fabios3

2022-06-27

I`m trying to prove that:
$\mu \left(A\right)=\sum _{n\in \mathbb{N}}{\delta }_{\frac{1}{n}}\left(A\right)$
is not a regular measure.
A measure is regular if:

for any measurable $A$.
I'm trying to find a counter example for that but can't come up with any.
I was thinking about a set which contains a finite number of 1/n but can only be covered by $\left(0,ϵ\right)$ for some $ϵ>0$, leading to a contradiction.

Expert

Take $A=\left\{0\right\}\subseteq \mathbb{R}$. For every open set $G$ containing $A$ there exists an $\epsilon >0$ such that $\right]-\epsilon ,\epsilon \left[\subseteq G$. This open interval contains infinitely many $\frac{1}{n}$. Thus $\mu \left(G\right)\ge \mu \left(\right]-\epsilon ,\epsilon \left[\right)=\mathrm{\infty }$. But $\mu \left(A\right)=0$. This shows that $\mu$ is not outer regular.