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 $m(A)&gt;0$, $m(B)&gt;0$.

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.
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 $m(U)&lt;2m(K)$. 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
$$\int m(A\cap xB^{-1}) \,dx = \int\int 1_A(y) 1_B(y^{-1}x)  \,dy\,dx = m(A)m(B)&gt;0$$
by Fubini's theorem, so there is some $x$ such that $m(A\cap xB^{-1})&gt;0$. 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 $x$. $\square$
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 $B=K^{-1}$.