$A$ and $B$ are two measurable subset of $ \mathbb{R}$, and $m$ is a Lebesgue measure on $\mathbb{R}$, if $m(A)>0$ and $m(B)>0$, then there exist a $ x\in \mathbb{R}$, such that $m(A \bigcap (x+B))>0$.
Generally, $G$ is a topological group, $A$ and $B$ are the subsets of $G$, we denote $AB$=$\{ab:a \in A, b\in B \}$.
Let $G$ be a locally compact Hausdorff topological group, $m$ is a left Haar measure on $G$, $A$ and $B$ are two Borel subsets(generated by open subsets) in $G$, $0<m(A)<+ \infty$, $0<m(B)<+ \infty$, then there exist a $ x\in G$,such that $m(A \bigcap xB)>0$.
How to prove it?Is there any reference book about this proposition?
And if $0<m(A) \leq + \infty$, $0<m(B) \leq+ \infty$, then there exist a $ x\in G$, such that $m(A \bigcap xB)>0$?
And if $0<m(A)< + \infty$, $0<m(B) <+ \infty$, then there exist a $ x\in G$, such that $m(A \bigcap Bx)>0$?
And if $0<m(A)< + \infty$, $0<m(B) <+ \infty$, then there exist a $ x\in G$, such that $m(A \bigcap xB^{-1})>0$?
Are these all right?
Thanks in advance.
By regularity of $m$, you may assume that $A$ and $B$ are both compact. Now a simple calculation shows that $$ m(A)m(B) = \int m(A\cap (x-B))dx \qquad (\ast) $$ Now with $A=B$ in $(\ast)$, you see that $$ m(-B) > 0 $$ And with $-B$ replacing $B$ in $(\ast)$ you see that $\exists x\in \mathbb{R}$ such that $$ m(A\cap (x+B)) > 0 $$