If $(B_t)$ is a Brownian motion, then $P(\sup_{t\in [0,\infty )}B_t=0)=P(\sup_{t\in [0,\infty )}B_t<\infty )$.

Question

If $(B_t)$ is a Brownian motion, how can I prove that $$P(\sup_{t\in [0,\infty )}B_t=0)=P(\sup_{t\in [0,\infty )}B_t<\infty )\ \ ?\tag{1}$$

I can't use the fact that $P(\sup_{t\in [0,\infty )}B_t=\infty)=1 $ (because we use $(1)$ to prove this). The fact that $$P(\sup_{t\in [0,\infty )}B_t=0)\leq P(\sup_{t\in [0,\infty )}B_t<\infty ) $$ is fine. For the converse inequality, I tried to use that $(tW_{1/t})$ is a Brownian motion, but I can't conclude. any idea ? :-)

Answer 1

Let $X:=\sup B_t$, for all $0<a<b$, by scaling property, we imply that: $$P(X<a)=P(X<b)$$ Hence, conclusion