# Prove that $\lim_{L\rightarrow\infty} P\left(\sup_{0\leq s\leq t}|B(s)|>L\right)=0$, for each $t\geq0$, where $B$ standard Brownian motion.

Let $$B(t)$$, $$t\geq0$$, be a standard Brownian motion. I would like to prove that $$\lim_{L\rightarrow\infty} P\left(\sup_{0\leq s\leq t}|B(s)|>L\right)=0,$$ for each $$t\geq0$$.

In my class notes, I have the following proof: since $$\sup_{0\leq s\leq t}|B(s)|$$ has the same probability distribution as $$|B(t)|$$, then $$P\left(\sup_{0\leq s\leq t}|B(s)|>L\right)=P(|B(t)|>L),$$ and by Chebyshev's inequality, that probability is bounded by $$L^{-1} E(|B(t)|)$$, which tends to $$0$$ as $$L\rightarrow\infty$$.

I do not understand why $$\sup_{0\leq s\leq t}|B(s)|$$ has the same probability distribution as $$|B(t)|$$. I know that $$\sup_{0\leq s\leq t}B(s)$$ (with no absolute value) possesses the same probability distribution as $$|B(t)|$$, but this is not the same statement. Could you please elaborate on this? Thank you!

You are correct in your doubts. $$\sup_{s does not have the same distribution as $$|B_t|$$, so your notes have a mistake.
One way around this; you can write $$\sup_{0 and then the event the LHS is more than $$L$$ is the union of the events that both arguments of the RHS are more than $$L$$. Therefore, $$P(\sup_{0L)\le 2P(|B_t|>L)$$.
However, there is an even easier proof. For any (finite) random variable $$X$$, we have $$\lim_{L\to\infty}P(X>L)=0$$. This follows from continuity of probability, since the events $$\{X>L\}$$ decrease to the empty set as $$L\to\infty$$.