Showing that a hitting time is $\mathbb P-\text{a.e.}$-finite

Question

Let be $\alpha, \beta \in \mathbb R$ such that $\alpha < \beta $ and $x \in [\alpha, \beta ]$. Consider the random time

$$T_x = \inf \{ t\geq 0 : x+ B_t \notin [\alpha, \beta]\},$$ where $B=(B_t)_{t\geq 0}$ is a standard brownian motion in $\mathbb R$ starting from zero.

To show that $T_x < + \infty \ \mathbb P -\text{a.e.}$, one wrote the line

\begin{align}\mathbb P \left( T_x = +\infty \right) &=\mathbb P \left(\{ \forall t \geq 0, \alpha \leq x+B_t \leq \beta\} \right) \\&\leq \mathbb P \left(\{0\leq \liminf _{t \rightarrow \infty}\frac {B_t} {\sqrt{2t\log(log(t))}} \} \cup \{ \limsup _{t \rightarrow \infty}\frac {B_t} {\sqrt{2t\log(log(t)) }}\leq 0 \} \right) = 0 \end{align}

where the last equality comes by the law of the iterated logarithm.

How to justify the inequality ? Ideally by writting some intermediate line(s).

I would appreciate any advise.

Thanks in advance.

Answer 1

You do not say who appeals to the law of iterated logarithm to prove this (and indeed, as explained in a comment, once one knows the LIL the result follows) but one should probably mention that much less elaborate approaches exist.

To wit, for every $n$, consider $A_n=[|B_{4^n}|\leqslant c]$, for some $c\gt0$. Then, $B_{4^n}$ is distributed like $2^nZ$ where $Z$ is standard normal hence $P[A_n]=P[|Z|\leqslant2^{-n}c]\sim2^{-n}c\sqrt{2/\pi}$.

The series $\sum\limits_nP[A_n]$ converges hence, by Borel-Cantelli, at most finitely many $A_n$ happen. In particular, at least one event $A_n$ does not happen, that is, $|B_{4^N}|\gt c$ for some finite random $N$, almost surely.

Choosing $c$ large enough, namely, $c=\max\{(\beta-x)^+,(x-\alpha)^+\}$, this implies that $T_x\leqslant 4^N$ hence $T_x$ is almost surely finite.

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Edit: The inclusion itself follows from the following deterministic result.

Let $u$ and $v$ denote functions defined on $[0,+\infty)$ such that $u$ is bounded and $\lim\limits_{t\to\infty}v(t)=\infty$, then $\lim\limits_{t\to\infty}\frac{u(t)}{v(t)}=0$ hence $\limsup\limits_{t\to\infty}\frac{u(t)}{v(t)}=\liminf\limits_{t\to\infty}\frac{u(t)}{v(t)}=0.$

Perhaps getting rid of the probabilistic apparatus will make apparent that the statement holds trivially. Then apply it to $u(t)=x+B_t$ and $v(t)=\sqrt{2t\log\log t}$ on the event $[T_x\ \text{infinite}]$.