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Suppose we have a discrete group $G$ acting on a compact set $X \subseteq \mathbb{R}^d$ via measure-preserving homeomorphisms, and suppose we have a point $x$ whose orbit $Gx$ is finite (say $|Gx| = n$). Does the setup induce a probability measure on $Gx$?

Of course $Gx$ has measure 0 (outside degenerate cases) so in the most naive sense (i.e. conditional probability) the answer is "You're asking for too much." But since we have both measure and topology in our setup, it seems to me that there might be a principled way of deriving a measure on $Gx$. In particular, when $G$ is amenable, I think the stabilizer of $x$ should be a subset of $G$ with well-defined density, specifically density $1/n$. I have a sense of how this might work when you can put a neighborhood $U$ around $x$ with the property that each $y \in U$ has $|Gy| = n$, but it's less clear to me what to do when this condition fails.

I have even less intuition for the case in which $G$ is not amenable.

Are there situations in which the action of $G$ on $X$ induces a probability measure on the finite set $Gx$ that is NOT uniform? E.g., could there be an $G$-orbit of size 2 (call it $\{x,x'\}$) such that in a well-defined sense 1/3 or 2/3 (as opposed to 1/2) of the elements of $G$ fix $x$? (Here I am thinking of the "Banach-Tarski paradox for the hyperbolic plane", concerning a set $S \subset \mathbb{H}^2$ with the property that $\mathbb{H}^2$ can be expressed as the union of 2 images of $S$ or 3 images of $S$, using different symmetries of the hyperbolic plane. I would include a link but I had trouble finding one that was suitable. Maybe someone else can supply one?)

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    $\begingroup$ Measure-preserving means preserving Lebesgue measure on $X$? note that this is an empty condition if $X$ has zero Lebesgue measure. Else you might mean that there's an invariant probability measure on $X$. I don't guess. In any case on a finite set you always have the uniform probability... it's hard to guess what you mean by "induces". $\endgroup$
    – YCor
    Sep 30, 2020 at 21:56
  • $\begingroup$ My $X$ is homeomorphic to a $d$-ball so it has positive Lebesgue measure; I should have said that. The invariant measure is Lebesgue measure. I should also have said that $G$ is infinite. I'm being intentionally vague about what "induces" means because I don't know what the right definition is. It should "morally" be a kind of conditional probability, as if there were a uniform probability measure on $G$ (which of course there isn't!). $\endgroup$ Oct 1, 2020 at 2:58

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The standard way of defining measures on orbits of discrete groups consists in using the Radon-Nikodym cocycles (it is used a lot in the orbit equivalence theory, when talking about measures invariant or quasi-invariant with respect to equivalence relations, etc.) . Of course, strictly speaking the RN cocycles are only defined mod 0, but still there are many situations (for instance, when the RN derivatives are continuous) in which this definition makes sense for individual orbits as well. I very strongly doubt that there is a reasonable way to obtain non-uniform distributions for actions with a finite invariant measure.

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In other contexts, it'd be natural to count points on the boundary of the compact set $K$ as having less weight than points in the interior; specifically, to assign them weight proportional to $\lim_{\epsilon \rightarrow 0} \mbox{vol}(K \cap B(\epsilon))/\mbox{vol}(B(\epsilon))$, where $B(\epsilon)$ is a ball of radius $\epsilon > 0$ centered on the point in question, and to use a probability distribution in which the probability of a point is proportional to its weight. But this doesn't use the dynamics at all, so it may not be relevant.

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