Let $(\mathbb{B}, \nu)$ be the binary space with probability measure $\nu(0) = \nu(1) = \frac{1}{2}$. The map $T : \mathbb{B}^{\mathbb{N}} \to \mathbb{B}^{\mathbb{N}}$ is defined as left addition with carry, i.e. if $x = (1, \ldots, 1, 0, x_k, \ldots)$ then $T(x) = (0, \ldots, 0, 1, x_k, \ldots)$. This map is a measure preserving Borel isomorphism. We can define two actions on the product space $(\mathbb{B}^{\mathbb{N}}, \nu^{\otimes \mathbb{N}})$ as follows:
- $\mathbb{B}^{\mathbb{N}} \curvearrowright \mathbb{B}^{\mathbb{N}}$ through componentwise addition modulo 2.
- $\mathbb{Z} \curvearrowright \mathbb{B}^{\mathbb{N}}$ through $k \cdot x := T^k(x)$.
Show that these actions are orbit equivalent, in other words, there exists a Borel isomorphism $\varphi : \mathbb{B}^{\mathbb{N}} \to \mathbb{B}^{\mathbb{N}}$ which is measure preserving and satisfies $\varphi(\mathbb{B}^{\mathbb{N}} \cdot x) = \mathbb{Z} \cdot \varphi(x)$ for almost every $x \in \mathbb{B}^{\mathbb{N}}$.
I'm pretty stumped on this problem. If I'm not mistaken the orbit $\mathbb{B}^{\mathbb{N}} \cdot x$ should be the entire space $\mathbb{B}^{\mathbb{N}}$ itself. Since $\varphi$ is supposed to be an isomorphism I assume this means $\mathbb{Z} \cdot \varphi(x) = \mathbb{B}^{\mathbb{N}}$ as well. However the former seems to be a countable set, while the latter isn't? Is there something obvious I'm seeing wrong here? This problem is taken from the paper A. Ioana's lecture notes "Orbit Equivalence of Ergodic Group Actions", Exercise 1.21.
Edit: So I thought of a potential solution, but I'm not sure I can take it all the way to its end, so here goes. For any $x \in \mathbb{B}^\mathbb{N}$, it seems clear on one hand that $\mathbb{B}^\mathbb{N} \cdot x = \mathbb{B}^\mathbb{N}$. On the other hand, the set $\mathbb{Z} \cdot x$ contains all sequences that eventually have the same tail end as $x$, in other words it contains all the following sets:
$$ \begin{align} \{x_0\} \times \{x_1\} \times \{x_2\} &\times \cdots \\ \mathbb{B} \times \{x_1\} \times \{x_2\} &\times \cdots \\ \mathbb{B} \times \mathbb{B} \times \{x_2\} &\times \cdots \\ &\vdots \end{align} $$
Hence if we define the Borel sets $Z_k = \mathbb{B}^k \times \{x_k\} \times \{x_{k+1}\} \times \cdots$ for all $k \in \mathbb{N}$ it follows from $Z_k \subseteq \mathbb{Z} \cdot x$ that $\bigcup^\infty_{k=0} Z_k \subseteq \mathbb{Z} \cdot x$. At this point I'd use the measure of the union as a lower bound for the measure of $\mathbb{Z} \cdot x$ but I now realize that's not as obvious as it first looked.