Let $ a \in \mathbb{R}^{N} $ and $ \delta_{a} $ be the corresponding Dirac measure. Which subsets of $ \mathbb{R}^{N} $ are measurable with respect to $ \delta_{a} $? Is $ \delta_{a}$ Borel-measure, Borel-regular, open-regular or even a Radon measure?
The dirac measure is defined as follows: $$\delta_{a}(A)=1_{A}(a)=\left\{\begin{array}{ll}0, & a \notin A \\ 1, & a \in A\end{array}\right.$$ Unfortunately, I don't really have an idea how to show which subsets of $ \mathbb{R}^{N} $ are measurable with respect to $ \delta_{a} $.
I have the following ideas for the other sub-questions:
To show that $ \delta_{a} $ is a Borel measure, I think I would have to show that $ \delta_{a} $ an outor measure, so that all Borel sets are $ \delta_{a} $-measurable.
$ \delta_{a} $ is called Borel regular if additionally each subset $ A \subset \mathbb{R}^{N} $ has a Borel superset $B$ of the same dimension.
$ \delta_{a} $ is called open-regular if $B$ in the case $ \delta_{a}(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$.
$ \delta_{a} $ is called a Radon measure if $ \delta_{a} $ is an open-regular Borel measure for which $ \delta_{a}(K)<\infty$ is compact for all $ K \subset \mathbb{R}^{N}, K$
But how can this be concretised and proven?
All subsets of $\mathbb{R}^{N}$ are $\delta_{a}$-measurable. Let $A, T \subset \mathbb{R}^{N}$ be arbitrary. If $a \notin T$, then $a \notin T \cap A$ and $a \notin T \backslash A$. If $a \in T$, then either $a \in T \cap A$ or $a \in T \backslash A$. In both cases, the following applies $$\delta_{a}(T)=\delta_{a}(T \cap A)+\delta_{a}(T \backslash A).$$ Obviously, all Borel quantities in particular are then measurable, i.e. $ \delta_{a}$ is a Borel measure. Furthermore, for each subset $A \subset \mathbb{R}^{N}$ an open superset with the same measure can be selected, namely $\mathbb{R}^{N} \backslash\{a\}$ if $a \notin A$ and $\mathbb{R}^{N}$ if $a \in A$. Thus $\delta_{a}$ is open-regular. Finally, $\delta_{a}$ is a Radon measure, since $\delta_{a}(K) \leq \delta_{a}\left(\mathbb{R}^{N}\right)=1$ applies to all compacta $K \subset \mathbb{R}^{N}$.