Why must a discrete atomic measure admit a decomposition into Dirac measures? Moreover, what is "an atomic class"? $\newcommand{\A}{\mathcal{A}}$EDIT: Half of this question may be answered by this post, which remarks that the definitions of atomic space indeed are contradictory in general, so my suspicion that Wikipedia was being lazy was correct. I still would like clarification on the discrete measure and atomic class definitions.
OP:
All of this is a series of quibbles with the very briefly written Wikipedia article on measure-theoretic atoms. Let $(X,\Sigma,\mu)$ be a measure space. An atom is a set $\A\in\Sigma$ for which $0\lt\mu(\A)$ and for every measurable $B\subseteq\A$, either $\mu(B)=0$ or $\mu(B)=\mu(\A)$. The space is defined to be atomic, or purely atomic, if every measurable set of strictly positive measure contains an atom.

Suppose $X$ is $\sigma$-finite. Then if $X$ is atomic, there are disjoint atoms $\{\A_n:n\in\Bbb N\}$ such that $X=\bigsqcup_{n\in\Bbb N}\A_n$.

So, $\sigma$-finitude implies there are disjoint measurable $\{E_n:n\in\Bbb N\}$ with $X=\bigsqcup_{n\in\Bbb N}E_n$, $\mu(E_n)\lt\infty$ for all $n$. The fact that $X$ is atomic implies every single one of the $E_n$ is either null or containing an atom. I note that if any of the $E_n$ are null, this makes the proof really very difficult since no null set can contain an atom. I am not even sure if the statement remains true in this case! Assuming all the $E_n$ are $\mu$-positive, they then all contain atoms $\A_n$.
I cannot see where the argument might lead from here. It is completely possible that $E_n\setminus\A_n\neq\emptyset$ for some $n$, and then in order for the result to hold we need to find a partition of $E_n\setminus\A_n$ into countably many disjoint atoms. However, it is again still possible that $\mu(E_n\setminus A_n)=0$ for the same $n$, at which point we have a problem - there is no set of atoms with $E_n$ as their union. I am dubious that the general result is true at all... how do we show this? I am beginning to think this is more a definition rather than a proof, but Wikipedia claims the two are formally equivalent statements. The reverse direction is not even true, since it possible for an atom to have infinite measure (at least, Wikipedia did not define any otherwise).
Now about discrete measures:

If $\A$ is an atom, all the subsets in the $\mu$-equivalence class of $[\A]$ of $\A$ are atoms, and $[\A]$ is called an atomic class.

I think Wikipedia means by this: the atomic class of $\A$ is the set of all atoms with the same measure as $\A$. However, they seem to think that the $\mu$-equivalence class - my only guess is that this means $E\sim F\iff\mu(E)=\mu(F)$ - of $\A$ will contain only atoms, but this is ridiculous so I think "$\mu$-equivalence" is referring to something else. Perhaps they mean the Nikodym metric: $A\sim B\iff\mu(A\Delta B)=0$ where $\Delta$ is symmetric difference.
So, there is a confusion on the definition of atomic class. Leaving that aside for now, they say that a discrete space is a $\sigma$-finite measure space in which all atomic classes have nonempty self-intersection:
$$\forall[\A],\,\bigcap_{A\in[\A]}A\neq\emptyset$$
But then they claim that this is equivalent to "$\mu$ is the countable weighted sum of Dirac measures", i.e.: $$\mu=\sum_{n=1}^\infty c_n\cdot\delta_{x_n}$$Where $x_n$ is some point in the intersection of each atomic class (there are countably many since they assert that in a $\sigma$-finite space there can only be countably many atomic classes - I am unsure of this because I am unsure of the definition of atomic class). However, to me this automatically implies that the singletons $\{x_n\}$ are all atoms since $\mu(x_n)=c_n$ by definition of $\delta$. This does not follow (in any way that I can see) from the intersection definition of atomic class, so I am suspicious. The reverse implication that discrete spaces are atomic is also not obvious to me, since if we suppose $\mu:=\sum_{n=1}^\infty c_n\delta_{x_n}$ there is no clear way to generate a partition of atoms $\A_n$ which make up $X$ from this.
Summary:

*

*Is $\sigma$-finite and "every set of strictly positive measure contains an atom" really equivalent to "there is a disjoint partition of $X$ into atoms"? Answer: no, not in general. Even if one assumes $\sigma$-finitude of the space, the result can only be proved if one allows for the partition to have up to a null set, or if one includes null sets as atoms (but this is silly to me, since there would no longer be any atomless measures) - see the linked post. I have edited the section of that article accordingly.

*What does Wikipedia really mean when they define atomic class?

*How is discrete - as defined by nonempty intersections of atomic classes - equivalent to the measure being a sum of Dirac measures?

