Measurability of two hitting times at the stopped $\sigma$-algebra

Question

Let $\mathcal{F}=(\mathcal{F}_t)_{t\ge 0}$ be the complete filtration generated by the Brownian motion $B $, and let $a<0<b$. Define the stopping times $\tau_a=\inf\{t\ge 0\mid B_t=a\}$ and $\tau_b=\inf\{t\ge 0\mid B_t=b\}$.

Let $1_A$ be the indicator function of the set $A$. Then is the random variable
$1_{\tau_b<\tau_a}$ 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?

Answer 1

Yes. By definition, we must show that for all $t \geq 0$, the event $A_t := \{\tau_b < \tau_a\} \cap \{\tau_b \leq t\}$ is $\mathcal F_t$ measurable. To do so, it suffices to show that $\mathbf 1_{A_t}$ can be written as a deterministic function of the path $\{B_s\}_{s \in [0, t]}$ up to time $t$. But we can write

$$ \mathbf 1_{A_t} = F(\{B_s\}_{s \in [0, t]}) := \begin{cases} 0, \text{ if } B_s \neq b \text{ for all } s \in [0, t] \text { or } B_s = b \text{ for some } s \in [0, t] \text{ with } M_s > a, \\ 1, \text { otherwise}. \end{cases} $$

where $M_s := \min_{t \in [0, s]} B_t.$