Escaping from a strip for a Brownian Motion Let $T_x=\inf\{t>0:B(t)=x\}$, i.e $T_x$ is the first hitting time the Brownian motion $B(t)$ hits the point $x$.

With $B(0)=0$, the first time a standard Brownian motion escapes from
  strip $[a,b]$ is $T_{ab}=\min\{T_a,T_b\}$.

However I don't understand this definition. This is something necessary for the Brownian motion to escape, not a sufficient one, since $B(t)$
 could hit $a$ or $b$ at some time, and then still remain in the strip...
Any help would be appreciated.
Edit: Based on Did's comment,
We want $P(\tau= 0) = 1$, where $\tau=\inf\{t>0:B(t)>0\}$ and his hint is to use 
$T_x=\inf \{t>0:B(t)=x\}$. 
(...) 
Knowing that $P(\tau = 0) = 1$ for a Standard Brownian Motion (SBM),and without loss of generality assuming that $T_a < T_b$, consider the process $B^*(s)=\{B(T_a+s)-B(T_a): s \geq 0\}$, which is also a SBM . 
So $P(\tau=0)=1$ which is equivalent to $P(\inf\{t>0:B(T_a+t)>B(T_a)\}=0)=1$, i.e. exit time of $[b,a]$ which, with our assumptions, is $\inf\{ t>0: B(t)>a \}$ is a.s. $T_a$, which is also the exit time of $(b,a)$.
 A: Since Brownian motion has continuous sample paths, we know that the hitting time of $\{a,b\}$ equals the first exit time from the open interval $(a,b)$. It remains to show that the first exit time from $(a,b)$ equals almost surely the first exit time from $[a,b]$, and for this it is enough to prove that the stopping time
$$\tau := \inf\{t>0; B_t >0\}$$
satisfies
$$\mathbb{P}(\tau=0)=1. \tag{1}$$
To this end, note that
$$\mathbb{P}(\tau>0) = \lim_{n \to \infty} \mathbb{P}(\tau>1/n).$$
By the very definition of $\tau$, we have $B_t(\omega) \leq 0$ for all $t \leq \tau(\omega)$ and so
$$\mathbb{P}(\tau>1/n) \leq \mathbb{P} \left( \sup_{t \leq 1/n} B_t = 0 \right).$$
The reflection principle implies that
$$\sup_{t \leq T} B_t \sim |B_T|$$
(see e.g. Brownian motion by Schilling & Partzsch, Chapter 6), and therefore
$$\mathbb{P}(\tau>1/n) \leq \mathbb{P}(|B_{1/n}|=0)=0$$
for all $n \in \mathbb{N}$. Thus,
$$\mathbb{P}(\tau>0) = \lim_{n \to \infty} \mathbb{P}(\tau>1/n)=0,$$
and this proves $(1)$.
