TITLE: Let $L = \sup\{t\le 1, B_t = 0\}$, does there exist $t_n<s_n<L$ a.s. s.t. $t_n\rightarrow L$, $B_{t_n}<0$ and $B_{s_n}>0$?
QUESTION [1 upvotes]: There are some contexts for this problem. I have shown that if $T= \inf\{t\ge 0, B_t =0\}$, the following holds.
$$P(L\ge t)= \int p(t, 0, dy) P_y(T>1-t)$$
Where $p(t,0,dy)$ is the Markov transition kernel of $B_t$.
If I can use this conclusion to construct $s_n$ and $t_n$

REPLY [1 votes]: Take $W_t=B(1-t)$ for $t\in [0,1)$, then $W_t$ is a Brownian motion. Notice that $L=1-\tau _0$ where $\tau_0 = \inf \{s|W_s=0    \}$.
Since, $\tau _0$ is a stopping time the difference $W_{\tau+t} -W_{\tau}$ is a Brownian motion starting at zero. Using Bluementhal's $0$-$1$ law, for a Brownian motion $B_t$, we obtain $$\mathbb{P_0}( \exists\epsilon\, \text{s.t.}\, \, B_t>0 \,\mathrm{for\,all}\, t \in (0,\epsilon)   )=0$$ 
Therefore, there must be $ s_n, t_n \rightarrow \tau^+$ where $B(t_n)<0$ and $B(s_n)>0$ almost surely.
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