A question about discrete measure. I have just started to read measure theory from very initial level. The note I am following refers to an example:
Atom: If $(X, \mathcal{X}, \mu)$ is a measure space, an atom of $\mu$ is a subset $A$ of $\mathcal{X}$ such that
$(i)$ $\mu(A)>0$.
$(ii)$ if $B\subset A$, then either $\mu(B)=\mu(A)$ or $\mu(B)=0$.
The measure space $(X, \mathcal{X}, \mu)$ is called discrete, if
$$X=Z\sqcup \bigsqcup\limits_{n=0}^\infty A_n,$$
where $\mu(Z)=0$ and $\{A_n\}_{n=0}^\infty$ is a collection of atoms.
Now the note asks to prove the following
If $X$ is countable, then for any measure $\mu$, the space $(X,\mathcal{X}, \mu)$ is discrete.
I have no idea, how to proceed! Thanks in advance.
 A: I'll use the convention $\bigcap\emptyset=X$. For $x\in X$, define:
$$A(x):=\bigcap\{A\in\mathcal X|x\in A\}$$
For any $y\in X\setminus A(x)$, there is $A_y\in\mathcal X$ such that $x\in A_y$ and $y\notin A_y$. Using the axiom of choice, this yields a map $X\setminus A(x)\to\mathcal X, y\mapsto A_y$. The following holds:
$$A(x)\subset\bigcap_{y\in X\setminus A(x)} A_y\subset\bigcap_{y\in X\setminus A(x)} X\setminus\{y\}=X\setminus\bigcup_{y\in X\setminus A(x)} \{y\}=X\setminus (X\setminus A(x))=A(x)$$
Since $X$ is countable, so is $X\setminus A(x)$. Therefore, $A(x)=\bigcap_{y\in X\setminus A(x)} A_y\in\mathcal X$.
If $A\subset A(x)$ is measurable and $x\in A$, then $A(x)\subset A$, if $x\notin A$, then $A(x)\subset A(x)\setminus A$. Therefore, the only measurable subsets of $A(x)$ are $\emptyset$ and $A(x)$. In particular, $A(x)$ is either a null set or an atom. Furthermore, for any $x,y\in X$, either $A(x)=A(y)$ or $A(x)\cap A(y)=\emptyset$.
Define the following set:
$$\mathcal A:=\{A(x)|x\in X\}$$
Since $A:X\to\mathcal A$ is surjective, $\mathcal A$ is countable. The following holds:
$$X=\bigsqcup\mathcal A=\underbrace{\bigsqcup\{A\in\mathcal A|\mu(A)=0\}}_{=:Z}\sqcup \bigsqcup\{A\in\mathcal A|A\text{ atom}\}$$
