Suppose a finite group $G$ acts on a standard probability space $(X, \mu)$ by measurepreserving 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 $g(D)$?

1$\begingroup$ Do you know the answer in the case where $(X,\mu)$ is the unit interval with Lebesgue measure? $\endgroup$ – Yemon Choi Mar 6 '10 at 0:32

$\begingroup$ no I do not. In fact by standard probability space I meant [0, 1] in Lebesgue measure $\endgroup$ – Mike Hartglass Mar 6 '10 at 1:35

1$\begingroup$ The measure preserving property is not relevant, because you can always replace mu(A) by sum_g mu(gA) $\endgroup$ – George Lowther Mar 6 '10 at 2:46

1$\begingroup$ Mike Hartglass, will you please include the specification of your measure space in the question? I think many of us don't use the terminology "standard probability space" to mean [0,1] with Lebesgue measure. $\endgroup$ – Jonas Meyer Mar 6 '10 at 6:54
I believe the answer is yes. You can first show, that once you have a set $A$ of nonzero measure, such that $g(A) = A$ for any $g \in G$, there is some $B \subset A$ of nonzero measure, such that $g(B) \cap h(B) = \emptyset$ for $g \ne h$. If this is shown, then you can prove your statement by Zornlike 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 nonzero 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.

$\begingroup$ The second paragraph has some gaps. What properties of $X$ are you using to construct $B_0$, since such a set doesn't have to exist for a more general probability space? You also want $B_1$ to have positive measure. How do you ensure that? $\endgroup$ – Douglas Zare Mar 6 '10 at 5:04

1$\begingroup$ I use separability of $X$. There is a theorem, usually attributed to von Neumann, that every measure preserving automorphisms of a separable measure algebra has a unique point realization on a standard Lebesgue space. From this existence of $B_0$ follows. By $g(B_0) \ne B_0$ I mean $\mu(g(B_0) \setminus B_0) \ne 0$, so $B_1$ has positive measure. $\endgroup$ – Konstantin Slutsky Mar 6 '10 at 5:59
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). 107150
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), 57128.
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.
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 $A\cupA$ 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 measuredisjoint. 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 $G$.

1$\begingroup$ By Standard probability space, I meant the standard diffuse nonatomic Borel probability space. This is [0, 1] in Lebesgue measure. I am curious if the answer is true even in this case $\endgroup$ – Mike Hartglass Mar 6 '10 at 1:37

1$\begingroup$ Ah, thanks for the clarification. I think the condition I gave in the last paragraph is still relevant for that. $\endgroup$ – Douglas Zare Mar 6 '10 at 4:16