ergodic theory - Fundamental domains of measure preserving actions

No, you need some other condition, since you aren't guaranteed to have many measurable sets.

For example, a probability pace consisting of a single atom (which need not be a set with one element) with measure $1$ has no measurable subsets of probability $|G|^{-1}$. You can let the space be $(-1/2,0) cup (0,1/2)$, let $C_2$ act by reversing signs, and let only the countable and cocountable subsets be measurable, with measures $0$ and $1$, respectively. Then no set has measure $1/2$.

Ok, assume $(X,mu)$ has no atoms. You still aren't guaranteed that there are enough measurable sets. Let the space be $(-1/2,0) cup (0,1/2)$, let $C_2$ act by reversing sign, and let the measurable sets be $Acup-A$ where $Asubset(0,1/2)$ is Lebesgue measurable. Then all measurable sets are fixed by the action of $C_2$, so there can't be a fundamental domain. As far as measure theory is concerned, this is still the trivial action, even though nothing is fixed by the nontrivial element.

You need some condition like that any set of positive measure contains subsets of positive measure which are moved by the action, whose images are measure-disjoint. Then (at least using some level of choice, but perhaps this isn't necessary) you can build a fundamental domain as a maximal set which has measure $0$ intersection with its images by nontrivial elements of $G$.