Consider $\mathbb{Z}^d$ as a graph with edges connecting each $x, y \in \mathbb{Z}^d$ with $|x-y|=1$. Let $\tau (e)$ be i.i.d random variables assigned to each edge $e$. $\tau (e)$ represents the time required to traverse the edge in in either direction. For $x, y \in \mathbb{Z}^d$, if $x_0 = x, x_1, \dots, x_n = y$ is a path from $x$ to $y$, the travel time along the path is defined as $\tau(x_0, x_1) +\tau(x_1, x_2) + \cdots + \tau(x_{n-1}, x_n)$. Define the passage time from $x$ to $y$ as $t(x,y)=$ the infimum of the travel times over all paths from $x$ to $y$.
Let $X_{m,n} = t(mu, nu)$, where $u=(1,0,\dots,0)\in \mathbb{Z}^d$.
This seems to be one of the famous examples of applications of the subadditive ergodic theorem. I took the description above from Durrett's Probability (See Example 6.5.4).
I have no problem understanding why $X_{m,n}$ satisfies conditions of the subadditive ergodic theorem (Liggett's version) except for the following condition, which is critical to show that $X_{0,n}/n$ converges to a constant almost surely.
Condition: $\{X_{nk, (n+1)k}, n\geq 1\}$ is ergodic.
Durrett says it can be reduced to the tail $\sigma$-field of the i.i.d. sequence $\{\tau (e)\}$. I don't see why that is the case. Other literature, such as this, says the ergodicity follows directly from the fact that $\{\tau (e)\}$ is i.i.d. But, still, it doesn't seem trivial to me.
If $X$ can be reduced to the tail $\sigma$-field of $\{\tau (e)\}$ as Durrett says, $X_{m,n}$ should be determined by a part of $\{\tau(e)\}$. However, it doesn't seem obvious to me.
Any assistance or suggestion would be very much appreciated!
Self-resolved:Realized that you can show that $X= \lim_{n\to \infty}X_{0,n}/n$ is constant by showing $X$ is measurable with respect to the tail $\sigma$-field of $\{\tau(e)\}$ instead of showing ergodicity of the sequence $\{X_{nk, (n+1)k}, k\geq 1\}$.
