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.


Too long for a comment.

Disclaimer: It is a deep night here and I’m too tired even to check my claims, so I can wrote something stupid.

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.

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    $\begingroup$ 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) $\endgroup$ – LBJFS Mar 8 at 20:32
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    $\begingroup$ @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. $\endgroup$ – Alex Ravsky Mar 8 at 21:24

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