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.
Best Answer
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.