I consider $B_t$ a brownian motion and $\mathcal{F}_t$ its natural filtration. Then I would like to show that
$$ \mathbb{P}\left(\left\{\omega : \sup_{t\geq 0} B_t(\omega)\geq 0\right\}\right)=1 $$
Here is my attempt :
We notice that
$$ \left\{\omega : \forall\epsilon>0 \sup_{0\leq t\leq\epsilon}B_t(\omega)\geq 0\right\} \subset \left\{\omega : \sup_{0\leq t}B_t(\omega)\geq 0\right\} $$
Now the idea would be to use the $0-1$ law for brownian motion. We notice that
$$ \mathbb{P}\left(\left\{\omega : \forall\epsilon>0 \sup_{0\leq t\leq\epsilon}B_t(\omega)\geq 0\right\}\right)=\mathbb{P}\left(\cap_{n\geq1}\left\{\omega : \sup_{0\leq t\leq\ 1/n}B_t(\omega)\geq 0\right\}\right)= \lim_{n\to\infty}\mathbb{P}\left(\left\{\omega : \sup_{0\leq t\leq\ 1/n}B_t(\omega)\geq 0\right\}\right) $$
The second equality shows that the set $\left\{\omega : \forall\epsilon>0 \sup_{0\leq t\leq\epsilon}B_t(\omega)\geq 0\right\}$ is in $\mathcal{F}_{0^{+}}$ so it has probability $1$ or $0$. However we notice that for all $n\geq 1$ we have
$$ \mathbb{P}\left(\left\{\omega : \sup_{0\leq t\leq\ 1/n}B_t(\omega)\geq 0\right\}\right)\geq\mathbb{P}\left(\left\{\omega : B_{1/n}(\omega)\geq 0\right\}\right) = 1/2 $$
So we conclude that
$$ \mathbb{P}\left(\left\{\omega : \sup_{0\leq t}B_t(\omega)\geq 0\right\}\right) = \mathbb{P}\left(\left\{\omega : \sup_{0\leq t\leq\ 1/n}B_t(\omega)\geq 0\right\}\right) = 1 $$
Is this seems correct to you and do you see another way to prove it whithout using the $0-1$ law ?
Thank you a lot !