Existence of "smallest" non-empty measurable set

Question

I am wondering if, for any $\sigma$-algebra $F$, there always exists some set $s\in F$ that satisfies the following properties,

1. $s\ne \emptyset$.
2. Except for $\emptyset$ and $s$ itself, there does not exist another set $t\in F$ that $t\subset s$.

It is easy to see that this is true if the $\sigma$-algebra is induced by the topology induced by some metric because a set consisting of a single point satisfies this. But does it hold in general?

Answer 1

Such a set is called an atom of the $\sigma$-algebra. There exist $\sigma$-algebras with no atoms; see this post on MO for some details.