I want to show that the counting measure $\zeta$ on $\mathbb{R}^{N}$ is Borel regular.
The counting measure is defined as follows:
In addition, I know that the counting measure $\zeta$ is not $\sigma$-finite on $ \mathbb{R}^{N}$.
An outer measure $\mu$ is called Borel measure if all open and thus all Borel sets $\mu$-are measurable. $\mu$ is called Borel-regular if in addition each subset $A \subset \mathbb{R}^{N}$ has a Borel superset $B$ with the same measure. $\mu$ is called open-regular if $B$ in the case $\mu(A)<\infty$ can be chosen in the form $ \bigcap_{n=1}^{\infty} O_{n}, O_{n}$ open with $A \subset O_{n}, \mu\left(O_{n}\right)<\infty$.
How can I now clearly show that the counting measure $\zeta$ is on $\mathbb{R}^{N}$ Borel regular. Is $\zeta$ also open-regular?
"$\zeta$ is Borel-regular" - All sets are $\zeta$-measurable, because $\zeta_{\mathbb{R}^{N}}$ is unrestrictedly additive. A set of finite measure is closed, i.e. a Borel set. The counting measure is not open-regular, since all non-empty open subsets of $\mathbb{R}^{N}$ have infinite measure and there are non-empty sets $A \subset \mathbb{R}^{N}$ with $\zeta(A)<\infty$. In the definition of open regularity, however, the existence of open supersets $O \supset A$ with finite measure is required in particular.