How to prove $τ=\inf\{t:B(t)=\max\limits_{0<s<1}B(s)\}$ is a random variable？
$B(t)$ is brownian motion process
only random variable is needed，don‘t have to prove it not stopping time.
i tried to rewrite τ as inf{t:B(t)-M(t)=0}， where M（t）=max B(s) , 0< s< 1， there is a collateray noting that B(t)-M(t)===|B(t)|===M（t） in distribution. then τ=inf{t:M(t)=0}，i can only reach here.
Let $Y(t)=\sup \{B(u):0 \leq u\leq t\}$. Let us show that $\tau \leq t$ if and only of $Y(t)=Y(1)$. If $Y(t)=Y(1)$ then (since continuous functions attain their maximum on any compact set) there exists $s \in [0,t]$ such that $B(s)=Y(t)=Y(1)$ and hence $B(s)=\sup \{B(u):0 \leq u\leq 1\}$ which implies $\tau \leq s$ by definition of $\tau$. Hence $\tau \leq t$. Conversely, suppose $\tau \leq t$. Suppose, if possible, $Y(t) \neq Y(1)$. Since $Y(t) \leq Y(1)$ we get $Y(t)=\sup \{B(u):0 \leq u\leq t\} <Y(1)$. By continuity of Brownian paths this implies $Y(t+\epsilon)<Y(1)$for some $\epsilon >0$. This implies that the set $\{v\in [0,1]: Y(v)=Y(1)\}$ contains no point of the interval $[0,t+\epsilon]$ so the infimum of this set is $\geq t+\epsilon $. By definition of $\tau$ this means $\tau \geq t+\epsilon$ which is a contradiction. We have proved the claim that $\tau \leq t$ if and only of $Y(t)=Y(1)$. It remains to show that $\{Y(t)=Y(1)\}$ is measurable. For this it is enough to show that $Y(t)$ is a measurable function for each $t$. But $Y(t)=\sup \{B(\frac i {2^{n}}):0\leq i\leq 2^{n} t,n \geq 1\}$ which is meaurable.