Brownian motion does not diverge in finite time Let $B_t$ be a $1$-dimensional standard Brownian motion and the stopping time $\tau_a$ is $\tau_a:=\inf\{t>0:|B_t|\geq a\}$ for $a>0$. Then,$\lim_{a\to\infty}\tau_a=\infty$ almost surely. 
Why is this? I think this is definitely fundamental, but I can not prove. 
Can someone please give me some clues? Thanks!
 A: By the very definition of the stopping times, $\tau_a(\omega)$ is increasing in $a$ for each fixed $\omega$. This means that
$$\lim_{a \to \infty} \tau_a(\omega)=\sup_{a>0} \tau_a(\omega)$$
exists for all $\omega \in \Omega$. If the supremum were finite for some $\omega \in \Omega$, say, $\sup_{a>0} \tau_a(\omega)=T<\infty$, then $$\sup_{t \in [0,T]} |B_t(\omega)| = \infty.$$ This means that $t \mapsto B_t(\omega)$ cannot be continuous (otherwise it would be bounded on the compact set $[0,T]$). Since the sample paths of Brownian motion are continuous (with probability $1$), this means that $\sup_{a>0} \tau_a(\omega)<\infty$ can only happen on a null set. Hence, $\sup_{a>0} \tau_a(\omega)=\infty$ almost surely.
Finally, to prove that $\tau_a(\omega)<\infty$ with probability $1$, we can use a martingale argument. Since $(B_t^2-t)_{t \geq 0}$ is a martingale, it follows from the optional stopping theorem that
$$\mathbb{E}(B_{t \wedge \tau_a}^2-(t \wedge \tau_a))=0.$$
Using $|B_{t \wedge \tau_a}| \leq a$ we get
$$\mathbb{E}(\tau_a \wedge t) \leq a^2.$$
By monotone convergence,
$$\mathbb{E}(\tau_a) \leq a^2 < \infty,$$
and in particular $\tau_a<\infty$ almost surely.
