Let $T$ be a measure-preserving action of a group $G$ on a Lebesgue space $X$. That means that $T$ associates an automorphism (*i.e.* an invertible measure-preserving transformation) $T^g$ of $X$ to every $g \in G$, and the map $g \mapsto T^g$ is a group homomorphism from $G$ to the group of automorphisms of $X$.

Given a finite or countable partition $P=\{B_i\}_{i \in A}$ of $X$, denote by $P(x) \in A$ the index $i$ of the block $B_i$ to which a point $x \in X$ belongs. This provides a map $\phi\colon X \to A^G$ by defining $\phi(x)$ to be the function $f_x\colon G \to A$, $g \mapsto P(T^g x)$. The set $A^G$ is equipped with the product topology and the Borel $\sigma$-algebra, and the map $\phi$ carries the probability measure $\mu$ on $X$ to a probability measure $\nu$ on $A^G$.

Now assume $G$ is countable, hence $A^G$ is a Lebesgue space.
Say that $P$ is a *generating partition* if the map $\phi$ is an isomorphism of Lebesgue spaces. This is equivalent to require that the $\sigma$-algebra $\bigvee_{g \in G} \sigma(T^g P)$ is the full $\sigma$-algebra on $X$, up to negligible sets.

When $G=\bigoplus_{n=0}^\infty \mathbb{Z}/2\mathbb{Z}$ and the action is ergodic (*i.e.* $\mu(T^gB \Delta B)=0 \, \forall g$ only if $\mu(B)=0$ or $1$), does there always exist a generating countable partition ?

## UPDATE

According to the claim in the proof of Corollary 2.7 in this paper, Theorem 5.4 of "Countable Borel Equivalence Relations" by Jackson, Kechris, and Louveau states that any aperiodic Borel action of a countable group has a countable generating partition.
I have not checked that this applies to the situation of my question, I only guess it does. By the way I forgot to say I am interested in *free* actions of $G$, and this fits the statement of this Corollary 2.7 (I also guess that $G$ is amenable, I don't remember the definition of amenability).