an existence question from topological groups

Question

$G$ 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.

Answer 1

Here is a counterexample when you allow the sets to have infinite measure. Let $H$ be an uncountable discrete group and $G=\mathbb{R}\times H$. If $B$ is any set such that $B\cap \mathbb{R}\times \{h\}$ is closed for all $h\in H$ and nonempty for uncountably many $h\in H$, then $B$ is closed and $m(B)=\infty$ by outer regularity (any open neighborhood of $B$ must have non-null intersection with uncountably many $\mathbb{R}\times\{h\}$ and hence have infinite measure). Now let $A=[0,1]\times \{1\}$ and let $B$ be the graph of some bijection between an uncountable subset of $\mathbb{R}$ and an uncountable subset of $H$. Then $m(A)=1$ and $m(B)=\infty$, but any translate of $B$ will intersect $A$ in at most one point.

It is no coincidence that this example involves uncountable pathology: this can't happen if $G$ is $\sigma$-compact. Indeed if $G$ is $\sigma$-compact, then its Haar measure is $\sigma$-finite, so if $B$ has positive measure, countably many translates of $B$ must cover almost all of $G$. By countable additivity, every set of positive measure must have positive-measure intersection with one of these translates.

This also can't happen if $A$ and $B$ have finite measure. Indeed, I claim that any Borel set $C$ of finite measure is contained in a $\sigma$-compact open subgroup $G_0$, so you can apply the argument of the previous paragraph to such a $G_0$ containing $C=A\cup B$. To show this, let $K$ be any compact neighborhood of $1$; the subgroup $H$ generated by $K$ is then a $\sigma$-compact open subgroup of $G$. By outer regularity, any Borel set $C$ of finite measure can only intersect countably many cosets of $H$ (since any neighborhood of $C$ must have non-null intersection with every coset of $H$ that $C$ intersects). We can then take $G_0$ to be the subgroup generated by these countably many cosets.

This argument also gives a positive answer to your last two questions, since we can work in $G_0$ and the sets $Bx$ and $xB^{-1}$ still all have positive measure and so countably many of them cover almost all of $G_0$.

Answer 2

Let $1_A$ be the indicator function of $A$ and $1_{B^{-1}}$ the one of $B^{-1}$. Then $$ 1_A*1_{B^{-1}}(x)=m(A\cap xB). $$ But for the L^1-norm we have $\parallel 1_A*1_{B^{-1}}\parallel=\parallel 1_A\parallel \ \parallel 1_{B^{-1}}\parallel=m(A)m(B^{-1})$. Now if $B$ has positive measure, ten so has $B^{-1}$ and therefore there exists $x$ with $m(A\cap xB)\ne 0$.