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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 $X$.

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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.

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