# Generator of a $\bigoplus_{n=0}^\infty \mathbb{Z}/2\mathbb{Z}$-action

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).

• Dear Stephane, to help the reader, you might want to define the term "countable generator". – André Henriques Jan 5 '16 at 22:41
• @AndréHenriques Hmm yes. In fact, I want a definition that implies 2 :-) I thought this is a usual notion, because it is not defined in papers about group actions, such as this one. Anyway, my question was precipitate, and I have even not look at these papers. I will not be there that day, I will edit or perhaps delete this question when I'll be back. – Stéphane Laurent Jan 6 '16 at 6:29
• If you figure out the answer to your question I suggest that, instead of deleting the question, you write up your own answer here on MO for others to read. – André Henriques Jan 6 '16 at 10:17
• @AndréHenriques I totally rephrased my post. – Stéphane Laurent Jan 8 '16 at 17:17
• I was quite confused by your last sentence of the update "I also guess that $G$ is amenable...". I had to open the link to decipher it. So: the point is that Corollary 2.7 takes as an assumption that the acting group is amenable. And yes, abelian groups are amenable groups. – YCor Jan 13 '16 at 23:42