Steinhaus theorem for topological groups $G$ 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.

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 $B=K^{-1}$.
