Fundamental domains of measure preserving actions

Question

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 $g(D)$?

Answer 1

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.

Answer 2

Yes, take a look at the old papers by V. Rohlin:

MR0030584 (11,18f) Rohlin, V. A. On the fundamental ideas of measure theory. (Russian) Mat. Sbornik N.S. 25(67), (1949). 107--150

MR0030710 (11,40b) Rohlin, V. A. Selected topics from the metric theory of dynamical systems. (Russian) Uspehi Matem. Nauk (N.S.) 4, (1949). no. 2(30), 57--128.

Answer 3

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.

Answer 4

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 $A\cup-A$ where $A\subset(0,1/2)$ 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 $G$.