Steinhaus theorem for topological groups

Question

$G$ is a locally compact Hausdorff topological group, $m$ is a (left) Haar measure on $X$, $A$ and $B$ are two finite positive measure in $G$, that is $m(A)>0$, $m(B)>0$.

My question is:

Can we conclude that $AB= \{ab, a\in A, b\in B\}$ contains some non-empty open set of G?

Is this question right? Or is this right just for $G=R^n$, $R^n$ is the Euclid space, and $m$ is the Lebesgue measure on $R^n$. If so, how to prove it?

Thanks a lot.

Answer 1

Here is another proof, using regularity of the measure instead of convolution.

Claim: The result holds when $B=A^{-1}$.

Proof: By regularity there is a compact set $K$ and an open set $U$ such that $K\subset A\subset U$ and such that $m(U)<2m(K)$. The multiplication map sends $\{1\}\times K$ into $U$, so by continuity of multiplication and compactness of $K$ there is a neighbourhood $V$ of $1$ such that multiplication sends $V\times K$ into $U$. But then if $x\in V$ the sets $K$ and $xK$ are each more than half of $U$, so $K\cap xK$ is nonempty, so $x\in KK^{-1}$. Thus $KK^{-1}$ contains a neighbourhood $V$ of $1$. $\square$

Claim: The result holds in general.

Proof: By regularity we may assume both $A$ and $B$ are compact. For $x$ running over $G$ we have $$\int m(A\cap xB^{-1}) \,dx = \int\int 1_A(y) 1_B(y^{-1}x) \,dy\,dx = m(A)m(B)>0$$ by Fubini's theorem, so there is some $x$ such that $m(A\cap xB^{-1})>0$. Now apply the previous result to $A\cap xB^{-1}$. Since $$(A\cap xB^{-1})(A\cap xB^{-1})^{-1} \subset ABx^{-1},$$ we deduce that $AB$ contains a neighbourhood of $x$. $\square$

Answer 2

You can prove this by considering the convolution $f(x) = \int_G \chi_{A}(xy^{-1}) \chi_{B}(y) dy$ of the characteristic functions of $A, B$. WLOG, we can take $A, B$ to be compact sets of positive measure. So $f$ is continuous and has positive integral. That $AB$ has interior immediately follows.

Alternatively, you can also use the regularity of Haar measure to show this directly.

Answer 3

You can take this example if the topological group $G$ has finite Haar measure, then you can construct a compact set $K$ satisfies for every $x \in G$, you have $K \cap xK \neq \phi$ see proposition 1.4.5 in the book of principles of Harmonic analysis of Deitmar and Echterhoof. Then you can easily show that $G=KK^{-1}$ then you can take $A=K$ and $B=K^{-1}$.