Finite atomic measure space can be a countable disjoint union of atoms An atom A in a measure space is a measurable set with positive measure such that every measurable subset of A has the same measure of A or 0.
An atomic measure is a measure space such that in every measurable set there is an atom.
Let's say that I have a positive atomic measure space $(X, \Sigma, \mu)$ such that $\mu(X)<\infty$.
I want to prove that X is a countable disjoint union of atoms and a set with zero measure.
My attempt:
X is a measurable set therefore exist an atom $A_1 \subseteq X$, if $\mu(X/A_1) = 0$ we are done, if not we define $A_2$ as the sub atom of $X/A_1$ and we precede this way for every n. What I need to prove is that $\mu(X/\bigcup_{n=1}^{\infty}A_n) = 0$.
 A: Am not sure if you can prove that $\mu\bigl(X\setminus \bigcup_{n=1}^{\infty}A_n\bigr)=0$ since you dont have any condition in your selection of the $A_n$'s, but this can be done if you strenghthen your recursive construction of the $A_n$'s.
The above argument is an example of an "exhaustion argument", we will "exhaust" all the of the atoms of $X$.
Let $\mathcal{A_1}=\{A\in \Sigma:\, A\subseteq A,\ A\ \text{is an  atom}\}$ and $\alpha_1=\sup_{A\in \mathcal{A_1}}\mu(A)>0$. Then, we find an atom $A_1\subseteq X$ such that $\mu(A_1)\geq 2^{-1}\alpha_1$ (this is our condition). As you said, if for every $B\subseteq X\setminus A_1$ we have $\mu(B)=0$ then we write $X=A_1\cup B$ and we are done. Suppose now, that $X$ cannot be written as finite disjoint union of atoms and a set of zero measure. Then continuing as before, recursively we find a sequence $A_n$ of atoms such that
$1)$ $\mu(A_{n+1})\geq 2^{-1}\alpha_{n+1}$
$2)$ $\alpha_{n+1}=\sup_{A\in \mathcal{A_{n+1}}}\mu(A)$
$3)$ $\mathcal{A_{n+1}}=\{A\in \Sigma:\, A\subseteq X\setminus(A_1\cup...\cup A_{n}),\,\ $A$\, \text{is an atom}\}$
Now, if $A=\bigcup_{n=1}^{\infty}A_n$ we will show that $\mu(X\setminus A)=0$. Since, the $A_n$'s are disjoint by $(1)$ we have
$$\mu(A)=\sum_{n=1}^{\infty}\mu(A_n)\geq \sum_{n=1}^{\infty}\frac{\alpha_n}{2}$$
Now, $\mu$ being finite implies that $\alpha_n\to 0$ as $n\to \infty$. Suppose now that $X\setminus A$ has positive measure. Then, $X\setminus A$ would contain an atom, say $B$. But $B\subseteq X\setminus A$ implies that $B\subseteq X\setminus (A_1\cup ...\cup A_{n})$ for every $n$. So, since $B$ is an atom it follows that $B\in \mathcal{A_{n+1}}$. Hence, by the definition of the $\alpha_n's$ we must have $\mu(B)\leq \alpha_{n+1}$ for every $n$. So, $B$ must have zero measure, which contradicts the fact that $B$ is an atom and must have a positive measure.
