# Can each non-open analytic subgroup of a Polish abelian group be covered by countably many closed Haar null subsets?

By a result of Laczkovich ('Analytic subgroups of the reals' Proc AMS Vol 126 (1998)), any non-open analytic subgroup of a Polish locally compact group can be covered by countably many closed Haar null sets.

Is this result of Laczkovich true for any Polish (not necessarily locally compact) groups? More precisely:

Problem 1: Does each non-open analytic subgroup $H$ of a Polish abelian group $G$ belong to the $\sigma$-ideal $\mathcal E$ generated by closed Haar null subsets of $G$?

A Borel subset $A$ of a Polish group $G$ is Haar null of there exists a Borel probability measure $\mu$ on $G$ such that $\mu(xAy)=0$ for all $x,y\in G$.

A related weaker question also seems to be open.

Problem 2: Does each non-open analytic subgroup $H$ of a Polish abelian group $G$ belong to the $\sigma$-ideal $\mathcal M_h$ generated by closed Haar-meager subsets of $G$?

A Borel subset $A$ of a topological group $G$ is called Haar-meager if there exists a continuous map $f:K\to G$ defined on a compact metrizable space $K$ such that $f^{-1}(xAy)$ is meager in $G$ for all $x,y\in G$. It can be shown that a closed subset $A$ of a topological group $G$ is Haar-meager if and only if there exists a compact subset $K\subset G$ such that for every $x,y\in G$ the intersection $K\cap xAy$ is nowhere dense in $K$. It is easy to see that each closed Haar-null set is Haar-meager. More information on Haar meager sets can be found in the paper "On Haar meager sets" by Darji.

• It seems that the answer to this problem is negative: the Polish group $G=\mathbb Z^\omega$ contains a meager analytic subgroup $H$ such that for every $F_\sigma$-set $\bigcup_{n\in\omega}A_n$ containing $H$ some set $A_n$ is prethick in the sense that any compact set $K$ in $G$ is contained in the sum $F+A_n$ for some finite set $F$ in $G$. It is clear that the set $A_n$ is neither Haar-null nor Haar-meager in $G$. So, $H$ does not belong to the ideal $\mathcal E$. The construction of $H$ is not entirely trivial (it uses some results on filter games). I will write a proof and post a link. – Taras Banakh Sep 23 '15 at 7:20

The answer to both problems (1 and 2) is negative: the Polish group $G=\mathbb Z^\omega$ contains a dense meager Borel subgroup $H$ (which can be written as the difference $H=A\setminus B$ of two $F_\sigma$-sets in $G$) which cannot be covered by countably many closed Haar-meager subsets of $G$. Such subgroup $H$ is constructed in Example 6.3 of this paper.
The subgroup $H$ also gives a (partial) counterexample to the question Meager subgroups of compact groups .
Indeed, for any countable cover $\{A_n\}_{n\in\omega}$ of the group $H$ the closure $\bar A_n$ of some set $A_n$ is not Haar-meager (and hence not Haar-null) in $G$ and hence $\bar A_n\bar A_n^{-1}$ is a neighborhood of zero, which implies that $A_nA_n^{-1}$ is not nowhere dense in $H$.