in folland-real analysis,chapter 11.1, exercise $9$ have been come that:
if the topology of locally compact topological group G is not discrete, then haar measure of singlton is zero
meaning that $\mu(\{x\})=0$ for all $x\in G$.
$\mathbf{EDIT}$: we know that if topology of $G$ is discrete and if $x\in G$, then since $\{x\}$ is compact, then $\mu(\{x\})<\infty$ and since $\{x\}$ is open, then $\mu(\{x\})>0$. hence $0<\mu(\{x\})<\infty$ and translation invariant property of haar measure implies that $\mu(\{x\})=\mu(\{e\})$ for every $x\in G$, where $e$ is the identity and in this case, haar measure is the counting measure. know what happens for $G$ if the topology of $G$ is not discrete, is that $\mu(\{x\})=0$ for every $x\in G$.
can anyone give me a proof?
we know compactness of $\{x\}$ implies that $\mu(\{x\})<\infty$, but how does it become zero?
since every compact set have finite haar measure, if we built an infinite compact set $E$ , then since haar measure is decomposable (definition of decomposable measure is here Haar measures are decomposable) if $\mu(\{x\})>0$, then since compact sets have finite Haar measure, then we have
$\infty>\mu(E)=\underset{x\in E}{\sum}\mu(\{x\})=\infty$
that is contradiction and it's necessary that $\mu(\{x\})=0$, for every $x\in G$, but I can't built this infinite compact set!
can anyone help me?
