# A σ-algebra may have no non-empty atoms at all?

How to prove the fact that a $$\sigma-$$algebra may have no non-empty atoms at all?

Definition: An atom of a $$\sigma$$-algebra $$\mathscr{A}$$ is a non-void set $$\emptyset \neq A \in \mathscr{A}$$ that contains no other set of $$\mathscr{A}$$.

• Maybe the Borel $\sigma$-algebra formed by open sets of $\mathbb{R}$? It has been a while. Potentially relevant might be math.stackexchange.com/questions/3817786/… Feb 18, 2021 at 2:46
• @leslietownes The open sets of $\mathbb R$ do not form a $\sigma$-algebra; they are not closed under complementation or countable intersection. The $\sigma$-algebra generated by the open sets of $\mathbb R$ consists of all the Borel sets, so it contains singletons.
– bof
Feb 18, 2021 at 2:54
• What about the quotient of the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb R$ by the ideal of Lebesgue measure zero sets?
– bof
Feb 18, 2021 at 3:00
• @bof thanks. Your example seems like it might work. Feb 18, 2021 at 4:08

I'm copying this very nice and elementary (but somewhat buried) answer from the following thread: https://mathoverflow.net/questions/22477/sigma-algebra-without-atoms

"A good source for results on atomless $$\sigma$$-algberas is chapter 3 of Borel Spaces (1981) by Rao and Rao (Diss. Math.). Here are two remarks, taken from this text:

1. There is no atomless countably generated $$\sigma$$-algebra.

Let $$\mathcal{C}$$ be a countable set of generators for a $$\sigma$$-algebra on $$X$$. W.l.o.g., we can assume that $$\mathcal{C}$$ is closed under complements. Then for every $$x\in X$$, the set $$A(x)=\bigcap \{C:x\in C,C\in\mathcal{C}\}$$ is measurable in $$\sigma(\mathcal{C})$$ as a countable intersection of measurable sets. Since one cannot separate points by $$\sigma(\mathcal{C})$$ that one cannot separate by $$\mathcal{C}$$, for each $$x$$, $$A(x)$$ is an atom of $$\sigma(\mathcal{C})$$

1. On every uncountable set, there exists an atomless $$\sigma$$-algebra that separates points.

Let $$X$$ be an uncountable set. Clearly, it suffices to show that an atomless $$\sigma$$-algebra that separates points exists on a set with the same cardinality as $$X$$. The set $$Y$$ of all finite subsets of $$X$$ has the same cardinality as $$X$$, so we work with $$Y$$. We verify that the $$\sigma$$-algebra on $$Y$$ generated by elements of the form $$G_x=\{F:x\in F\}$$ for some $$x\in X$$ does the job.

It is obvious that this $$\sigma$$-algebra separates the elements of $$Y$$. So if there were any atom, it would be a singleton $$\{F\}$$ and it would be generated by countably many of the $$G_x$$. So let $$C$$ be a countable subset of $$X$$ such that $$\{F\}\in\sigma\{G_c:c\in C\}$$. For the reason pointed out in 1., $$\{F\}$$ would be the intersection of all elements in $$\{G_c:c\in C\}\cup\{G_c^C:c\in C\}$$ that contain $$F$$. Now $$F\in G_c$$ only if $$c\in F$$. So let $$x\notin F\cup C$$. Then $$F\cup\{x\}\in G_c$$ for all $$c\in F$$ and $$F\cup\{x\}\in G_c^C$$ for $$c\notin F$$. So the intersection doesn't contain only $$F$$, which is a contradiction."

• Because there are other answers voted higher in that MO thread (but which are way more advanced than the question here calls for), I thought it was better to drop this here than just to claim 'duplicate'. Feb 18, 2021 at 3:22