap_{n \in \mathbb{N}}\mathscr{F}_{t+1/n}$. This finally implies $$\{\tau_b\geq \tau_a\}\in \mathscr{F}_{\tau_b^+}:=\{A\in \mathscr{F}_\infty:A\cap \{\tau_b\leq t\}\in \mathscr{F}_{t^+}\}$$ However, if $(\mathscr{F}_t)_{t \geq 0}$ is assumed to be a right-continuous filtration (i.e. $\mathscr{F}_{t^+}=\mathscr{F}_t,\forall t\geq 0$) then $\{\tau_b\geq \tau_a\}\in \mathscr{F}_{\tau_b}$.