Is a family of disjoints atoms in $\sigma$-finite neasurable space at most countable?

Question

Let $\mu$ be a positive measure on a $\sigma$-algebra $S$ in $X$.

A subset $A\subset X$ is called an atom if $\mu(A)>0$ and for each $E\subset A$ is $\mu(E)=0$ or $\mu(E)=\mu(A)$.

Let's assume that $\mu$ is $\sigma$-finite and consider a family $\cal F$ of pairwise disjoint atoms in $(X, S, \mu)$. Is $\cal F$ at most countable?

Remarks.

Is it not true without $\sigma$-finitness of a measure, for example if $X$ is uncountable, $S=2^X$, $\mu(A)=+\infty$ if $A\neq \emptyset$, $\mu(\emptyset)=0$ and ${\cal F}= \{ \{x\}: x\in X \} $.

It is true if $\mu $ is finite; in this case for each $c>0$ the family ${\cal F}_c=\{A\in {\cal F}: \mu(A)>c \}$ is finite and consequently $\cal F=\bigcup_{n\in \mathbb N} {\cal F}_{\frac{1}{n}}$ is at most countable.

Answer 1

Let $X = \cup_n E_n$, where $\mu(E_n) <+\infty$. Then $E_n$ being a finite measure space, can only intersect countably many of the atoms, by your own argument for finite measure spaces. So $X$ can also have at most countably many atoms in a family.