Show that $\mu$ is a measure where $\mu(A)$ is the Dirac measure at $x=0$

Question

Let $M$ consist of Borel subsets $A$ of $\mathbb{R}$. Furthermore, define the measure $\mu : M \rightarrow \mathbb{R}$ where $\mu(A) = \chi_A(0)$ and $\chi_A$ is the characteristic function on $A$

\begin{align*} \chi_A(0) = \begin{cases} 1, 0 \in A \\ 0, 0 \notin A \end{cases} \end{align*}

In other words, this is the Dirac measure at $x=0$.

How does one proceed to verify that $\mu$ is a measure?

Answer 1

This $\mu$ is called the dirac measure at $0$.

It clearly satisfies $\mu(\emptyset)=0$. For a countable union of disjoint sets, $A_n$, if they all don't contain $0$, then $0 \notin \displaystyle\bigcup A_n$ so $\mu\left(\displaystyle\bigcup A_n\right)=0 = \sum \mu(A_n)$. If $0 \in A_k$ for some $k$ then by the pairwise-disjoint assumption we know that $\mu(A_n) =0$ for all $n\neq k$. So $$\mu\left(\bigcup A_n\right) = 1 = \sum_n \mu(A_n) = \mu(A_k).$$

Hence the measure is countably additive.