I am aware that the asking of multiple questions is discouraged but these are individually all fairly brief to answer and they are so related that it would feel strange to ask them separately.
 A: $\newcommand{\A}{\mathcal{A}}\newcommand{\M}{\mathcal{M}}\newcommand{\N}{\mathcal{N}}$I was able to fill in the blanks myself:
Let $(X,\M,\mu)$ be a $\sigma$-finite atomic space. By the linked post, we know that $X=\N\cup\bigsqcup_{n\in F}\A_n$ where $\N$ is null (and disjoint from the atoms), $F\subseteq\Bbb N$ and each $\A_n$ is an atom of finite measure disjoint from all the others. I will show both of Wikipedia's claims - there are countably many atomic classes, and if each of the atomic classes has nonempty intersection then $\mu$ is a discrete measure (a sum of Dirac measures).
The $\mu$-equivalence of class of each $\A$ is the set of all $E\in\M$ for which $\mu(\A\Delta E)=\mu(\A\setminus E)+\mu(E\setminus\A)=0$; then $\A\setminus E,E\setminus\A$ are both null.

The equivalence class of an atom consists only of atoms.

Proof:

Let $\A\in\M$, where $\A$ is an atom, and $B\in[\A]$. Then $B=(\A\cap B)\sqcup(B\setminus\A)$ and $\mu(B\Delta\A)=0$ implies that $B\setminus\A$ is null, thus $\mu(\A\cap B)=\mu(B)=\mu(A)\gt 0$ since $B$ is equivalent in measure to $\A$. Let $E\subseteq B$ be measurable. $E=(\A\cap E)\sqcup(E\setminus\A)$; $E\setminus\A\subseteq B\setminus\A$ which is null, hence $\mu(E)=\mu(\A\cap E)$. However, $\A\cap E$ is a subset of the atom $\A$ and either $\mu(E)=0$ or $\mu(E)=\mu(\A)=\mu(B)$. Hence $B$ is an atom.

Claim 1:

A $\sigma$-finite atomic space has countably many atomic classes

Proof:

Let the space be $(X,\M,\mu)$ and have the same decomposition as above. Let $\A\in\M$ be an arbitrary atom. We know that $0\lt\mu(\A)=\sum_{n\in F}\mu(\A\cap\A_n)$, since $\N$ is null, and by atomicity of $\A$, exactly one of the intersections is not null - let then $\mu(\A\cap\A_n)\gt0$ for that unique $n$. For $F\ni m\neq n$, if $\A\in[\A_m]$ then $\A\setminus\A_m$ would be null, hence $\mu(\A)=\mu(\A\cap\A_m)=0$, a contradiction. Thus $\A$ can only be in the atomic class of $[\A_n]$ or in a different one of its own. However, note that $\mu(\A\cap\A_n)\neq0\implies\mu(\A\cap\A_n)=\mu(\A)=\mu(\A_n)$ as $\A_n$ is an atom, and accordingly $\mu(\A\Delta\A_n)=0$ and $\A\in[\A_n]$. Thus the countably many atomic classes $\{[\A_n]:n\in F\}$ account for all atoms of $X$.

Now onto discreteness:

A $\sigma$-finite atomic space for which every atomic class has nonempty intersection is discrete.

Proof:

Let $(X,\M,\mu)$ and $\A_n,\N$ be as above. We have assumed: $$\forall n\in F:\exists x_n\in\bigcap_{\A\in[\A_n]}\A$$
Now let $E\in\M$ be arbitrary. $\mu(E\cap\N)=0$ as $\N$ is null, hence: $$\mu(E)=\sum_{n\in F}\mu(E\cap\A_n)$$
Let $n\in F$ be such that $E\cap\A_n$ is null. Since $\A_n=(E\cap\A_n)\sqcup(\A_n\setminus E)$ we have that $(\A_n\setminus E)\subseteq\A_n$ is not null, and inherits atomicity from $\A_n$; $\mu(\A_n\Delta(\A_n\setminus E))=\mu(\A_n\cap E)=0$ so $\A_n\setminus E\in[\A_n]$. Then by definition: $x_n\in\A_n\setminus E\in[\A_n]$ so $x_n\notin E$.
Now suppose $n\in F$ is such that $\A_n\cap E$ is not null. It follows by atomicity that $\mu(\A_n\cap E)=\mu(A),\,\mu(\A_n\setminus E)=0$ and $\A_n\cap E$ is an atom in $[\A_n]$ by the same arguments as before: then $x_n\in\A_n\cap E$ so in particular we have $x_n\in E$.
Since $x_n\in E\iff\mu(E\cap\A_n)=\mu(\A_n)\gt0$, we find that: $$\mu(E)=\sum_{n\in F:x_n\in E}\mu(\A_n)=\sum_{n\in F}\mu(\A_n)\cdot\delta_{x_n}(E)$$For all $E\in\M$. The proof is then complete, as $\mu$ is the weighted sum of Dirac measures and as such is discrete by the more conventional definitions.

