Let $(B_t)_{t\ge 0}$ be a one-dimensional Brownian motion and set $M_t = \sup_{s\le t} B_s$. Denote the $\xi_t$ the largest zero of $B_s$ before time $t$ and by $\eta_t$ the largest zero of $Y_s = M_s - B_s$ before time $t$. Show that $\xi_t \sim \eta_t$.
I know that $M-B$ is a Markov process with the same transition as $|B|$. Hence they have the same finite dimensional distributions.
But, how do we conclude from this that $\xi_t$ has the same distribution as $\eta_t$?
I am not fully convinced with the above argument because, e.g. $\xi_t = \sup\{s \le t: |B_s| = 0\}$, so to find out $\xi_t$ we need to know $|B_s|$ at all $s \in [0,t]$, which is uncountable. And similarly for $M_s - B_s$, $s\in [0,t]$. Hence, how does the equivalence of the distributions of $|B|$ and $M-B$ at all finite steps in $[0,t]$ conclude that $\xi_t$ and $\eta_t$, which require knowledge of continuous times at $[0,t]$, have the same distribution?
By definition, $$\xi_t(\omega) < s \iff \forall r \in [s,t]: |B_r(\omega)| \neq 0.$$ Since Brownian motion has (a.s.) continuous sample paths, it attains its minimum on compact sets. Thus, $$\xi_t(\omega)<s \iff \inf_{r \in [s,t]} |B_r(\omega)|>0 \iff \inf_{r \in [s,t] \cap \mathbb{Q}} |B_r(\omega)|>0.$$ Consequently, $$\mathbb{P}(\xi_t < s) = \mathbb{P} \left( \inf_{r \in [s,t] \cap \mathbb{Q}} |B_r| > 0 \right),$$ i.e. the distribution of $\xi_t$ is uniquely determined by the finite-dimensional distributions of $|B|$. An analogous reasoning works for $\eta_t$ and $M-B$, which gives, in conclusion, that $\xi_t \sim \eta_t$.