Suppose, $G$ is a locally compact topological group. $\mu$ is Haar measure on that group. $A$ and $B$ are Borel subsets of $G$, such that $\mu(A)$ and $\mu(B)$ are finite. Does the inequality $\mu(\{ab | a \in A, b \in B\}) \leq \mu(A)\mu(B)$ always hold?
The question seems to be correct as it is quite easy to prove by induction, that if $A$ and $B$ are Borel sets, then $\{ab | a \in A, b \in B\}$ is also a Borel set. However, I do not know, how to solve this problem.
Any help will be appreciated.
This does not hold. Take $G=\{0,1\}$ with addition modulo 2 the counting measure times $1/100$ (this is a Haar measure indeed). So the measure of $G$ is $1/50$ the measure of $\{0\}$ is $1/100$ and so on.
Now let $A=\{0\}$ and $B=\{1\}$. Then $\mu(A)\cdot \mu(B)=1/10,000$ and it's clearly less than the measure of $A+B = G$.
Edit: It turns out that you don't even have to multiply by $1/100$ any normalized measure would work because $A\subseteq A+B$ and $\mu(B)<1$.