Why is $T:=\inf\{t\geq0:B_t\leq at^p-b\}$ a stopping time?

Question

How can I prove that $T:=\inf\{t\geq0:B_t\leq at^p-b\}$ is a stopping time w.r.t. a natural filtration of $B$, where $B$ is a $BM$, $p>1/2$ and $a,b>0$?

I can introduce a new random process, $X_t:=e^{B_t-at^p}$, for which $T=\inf\{t\geq 0:X_t\leq e^{-b}\}$.

I started: $$\{T\leq t\}= \{\exists s\leq t:X_s\leq e^{-b}\}.$$ Is this then equal to $$\cup_{s\leq t}\{X_s\leq e^{-b}\} = \cup_{s\leq t\cap \mathbb{Q}}\{X_s\leq e^{-b}\}?$$

Is it now enough for me to say, as $X_s$ is measurable on the given filtration, that a countable union is also measurable and that gives us the stopping time?

Answer 1

We can use this proposition.

Proposition. Let $X=(X_t)_{t\geq0}$ be an adapted process with values in a metric space $(E, d)$. Assume that the sample paths of $X$ are continuous, and $F$ be a closed subset of $E$. Then $$T_F=\inf\{t\geq0: X_t \in F\}$$ is a stopping time.

Clearly $X_t$ has continuous sample paths.  Since $(-\infty, e^{-b}]$ is closed, we can conclude that $T$ is a stopping time.

Answer 2

Your argument is not valid. LHS of the set theoretic identity you have written need not be contained in RHS. For example $X_t \leq e^{-b}$ does not imply that there exists $s\leq t,s \in \mathbb Q$ such that $X_t \leq e^{-b}$. Instead, consider $\{T>t\}$. You can write this as union over $s \leq t, s \in \mathbb Q$ of $\{X_s>e^{-b}\}$ and this proves that $\{T>t\} \in \sigma \{B_u:u \leq t\}$.