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 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.
