In a one-sided Markov shift $(X, \mathcal{B}, \mu, T)$ where $X = \{1, 2, \cdots, n\}^{\mathbb{N}}$ is the sample space, $\mathcal{B}$ the $\sigma$-algebra generated by finite cylinder sets, $\mu$ a product measure and $T$ the shift transformation, first, we define a metric $d(x, y) = 2^{-n}$ where $n$ is the least integer such that $x[n]\neq y[n]$. Then, we define the following relation:
Given $x, y\in X$, $x\sim_r y$ iff there exists $n, m\in\mathbb{Z}_{\geq 0}$ such that $T^n x = T^m y$.
One can easily check $\sim_r$ is an equivalent relation. We set $R = \{(x, y)\in X\times X\,\vert\,x\sim_r y\}$ and, for each subset $A\subseteq X$, we set $R(A) = \bigcup_{a\in A}[a]_{\sim_r}$ to be the union of equivalent classes of each element in $A$.
My question is:
- Is $R\in\mathcal{B}\times\mathcal{B} = \{A\times B\,\vert\,A, B\in\mathcal{B}\}$?
- For each $B\in\mathcal{B}$, is it true that $\mu(B)=0$ iff $\mu[R(B)]=0$?
The first question is asking if $R$ is measurable and the second is asking if $R$ is non-singular. For the first one, it suffices to show that, for each $B\in\mathcal{B}, R(B)\in\mathcal{B}$. However, $R(B)$ could be properly contained in $\bigcup_{n\in\mathbb{N}}T^n(B)$ and then I do not know how to proceed. I also try to consider the function $(x, y)\mapsto \inf_{n, m}d(T^n x, T^m y)$. This function is continuous but it may obtain zero outside of $R$.
For the second question, I suppose whenever $\mu(B)>0$, $B$ will contain a finite cylinder set and hence so does $R(B)$. However, I do not how to deal with the other direction.
For the first question, the crucial fact is that $R$ is the union of the sets $$ R_{n, m}= \big \{(x, y)\in X\times X: T^n(x)=T^m(x)\big \}, $$ as $n$ and $m$ range in ${\mathbb N}$, each of which is closed in the product topology, and hence measurable. Consequently $R$ is an $F_\sigma $ set, therefore measurable.
For the second question, notice that if $B$ has measure zero, then so does $T(B)$, and hence also $T_m(B)$, for every $m$. Moreover, since $T$ preserves $\mu $ (meaning that $\mu (T^{-1}(E))=\mu (E)$ for every measurable set $E$), we have that $$ \mu (T^{-n}(T^m(B)))= \mu (T^m(B))= 0, $$ for every $n$. The fact that $R(B)$ has measure zero then follows from $$ R(B) = \bigcup_{n, m\in {\mathbb N}}T^{-n}(T^m(B)), $$