Convergence almost surely of Brownian Motion If $B_t \sim N(0,t)$ then, intuitively, for any fixed $\varepsilon$,  as $t \to \infty$, the probability that $B_t$ will be observed within the $[-\varepsilon, \varepsilon]$ interval should converge to $0$, due to the increasing variance. But since $B_t$ is a continuous rv, I'm not sure how to use Borel-Cantelli Lemma here.
First, if $B_t \sim N(0,t)$, then $Z_t = \frac{B_t}{\sqrt{t}} \sim N(0,1)$. Therefore,
\begin{align}
\lim_{t \to \infty}P(|B_t|>\varepsilon) &= \lim_{t \to \infty} P\bigg(\frac{|B_t|}{\sqrt{t}}>\frac{\varepsilon}{\sqrt{t}}\bigg) \\&= \lim_{t \to \infty}P\bigg(Z_t>\frac{\varepsilon}{\sqrt{t}}\bigg) + P\bigg(Z_t<-\frac{\varepsilon}{\sqrt{t}}\bigg) \\&= \lim_{t \to \infty}1- \Phi(\frac{\varepsilon}{\sqrt{t}}) + \Phi(-\frac{\varepsilon}{\sqrt{t}}) \\&= 1-\frac{1}{2}+\frac{1}{2} = 1. 
\end{align}
This, I believe, amounts the proof that $B_t$ diverges in probability, i.e.
$$
\text{plim}_{t \to \infty}B_t = \pm\infty
$$
But I'm not sure how to extend it to $\lim_{t \to \infty} P(\limsup B_t=\infty)=1$. I understand that $\mathbf{E}[B_{t+h}B_t]=t \neq 0$, so $B_t$ are not independent, hence only Borel-Cantelli Lemma-I will work here, so I somehow need to show that there exists a sequence of events $I_t = \{t:|B_t|<\varepsilon\}$, and then prove the sum converges, but not sure how to do it. Do I need to split the timeline into disjoint intervals?
I know this question was asked before, but I'm interested if the logic above is correct and can be extended to the proof if convergence a.s.
 A: You have shown that for any $\varepsilon > 0$ and $\delta > 0$, there is $T$ large enough that for any $t \geq T$, $P(|B_t| \leq \varepsilon) < \delta$. Presumably, you want to show that
$$P(\limsup_{t\rightarrow\infty} B_t = \infty) = 1,$$
not $P(\dots) = \infty$, as pointed out in a comment. Moreover, the statement  $\lim_{t\rightarrow\infty} B_t = \pm\infty$ is not true, because almost surely $B = \{B_t, t\geq0\}$ crosses $0$ infinitely often, and the limit does not exist.

*

*If you don't need to use Borel-Cantelli, the Law of Iterated Logarithm will give you $P(\limsup_t B_t = \infty) = 1$ directly.


*If you don't want to use the hammer of LIL but don't need to directly use Borel-Cantelli, the usual argument is the same as that for a random walk on the integers using a 0-1 law. First, note that, for any $0<A<\infty$,
$$\{ \limsup_n B_n \leq A\} \subset \liminf_n \{B_n \leq A\} = \cup_{m=1}^\infty \cap_{n=m}^\infty \{B_n \leq A\}.$$
The right-hand side is the liminf of sets, which means that the sequence of events occurs "all but finitely many times". Indeed,  if $\omega \in\{\limsup_n B_n(\omega) \leq A\}$ then certainly there is an $m<\infty$ large enough that for all $n \geq m$, $\omega \in \{B_n \leq A\}$. You've already shown that $P(B_n \leq A) < \delta < 1$ for all $n$, and since for any sequence of sets $E_n$,
$$P(\liminf_n E_n) \leq \liminf_n P(E_n),$$
you also have $P(\limsup_n B_n \leq A) \leq \delta < 1$. Since $\{\limsup_n B_n \leq A\}$ is a tail event, the probability is $0$. Since $0 < A< \infty$ is arbitrary, the result follows.


*If you insist on using Borel-Cantelli, the only argument I could think of uses some form of strong Markov property, in order to use independent events, or perhaps use a stronger (the strongest) version of Borel-Cantelli. Here is an argument using the reflection principle.
