# How to prove that the exit time of a Brownian motion is a stopping time?

Given the following setting:

Let $$\{W_t:t\geq 0\}$$ be a Brownian motion. for arbitrary $$a>0$$, define the exit time of the interval $$[-a,a]$$ as $$\tau=\inf\{t\geq 0:|W_t|>a\}$$

The question is to show that this is a stopping time, i.e. that for a filtration $$\mathscr{F}_t=\sigma(W_s,0\leq s\leq t)$$, we have that $$(\tau\leq t)\in\mathscr{F}_t$$.

This is an exercise that came after proving that a hitting time, i.e. $$\tau=\inf\{t\geq 0:W_t=a\}$$, is a stopping time. I know how to prove that this is a stopping time, but the method I used there cannot be applied here because we do not have that $$(\tau\leq t)\in\mathscr{F}_t$$, but instead we have that $$(\tau\leq t)\in\mathscr{F}_{t^+}$$.

I'm not even sure where to start with this exercise, but the hint given was that we can ''make'' $$\mathscr{F}_{t^+}=\mathscr{F}_{t}$$ by adding things of measure $$0$$ to the filtration. However, this hint does not bring me any closer to knowing where to start. What is meant by adding things of measure $$0$$ to the filtration? Why does this help? How do I proceed with solving this exercise?

Any help is appreciated.

By considering right-continuous $$(\mathcal{F}_t)$$ it suffices to show that $$\{\tau. Then since $$t\mapsto W_t$$ is continuous,
$$\{\taua\}.$$
The same applies to any open set $$A$$ and $$\tau:=\inf\{t\ge 0:W_t\in A\}$$.
The completed natural filtration of a Brownian motion, $$(\mathcal{F}_t\bigvee \mathcal{N})$$ ($$\mathcal{N}$$ are the $$\mathsf{P}$$-null sets of $$\mathcal{F}$$), is right-continuous (e.g. Theorem 8.2.2 on page 309 here).
• This shows only that $\tau$ is a.s. equal to an $(\mathscr F_t)$ stopping time. – John Dawkins Jan 26 '19 at 21:07