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 $g(D)$? 



Fundamental domains of measure preserving actions
I believe the answer is yes. You can first show, that once you have a set $A$ of non-zero measure, such that $g(A) = A$ for any $g \in G$, there is some $B \subset A$ of non-zero measure, such that $g(B) \cap h(B) = \emptyset$ for $g \ne h$. If this is shown, then you can prove your statement by Zorn-like argument, showing that the set $A$ of maximal measure, such that $g(A) \cap h(A) = \emptyset$ will be the fundamental domain. 


Here is a hint how to show the first claim. Take $g \in G$, $g \ne 1$, and a $G$-invariant set $A$. Since $g(x) \ne x$, there is some $B\_0 \subset A$ of non-zero measure such that $g(B\_0) \ne B\_0$, set $B\_1 = B\_0 \setminus (g(B\_0))$. Then $g(B\_1) \cap B\_1 = \emptyset$. Now do the same thing for other $h \in G$ starting from $B\_1$. 


Alternatively, once can pass to a Stone space and use continuity to get this statement, though this requires more machinery.


