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I have a standard Brownian motion, $B(t)$, $t\ge0$.

I understand that every $t > 0$ is an accumulation point (i. e., cluster point) of $(s: B(s) = B(t))$ from the right, with probability $1$.

That is, for any fixed $t > 0,$

$P[$inf$(s > t : B(s) = B(t)) = t] = 1$

Now, I am trying to find out whether $P[$inf$(s > t : B(s) = B(t)) = t$ for all $t \in [0, 1]] =$ $1$ $?$

My heuristic is no:

$P[$inf$(s > t : B(s) = B(t)) = t$ for all $t \in [0, 1]] \neq$ $1$

But I would like to prove this mathematically using the last time in $[0, 1]$ when the Brownian motion hits $0$, instead of only trying to imagine or convince myself why. Any help would be most appreciated. Thanks a lot.

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    $\begingroup$ Before asking if $P[$inf$(s > t : B(s) = B(t)) = t$ for all $t \in [0, 1]] \neq$ $1$ you should also worry about measurability of this event. $\endgroup$ Dec 6, 2023 at 9:07

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(I assume $B(0)=0$.) Your intuition is on point. Let $G:=\sup\{s\le 1: B(s) =0\}$ denote the last-exit time from $0$ before time $1$. Then $\Bbb P[G=0]=0$ by your initial "accumulation point" observation (which is true even for $t=0$). Also, $$ \left\{\inf(s>t: B(s) = B(t)) =t,\forall t\in[0,1]\right\}\subset\{G=1\}. $$ But $\Bbb P[G=1] = 0$ by continuity of $t\mapsto B(t)$ and the fact that $\Bbb P[B(1)=0]=0$.

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