# Stopping time and supremum of Brownian motion

We consider on a filtered probability space $$(\Omega,\mathcal{F},\mathcal{F}_{r \in \mathbb{R}_+},P)$$ with the usual conditions. Let $$B$$ be a $$(\mathcal{F}_r)_{r \geq 0}$$-Brownian motion. Let $$\theta:=\inf\{r \geq 0,r\sup_{u \in [0,r]}|B_u|>1\}.$$

Can we claim that $$\theta$$ is a stopping time?

Attempt: it's sufficient consider for $$v >0,\{\theta \geq v\} \in \mathcal{F}_v.$$

We have $$\{\theta \geq v\}=\bigcap_{r \in [0,v[}\{r \sup_{u \in [0,r]}|B_u| \leq 1\}$$ where we obtained an uncountable intersection.

What do you think?

Since $$r \sup_{u \in [0,r]} |B_u|$$ is a continuous function of $$r$$, $$\bigcap_{r \in [0,v]} \{r \sup_{u \in [0,r]} |B_u| \le 1\} = \bigcap_{r \in [0,v]\cap \mathbb{Q}} \{r \sup_{u \in [0,r]} |B_u| \le 1\}$$ is a countable intersection of $$\mathcal F_v$$-measurable events and is hence $$\mathcal F_v$$-measurable. Therefore, $$\theta$$ is a stopping time.
Given a continuous function $$f:[0,\infty) \rightarrow \mathbb{R}$$, the function $$g(t) := \sup_{s \in [0,t]} f(s)$$ is continuous. This can be seen by considering that, for $$g$$ to be discontinuous at a point $$t$$, $$f$$ must also be discontinuous at that point. For a formal proof, we'll show left-continuity; the proof of right-continuity is similar. Let $$(t_n) \uparrow t$$. Then $$\lim g(t_n) = \sup_{s \in [0,t)} f(s)$$, but by continuity of $$f$$, $$\sup_{s \in [0,t)} f(s) = \sup_{s \in [0,t]} f(s) = g(t)$$.
• How do you know that $\sup_{u\in[0,r]}|B_u|$ is continuous? Commented Feb 1 at 21:13
• By continuity of Brownian motion we can say that $r\to \sup_{u\in [0,r]}|B_u|$ is measurable, can you provide a proof for the continuity of the supremum? Commented Feb 1 at 21:20
• @mathex I added a proof that the running supremum of any continuous function is left-continuous. You can apply this to $|B_t|$ to get the result. Commented Feb 1 at 21:35
• You forget $\leq 1$ Commented Feb 1 at 21:41