While I was reading Williams' "Diffusions, Markov Processes and Martingales" I found the following fact:
Let $B_t$ be a Brownian Motion. Then $\mathbb{P}\left(\sup_tB_t=\infty\right)=1$.
In the proof of that fact, Williams stated that if $Z=\sup_tB_t$ by Brownian scaling for all $c>0$,
$Z=cZ$ in distribution. How this facts imply that the law of $Z$ is concentrated on $\{0, \infty\}$
Best Answer
For each $R,c\gt 0$, using the equality $cZ=Z$ in distribution, $$ \Pr\left(\sup_{t\geqslant 0}B_t\gt R\right)=\Pr\left(Z\gt R\right) =\Pr\left(cZ\gt cR\right)=\Pr\left(Z\gt cR\right). $$ Letting $c$ going to $0$, we find $$ \Pr\left(\sup_{t\geqslant 0}B_t\gt R\right)= \Pr\left(Z\gt 0\right). $$ Letting $R$ going to infinity, we find $$ \Pr\left(\sup_{t\geqslant 0}B_t=+\infty\right)=\Pr\left(Z\gt 0\right). $$ Adding the probability that $Z=0$ proves the fact.