Probability measure preserving action on a countable space has finite orbits

Question

I need a reference to the statement:

Let $G$ be a countable group acting on $(E,\mu)$ a probability space where $\mu$ is measure preserving. Suppose further that $E$ is countable. Then the orbits of the action of $G$ are finite.

Thank you!

Answer 1

I am assuming the group action is a homomorphism from the group to measure preserving bijections on $E$. I don't have a reference but I could think of the following argument when the measure $\mu$ has full support: Since $\mu$ has full support, $\mu(\{e\})>0$ for all $e \in E$. Assume the orbit of some $e \in E$ is countable. Thus, there must exist countably many distinct points $\{e_n\}_{n \in \mathbb{N}} \subset E $ such that $g \cdot e=e_n$ for some $g \in G$ and $\mu(\{e_n\})= \mu(g^{-1}\{e_n\})=\mu(\{e\})>0$. Thus we have that $\mu(E)\geq \mu(\{e_n\}_{n \in \mathbb{N}})=+\infty$. This is a contradiction.