Hitting time of Brownian motion is finite

Question

I have the following task: let $B$ be a standard Brownian motion and let $a<0$, $b>0$. We define the following stopping time

$T=\inf\{t\ge 0;\,B_t=a\,\vee\,B_t=a\}$

I would like to show that $P(T<\infty)=1$. It seems to be quite intuitive. Any help?

Answer 1

By continuity of paths $P(T=\infty)\leq P(a<B_t<b \, \forall t)\leq P(a<B_t<b)=P(\frac a {\sqrt t} <X<\frac b {\sqrt t})$ for any $t>0$ where $X$ has standard normal distribution. Clearly $P(\frac a {\sqrt t} <X<\frac b {\sqrt t}) \to 0$ as $t \to \infty$.

[I have used the fact that $B_t$ has same distribution as $\sqrt t X$].

Answer 2

Just to add another approach if you can show that $$P(\limsup _{t\to \infty }B_{t}= \infty) = P(\liminf _{t\to \infty }B_{t}= -\infty) = 1 -(*)$$ Now by continuity of Brownian motion sample paths and $B_{0}=0$ you will have that if $B_{t}$ hits a level $b$ then it will always hit any level $a$,$ \forall a \leq b$. Then (*) guarantees that the hitting time will be finite a.s.