Bounded martingales of infinite path length Let $(X_t)_{t \in \mathbb{N}}$ be a real-valued martingale that is bounded, i.e.,
there are $a, b \in \mathbb{R}$ such that $a \leq X_t \leq b$ for all $t$.
Define the path length $L$ of $(X_t)_{t \in \mathbb{N}}$ as the distance the martingale covers,
$$
L := \sum_{t=0}^\infty | X_t - X_{t+1} |.
$$
My questions:


*

*Is $L$ almost surely finite? I.e., can there be a set of positive measure on which $L = \infty$?

*If yes to question 1, is $\mathbb{E}[L]$ finite?

 A: If you have a random walk on $\{0,\ldots,N\}$, it's known that the expected
number of steps to hit the boundary starting from the middle is something like $N^2/4$. That is: you travel a distance of $N^2/4$ before hitting the boundary. 
You can exploit this to produce a martingale where the expected path length is infinite: Do absorbing random walk on $[0,1]$, where the step sizes are dyadic 
and depend on time. For example, do $2^{10}$ steps of size $2^{-10}$; then $2^{11}$ steps of size $2^{-11}$ etc. In the $2^{-k}$ phase, your average displacement is something like $2^{-k/2}$, so that it's likely that you'll never hit the boundary.
But any sample path that doesn't hit the boundary has infinite length.
A: I like Anthony's answer but here's another construction.
Evaluate a standard Brownian motion on the sequence $t_n = 1- 1/n$.   The process has infinite length almost surely by Levy's theorem on the modulus of continuity of Brownian motion.
It's not a bounded martingale but you can stop it the first time it gets larger than $1$ in absolute value.  By Doob's stopping theorem this is still a martingale.  And it still has infinite length whenever the original Brownian motion remains in $[-1,1]$ up to time $1$.
