TITLE: Fundamental domains of measure preserving actions
QUESTION [5 upvotes]: Suppose a finite group $G$ acts on a standard probability space $(X, \mu)$ by measure-preserving actions (i.e. $\mu(g(A)) = \mu(A)$ for all $g \in G$ and $A \subset X$ measurable).  In addition, suppose that for $g \in G$ and $g$ not the identity then $\mu(\{x:g(x) = x\})=0$.  I am wondering if it is always possible to find a fundamental domain $D$ of the action of $G$, i.e. is there a measurable $D \subset X$ such that $G$ is the disjoint union (up to measure 0) of the sets $g(D)$?

REPLY [2 votes]: Here is another way to find a fundamental domain.
First identify $X$ with $[0,1]$. You want to pick a single point in each orbit of the action. Just take the smallest one.
Let be more specific. 
Consider the set $A$ of points $x$ for which the number of points in the orbit of $x$ is equal to the cardinal of $G$. Your assumption insures that this set is of full measure. Take some Borel subset $B$ in $A$ of full measure in $A$, such that $G$ acts on $B$ through Borel transformations.
The fundamental domain $D$ is then defined as the image of $B$ by the map $x \rightarrow  \min\lbrace\ gx\ |\ g\in G\ \rbrace$.
Now the image of a Borel set by a Borel map is always measurable. Restricting again to a Borel subset of full measure, we get  a Borel fundamental domain for the action.