Set of atoms must be countable My task is to prove that if an atomic measure space is $\sigma$-finite, then the set of atoms must be countable.
This is my given definition of an atomic measure space:

Assume $(X,\mathcal{M},\mu)$ is a measure space with all single points being measurable. An atom is a point $x$ with
$\mu(\{x\}) > 0$. Letting $\mathcal{A}$ be the set of atoms,
$(X,\mathcal{M},\mu)$ is called atomic if $\mathcal{A}\in\mathcal{M}$
and $\mu(\mathcal{A^c}) = 0$.


I didn't know how to prove this at first, so I looked it up on stack exchange and found this answer: (I do not have enough reputation to comment on the original post)

Here's how to prove your claim, with the appropriate assumption. Let
$S\subset X$ be the set of atoms for some measure $\mu$ on $X$. Let
$\{U_i\}$ be a countable measurable partition of $X$. Then if $S$ is
uncountable, some $U_i$ contains an uncountable subset $S'$ of $S$,
and $\mu(U_i)\geq \sum_{x\in S'}\mu(x)=\infty$ since any uncountable
sum of positive numbers diverges. Thus $\mu$ is not $\sigma$-finite.

My question is why do we have that $\mu(U_i) \geq \sum_{x\in S'} \mu(x)$ ? I am assuming that this inequality comes from subadditivity of $\mu$
but as I have understood it subadditivity is defined for countable unions, not for uncountable unions so I am confused as to how we arrive at an uncountable sum in this step.
 A: That does appear to be a bit sloppy. But you can mend that by simply specifying the sum over any countable selection and taking the supremum of that. You can easily show: If you have an uncountable family of positive values, then for any countable finite sum there exists an even larger one, so this sup must be $\infty$ (for else take a sequence of such countable selections so that the sum converges to the sup. Then the union of all these selections is also a countable selection, and at least as large as the limit).
A: You may argue as follows. Since $U_i\supset S'$ for some $i\in\mathbb{N}$,
$$
\mu(U_i)\ge \mu(S'')\label{1}\tag{1}
$$
for any $S''\subseteq S'$. Let $\mathcal{S'}$ be the collection of all sequences of points in $S'$. Then,
$$
\mu(U_i)\ge \sup_{s\in\mathcal{S'}}\sum_{x\in s}\mu(\{x\})\equiv\sum_{x\in S'}\mu(\{x\})=\infty,
$$
where the inequality follows from \eqref{1} and countable additivity, and the last equality holds because there exists $\epsilon>0$ s.t. $\mu(\{x\})\ge \epsilon$ for uncountably many $x$'s in $S'$.
A: Here I will take "countable" to mean finite or countably infinite.
For $n\in\{1,2,3,\ldots\},$ if the set $A_n = \{ x : \mu(\{x\}) \ge 1/n \}$ were not countable then the space would not be $\sigma$-finite.
Since $(0,+\infty)= \bigcup_{n=1}^\infty \left(\frac 1 n, +\infty\right),$ the set $\bigcup_{n=1}^\infty A_n$ is the set of all atoms. This is a countable union of countable sets.
