Countable partition in atoms Let $\mu: \Sigma \to [0, \infty)$ a measure over $\Omega$. We say a set $A \in \Sigma$ is an atom if for all $B \in \Sigma$ with $B \subset A$, $\mu(B)=\mu(A)$ or $\mu(B)=0$. We say that $\mu$ is atomic if every set $A \in \Sigma$ with positive measure contains an atom with positive measure.
So, I'm trying to prove that if $\mu$ is atomic, there exists $\{ A_n\}_{n \in \mathbb{N}}$ pairwise disjoint atoms such that its union covers $\Omega$.
I can build a sequence this way:
Let $A \in \Sigma$ be a set with positive measure. It contains an atom $A_1$.Then I pick up $\Omega \setminus A_1$. Again, it contains an atom $A_2$ which is disjoint from $A_1$. 
However, I do not know how to continue. I would appreciate any help.
 A: For finite measures:
Proof: If $\mu(\Omega) = 0$ the result is trivial, because $\Omega$ itself is an atom. Assume $\mu(\Omega) > 0$. Since $\mu$ is atomic, take $B_0=\Omega$ and define by induction, for all $n\in \mathbb{N}$, if $\mu(B_n)>0$ then let $A_n \subseteq B_n$ be an atom with the largest possible measure among the atoms contained in $B_n$ (such choice of $A_n $ is always possible because the measure is finite) and let $B_{n+1}=B_n\setminus A_n$. 
One of two things may happen: 


*

*For some $n_0\in  \mathbb{N}$, $\mu(B_{n_0+1})=0$ and in this case, $B_{n_0+1}$ is an atom, $A_0, \dots, A_{n_0}, B_{n_0+1}$ are disjoint atoms  and we have 
$$\Omega = A_0 \cup \dots \cup A_{n_0} \cup B_{n_0+1}$$
which proves the result (for this case). 

*For all $n\in  \mathbb{N}$, $\mu(B_{n})>0$. In this case we have an infinite sequence $\{A_n\}_{n\in  \mathbb{N}}$ of disjoint atoms of positive measures. Let 
$$ A= \bigcup_{n\in  \mathbb{N}} A_n$$. 
Claim: $\mu(\Omega \setminus A)=0$ 
Proof of the claim: if $\mu(\Omega \setminus A)>0$ the there is an atom $E\subseteq \Omega \setminus A$ such that $\mu(E)>0$. By our choice of $A_n$ at each step, we have that, for all $n\in  \mathbb{N}$ $\mu(A_n) \geqslant \mu(E)$. So $$ \mu(A)=\sum_{n\in  \mathbb{N}}\mu(A_n) =\infty$$
Contradiction, since the measure is finite. 
So we proved  $\mu(\Omega \setminus A)=0$. So we have that $\Omega \setminus A$ is (trivially) an atom and we have 
$$\Omega =  \left ( \bigcup_{n\in  \mathbb{N}} A_n \right ) \cup (\Omega \setminus A)$$
So $\Omega$ is covered by a countable collection of atoms and this proves the result for this case and completes the proof.
Remark: if the measure is not finite, this result may not hold.
Consider $(\mathbb{R}, P(\mathbb{R}), \mu)$ where $\mu$ is the counting measure. Every singleton is an atom of positive measure. Any set with positive measure is not empty, so it contains a singleton, which means, it contains an atom of positive measure. So $\mu$ is atomic. 
Note that the only atoms are the singletons (and the empty set), and clearly $\mathbb{R}$ is not a countable union of singletons. 
Remark 2: Let us prove that, since $\mu$ is finite, the for any set $A$, if $A$ contain an atom, then there is an atom $E\subseteq A$ such that for any atom $F \subseteq A$, $\mu(F)\leqslant \mu(E)$.
Suppose  $A$ contain an atom. Let us prove the result by contradiction. Suppose that for all an atom $E\subseteq A$, there is an atom $F \subseteq A$, $\mu(F)> \mu(E)$. 
Let $E_0\subseteq A$ is an atom. Then there is sequence of atoms $E_n\subseteq A$ such that for all $n\in \mathbb{N}$ $\mu(E_n) < \mu(E_{n+1})$.
Given any $i<j$, since $E_j$ is an atom and $\mu(E_i\cap E_j)\leqslant\mu(E_i) < \mu(E_j)$, we have that  $\mu(E_i\cap E_j)=0$. So we have
$$\mu \left ( \bigcup_{n\in  \mathbb{N}}E_n\right)= \sum_{n\in  \mathbb{N}}\mu(E_n)=+\infty$$
Contradiction, because the measure is finite.
So we proved that there is an atom $E\subseteq A$ such that for any atom $F \subseteq A$, $\mu(F)\leqslant \mu(E)$.
A: Hints. First show that the set of atoms is countable. The show it must be (essentially) the whole space.
