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.
Best Answer
By considering right-continuous $(\mathcal{F}_t)$ it suffices to show that $\{\tau<t\}\in\mathcal{F}_t$. Then since $t\mapsto W_t$ is continuous,
$$ \{\tau<t\}=\bigcup_{q<t,q\in\mathbb{Q}}\{|W_q|>a\}. $$
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).