# When does a subset of a Polish space meet all the orbits?

While I was studying Borel actions of Polish groups on Polish spaces (I assume of measure $$1$$), I have tried to understand if a measure $$1$$ (hence dense) subset of this Polish space meets all the orbits; more precisely, the question naturally arose because I want to solve the following problem:

Assume $$X$$, $$Y$$ be standard Borel spaces (Polish space equipped with the $$\sigma$$-algebra generated by open sets), that $$\mu$$ be an (possibly ergodic) invariant Borel probability measure on $$X$$ and that $$f\colon A\subseteq X\to Y$$ be a Borel function.

If $$G$$ is a Polish locally compact group (I'm particularly interested in the case $$G=SL_m(\mathbb{Z})$$) acting (as a Borel map) on $$X$$ and $$A$$ has measure $$1$$ in $$X$$ (hence $$\mu(G.A)=1$$ by invariance), I would like to define a Borel function $$f'$$ on the whole space $$X$$ by means of a Borel function $$c\colon X\to A$$ (so that $$f'=f\circ c$$ is a Borel function from $$X$$ into $$Y$$) s.t. $$c(x)$$ belongs to the same equivalence class of $$x$$.

Here is my attempt: $$\mu(A)=1$$ implies that $$A$$ is dense in $$X$$ (because otherwise the existence of a non-empty open set disjoint from it leads to a contraddiction). Now, if $$A$$ meets every $$G$$-orbit (I don't know how to prove this), I guess I can define $$c\colon X\to A$$ by letting $$c(x)=$$one element in $$\mathrm{Orb}(x)\cap A$$. But I'm not sure this works also because in this way I need the Axiom of Choice to guarantee the existence of such a function (am I right?)

Any hint?

Thank you in advance for your help.

Maybe I miss something, but if $$c(x)$$ belongs to the same equivalence class of $$x$$ means that $$c(x)$$ belong to the orbit $$\mathrm{Orb}(x)$$ of $$x$$ then why we always have $$\mathrm{Orb}(x)\cap A$$, that is $$A$$ meets every $$G$$-orbit? For instance, if a group $$G$$ is countable (for instance, when a discrete group $$\Bbb Z$$ of $$\Bbb Q$$ acts on the unit circle by rotations) then we can pick any $$x\in X$$ and put $$A=X\setminus \mathrm{Orb}(x)$$, right?
On the other hand, assume that $$\mathrm{Orb}(x)\cap A\ne\varnothing$$ for any $$x\in X$$ and there exists a countable subset $$F=\{f_n:n\in\omega\}$$ (with $$f_0=e$$) of $$G$$ such that $$X=FA$$ (maybe we can show that $$F$$ exists, using that $$\mu(A)=1$$ and $$X$$ and $$G$$ are Polish). Define the function $$c$$ by putting for any $$x\in X$$, $$c(x)=f_n(x)$$ for the smallest $$n$$ such that $$f_n(x)\in A$$. Then I guess that for any Borel subset $$B$$ of $$A$$ we have that $$c^{-1}(B)=\bigcup_{i=0}^\infty \left(f_i^{-1}(B)\setminus \bigcup_{j=0}^{i-1} f_j^{-1}(A)\right)$$ is a Borel set.
• First of all, I apologize for answering just now, but I had no time. Here are some doubts: (1) I don't understand why $A$ meets every $G$-orbit (I guess you are arguing in the case $G$ is countable); (2) How can we prove the existence of such a $F$? In this particular case, using the well-order on $\omega$, you avoid AC (very interesting! :D) – LBJFS Mar 8 at 20:32
• @LBJFS (1) It is only an assumption, which, I think, does not always hold, see the proposed example. (2) If $\mathrm{Orb}(x)\cap A\ne\varnothing$ for any $x\in X$ and $G$ is countable then we can put $F=G$. When $G$ is uncountable there still is a hope that such a set $F$ exists, but this looks a much more complicated topic, and I also don’t have a lot of time now to finish my investigation. I’m going to write my ideas about it later. – Alex Ravsky Mar 8 at 21:24