# Why the probability is concentrated on $\{0, \infty\}$

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 $\{0, \infty\}$

• $P(Z>\epsilon) = P(Z>1/\epsilon)$ and $P(Z>\epsilon) - P(Z > 1/\epsilon) = P(Z\in (\epsilon, 1/\epsilon)) = 0$. Does this help? – Jo' Aug 14 '18 at 22:41
• Also, $Z$ cannot concentrate at 0. Brownian motion changes sign infinitely often in $[0,\epsilon)$, in particular, a.s. – James Yang Aug 14 '18 at 22:47
• @JamesYang Since $Z=\sup_t B_t$ and $B_0=0$ clearly $Z\geqslant0$ a.s. – Math1000 Aug 15 '18 at 3:58
• Of course!. Thanks @Jo'. That is the subtle observation. – Uriel Herrera Aug 15 '18 at 4:08
• @JamesYang. Yes, that its the second part of the proof in that book. Thank you! – Uriel Herrera Aug 15 '18 at 4:09

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.