I know it is a really stupid question and it should be quite easy, but how can I show that $\int_0^{\infty}|B_t|\mathbb{d}t=\infty$ a.s. with $B_t$ being a standard brownian motion?
I just don't get it.
What I tried so far: I know that $\limsup_{t\rightarrow\infty}|B_t|=\infty$ and the liminf being 0. Hence I know that for a given $K$ there is a sequence $t_n$ s.t. $B_{t_n}>K$. But by continuity I only know that for every $t_n$ there is a $\delta_n>0$ s.t. $B_t>K$ for $t\in t_n\pm\delta$. But I don't know that these $\delta_n$ won't converge to $0$...
Hint: For every $t$ there exists $s\geqslant t$ such that $|B_s|\geqslant2$. Then, the first hitting time of $1$ after time $s$ has a constant distribution. Then sum these contributions.
Another argument is to note that by scaling $(B_{2t})$ is distributed like $(\sqrt2B_t)$ hence the integral $K=\int_0^\infty|B_t|\mathrm dt$ is distributed like $2\sqrt2K$. Since $K\gt0$ almost surely, this implies that $K=+\infty$ almost surely.
Edit: About the last implication, consider some random variable $Z$ such that $P[Z\gt0]=1$ and $Z$ equals $aZ$ in distribution, with $a\gt1$. Then for every $z\gt0$, $P[Z\gt z]=P[aZ\gt z]$ hence $P[Z\gt z]=P[Z\gt z/a]=\cdots=P[Z\gt z/a^n]\to1$ when $n\to\infty$. This implies that $Z\gt z$ almost surely, for every $z\gt0$. Thus, ...