# Why $\mathbb P(B_t\geq L)=\mathbb P(B_t\geq \ell, \tau\leq t)$?

Let $$(B_t)$$ a Brownian motion. I want to prove that for all $$L\geq 0$$, $$\mathbb P(\sup_{0\leq s\leq t}B_s\geq L)=2\mathbb P(B_t\geq L).$$

The proof start by : let $$\tau=\inf\{t\geq 0\mid B_t= L\}$$. Then, $$\mathbb P(B_t\geq L)=\mathbb P(B_t\geq L, \tau\leq t),$$

but I don't understand this equality... For me, we have $$\mathbb P(B_t\geq L)\geq \mathbb P(B_t\geq L,\tau\leq t),$$ but I don't get the equality.

I guess that $$\mathbb P\{B_0=0\}=1$$ (otherwise your equality is not correct). By continuity of $$t\mapsto B_t$$, you have that $$\{B_t\geq L\}\subset \{\tau\leq t\}$$. Therefore $$\{B_t\geq L\}\cap \{\tau\leq t\}=\{B_t\geq L\},$$ and thus, your result.