Let $H$ be some Polish group and $X$ some standard Borel space. Assume that $H$ acts measurably on $X$, i.e. $(h,x) \mapsto hx$ is Borel. Let $H\_0 \subset H$ and $A \subset X$ be some Borel sets. Is it true that the set of all $x \in X$ such that $H\_0 x \subset A$ is Borel?


The following theorem of descriptive set theory is related to the question:
If $A$ is some Borel subset of $P \times X$, with $P$ a Polish space and $X$ a standard Borel space, such that, for any $x \in X$, the $x$-fiber $A\_x=\{p \in P ; (p,x) \in A\}$ is a $K\_\sigma$ (i.e. a countable union of compact sets), then $\pi\_X(A)$ is Borel, with $\pi\_X : P \times X$ the projection onto $X$.



Measurability of $\{ x \in X ; H_0 x \subset A \}$
No, it is not true.


It suffices to show that the complement is not necessarily Borel.
Let $B=X\setminus A$, which is a general Borel set. We have
$$
\{x:H\_0x\not\subseteq A\} = \{x:H\_0x\cap B\ne\varnothing\} = \{x: (\exists h\in H\_0)(hx\in B)\} = \{x: x\in H\_0^{-1}B\} = H\_0^{-1}B
$$
Let $G=H\_0^{-1}$ which is again a general Borel set since $G^{-1}=H\_0$.
Then we want to determine whether $GB$ must be Borel.
But let the group be addition on $\mathbb R$: [Erdös and Stone showed that the sum of Borel sets is not necessarily Borel](https://mathoverflow.net/questions/48571/borel-set-plus-a-closed-set-borel).


