3
$\begingroup$

A famous Steinhaus theorem says that if measurable subsets $A,B$ of a locally compact topological group $G$ have positive Haar measure, then the difference $AA^{-1}$ is a neighborhood of the unit and the sumset AB has non-empty interior in $G$.

The first part of this theorem was generalized to Polish groups by Christensen who proved that for a Haar positive Borel subset $A$ of a Polish abelian group $G$ the set $AA^{-1}$ is a neighborhood of zero in $G$.

A subset $A$ of a Polish group $G$ is Haar null if there exists a Borel probability measure $\mu$ such that $\mu(xAy)=0$ for all $x,y\in G$. A subset which is not Haar null is called Haar positive.

What about the second part of the Steinhaus Theorem? Can it be generalized to Polish groups? For example, is it true that for any Haar positive closed sets $A,B$ of a Polish abelian group $G$ the sumset $AB$ is not meager in $G$?

An affirmative answer to this problem would imply an affirmative answer to the problem (Can each non-open analytic subgroup of a Polish abelian group be covered by countably many closed Haar null subsets?).

$\endgroup$
3
$\begingroup$

I have just realized that this my question has a simple negative answer: Denote by $\mathbb R_+=[0,\infty)$ the half-line. Observe that the countable product of lines $G=\mathbb R^\omega$ is an Abelian Polish group and the subsemigroup $A=\mathbb R_+^\omega$ is not Haar null in $G$ (since for each compact subset $K$ of $G$ there is an element $g\in G$ such that $g+K\subset A$). Since $A$ is a nowhere dense subsemigroup of $G$ the sum $A+A=A$ is nowhere dense in $G$ (in spite of the fact that $A-A=G$). So, the strong version of the Steinhaus Theorem is false for non-locally compact Polish groups.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.