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.
 A: 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.
