Brownian Motion is almost surely unbounded, and a proof for the discrete Random Walk 
If $B_t$ is a Brownian Motion, why we have 
  $$
 P(\liminf_{t\to\infty} B_t = -\infty) = 1
$$
  and $P(\limsup_{t\to\infty} B_t = +\infty) = 1$?

I guess for the following "discrete version" of the "Brownian motion" (which isn't called Brownian Motion but Random Walk I guess) I have an argument. Let $B_n$ with $B_n \sim \mathcal N(0, n)$. Set
$$
 A_k := \{ B_n > k \mbox{ for infinitely many }n \}
            = \limsup_n ~ \{ B_n > k \}
$$
and $A := \bigcap_k A_k$, then $\omega \in A$ iff the path of $\omega$ is unbounded, i.e. $A = \{\limsup_n B_n = +\infty\}$. Each set $A_k$ is in the tail-$\sigma$-algebra as the limsup of sets, and so $A$ as a countable intersection is a tail-event, therefore by Kolmogoroff's 0-1-law we have $P(A) \in \{0,1\}$. As for each $n$ we have
$$ 
 P(B_n > \sqrt n) = 1 - \Phi\left(\frac{\sqrt n}{\sqrt n}\right) > 0.1
$$
where $\Phi$ denotes the cumulative distribution function of the normal distribution, and as
$$ 
 \bigcup_n \{ B_n > \sqrt n \} \subseteq A
$$
and by the above $0.1 < P( \bigcup_n \{ B_n > \sqrt n \} )$
we could not have $P(A) = 0$, therefore $P(A) = 1$. Similarly the case $P(\liminf_n B_n = -\infty) = 1$ might be establised.
In my approach I used that the process is discrete, but I do not know how to show it in the general case for continuous time. So any help would be greatly appreciated! Also if my solution for the discrete case is way to complicated I would also be thankful for feedback and alternative approaches.
 A: The continuous case follows from the discrete case! It's too bad you used "$B$" for both Brownian motion and the gaussians you're adding in the discrete case...
Say $B_t$ is brownian motion. For $n=1,2,\dots$ let $$X_n=B_n-B_{n-1}.$$The definition of brownian motion shows that the $X_n$ are iid gauussians. So if we let $$S_n=X_1+\dots+X_n$$then you've shown (I haven't looked at your proof) that $S_n$ is almost surely unbounded. But $S_n=B_n$.
EDIT: Looked at the OP's proof of the discrete case, at the his request.
I was misled by his use of the term "random walk". What the OP calls a random walk is not a random walk at all, as I understand the term. And I don't see how what he says about his "random walk" can be right, since there's no independence assumed. (Come to think of it, perhaps the $B_n$ at that point is still supposed to represent brownian motion; if so ignore this paragraph. I assummed we were redefining $B_n$ at that point because of the words "Let $B_n$...". Except no, $B_n$ can't be brownian motion at that point because that would make literally trivial to get the continuous case from the discrete case...)
For the record, it's very easy to show that the $S_n$ I define above is almost surely unbounded. Hint: If $|S_n|\le k$ and $|S_{n+1}|\le k$ then $|X_{n+1}|\le 2k$. The probability of this is $p<1$, so the probability that $|S_n|\le k$ for all $n$ is $0$.
