I read the following exercise:
Prove that if $G$ is a locally compact topological group with Haar measure $\mu$ and $A \subset G, \mu (A) >0$, then $AA^{-1}$ contains an open neighborhood of the identity $e\in G$. The hint says something about a convolution being a function of finite type associated to the regular representation of $G$. I don't know anything about functions of finite type. Surely there's a more elementary proof?
I have seen the following argumentation, we use inner and outer regularity of the Haar measure. By inner regularity, we may show the claim for a compact $ K \subset A $. Since $ K $ is compact $ \mu(K) $ is finite, so by outer regularity there is an open set $ U \supset K $ such that
$ \mu(K) \leq \mu(U) < 2 \mu(K) $
A basic topological group argument shows that there is a neighborhood of the identity $ W $ such that $ WK \subset U $. We have for all $ w \in W $ that $ wK \cap K \neq \emptyset $ (otherwise we get a contradiction in the above inequality). Thus we have $ W \subset K K^{-1} $.
