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Let $(B_t)$ a Brownian motion starting at $0$ and let $B(0,R)$ the open ball of radius $R$. Let $\tau=\inf\{t>0\mid B_t\notin B(0,R)\}$. Does $|B_\tau(\omega )|\leq R$ for all $\omega $ or we only have $|B_\tau|\leq R$ a.s. ?

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  • $\begingroup$ Is $B_t$ continuous, or merely continuous a.s.? Do we have $B_0 = 0$ or merely $B_0 = 0$ a.s.? I think: if your question depends on what happens with probability zero, then anything goes. $\endgroup$
    – GEdgar
    Commented May 20, 2019 at 12:03
  • $\begingroup$ @GEdgar: A Brownian motion is normally continuous a.s. and start at $0$ a.s., no ? $\endgroup$
    – user657324
    Commented May 20, 2019 at 16:23
  • $\begingroup$ I agree. But then we certainly can only conclude $|B_\tau|\le R$ a.s. and not $|B_\tau|\le R$. $\endgroup$
    – GEdgar
    Commented May 20, 2019 at 17:35

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$|B_{\tau}| \leq R$ holds at every point $\omega$ where $\tau (\omega) <\infty$. But we can only assert that $\tau (\omega) <\infty$ almost surely.

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  • $\begingroup$ Thank you, it's clear now :) $\endgroup$
    – user657324
    Commented May 20, 2019 at 9:31

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