Assume a standard (separable, metrizable, complete) probability space $(X, \mathcal{B}, \mu)$ with $\mu : \mathcal{B} \to [0, 1]$ a $\sigma$-additive measure s.t. $\mu(X) = 1$. Furthermore a countable group $\Gamma$ defines a p.m.p. action on $X$, meaning all orbits are countable, Borel and the action is measure preserving i.e. $\mu(g \cdot B) = \mu(B)$ for any $g \in \Gamma, B \in \mathcal{B}$.
- The action $\Gamma \curvearrowright X$ is called finite iff almost all $x \in X$ have a finite orbit for $\Gamma$.
- The action $\Gamma \curvearrowright X$ admits a fundamental domain iff there exists a Borel set $Y \subseteq X$ such that almost all $x \in X$ have an orbit $\Gamma x$ such that $\Gamma x \cap Y$ is a singleton.
Apparently these two statements are equivalent, which I have been trying to prove, but haven't gotten far. I'm gonna post what I have, so assume a fundamental domain $Y \subseteq X$, then per definition $\Gamma Y = X$ almost everywhere, or more precisely $\mu(\Gamma Y) = \mu(X) = 1$. Since $\Gamma$ is a countable group, $1 = \mu(\Gamma Y) \le \sum_{g \in \Gamma} \mu(g \cdot Y)$. Since the action is measure preserving, this is actually a countable sum over a constant value, which means at the very least $\mu(Y) > 0$.
After this, I wished to find a way to cover $X$ in cosets of $Y$, and then argue that such a cover must necessarily be finite. But these cosets are not open sets (I think), so compactness of the space cannot be applied directly (but is there a way around it?).
In another attempt I tried to prove that the countable sum must necessarily be finite by trying to find bounds to the measure $\mu(Y \cap g \cdot Y)$ or $\mu(g \cdot Y \cap h \cdot Y)$ in general. This didn't get me very far, beyond realizing that $x \in Y \cap g \cdot Y$ iff $gx = x$ almost always. This follows from the uniqueness of most representatives in the fundamental domain.
Source: A. Ioana: "Orbit Equivalence of Ergodic Group Actions" - Exercise 2.5.