# Prove: if a $\sigma$-algebra over $X$ contains one unique atom $A$, then $A=X$.

Let $$\mathcal{F}$$ be a $$\sigma$$-algebra over some non-empty $$X$$. Given that $$A$$ is the only atom of $$\mathcal{F}$$ (recall $$A \in \mathcal{F}$$ is an atom of $$\mathcal{F}$$ if it's non-empty and no proper subset of $$A$$ other than the empty set is in $$\mathcal{F}$$), show that $$A=X$$.

My attempt: It's clear $$A \subset X$$. To show $$X \subset A$$, suppose for contradiction there's some $$x \in X$$ and $$x \in A^c$$. Let $$B = \bigcap_{S \in \mathcal{F}: x \in S} S$$ (which is clearly non-empty as $$x \in A^c \in \mathcal{F}$$). My idea is to show that $$B$$ is also an atom of $$\mathcal{F}$$ (basically using this idea), and $$A\cap B = \emptyset \implies A \neq B$$, contradicting that $$A$$ is the unique atom. However, I'm stuck on showing $$B \in \mathcal{F}$$, because the intersection defining $$B$$ might not be countable. It's intuitively clear that if $$\mathcal{F}$$ has $$n \in \mathbb{N}$$ distinct atoms, then $$| \mathcal{F}| = 2^n$$ (or at least $$\mathcal{F}$$ should be finite), but I have yet to prove this (it's actually the last part of this problem, and I'm stuck on part 1)...

• Maybe I don't understand your question. Let $x \in X$ and $\{x\}$ be an atom. Then $\mathcal =\{\emptyset, {x}, X-\{x\}, X\}$ is a $\sigma$-algebra and $\{x\} \neq X.$
– UBM
Commented Jan 15, 2021 at 11:47
• @UBM: The catch is that if $\{x\}$ is the single unique atom of the sigma algebra, then it kind of forces $\{x\} = X$. Let $x=2, X=\{2, 3\}$ in your example, then your sigma algebra ends up with an extra atom $X - \{x\} = \{3\}$. Commented Jan 15, 2021 at 17:23
• yes, you are right. I misunderstood your definition of atom.
– UBM
Commented Jan 15, 2021 at 18:35

As I understand it, the problem as stated is false.

Let $$Y$$ be a set with an atomless $$\sigma$$-algebra $$\mathcal{G}$$ (see here for examples). Let $$X = Y \sqcup \{a\}$$ for some $$a$$. Define a $$\sigma$$-algebra $$\mathcal{F}$$ on $$X$$ by $$E \in \mathcal{F}$$ if and only if $$E \backslash \{a\} \in \mathcal{G}$$. In other words, $$E \in \mathcal{F}$$ exactly when $$E \in \mathcal{G}$$ or $$E$$ is the union of a set in $$\mathcal{G}$$ with $$\{a\}$$.

In this scenario, we have that $$\{a\}$$ is an atom. On the other hand, every other nonempty set $$E \in \mathcal{F}$$ satisfies $$E \backslash \{a\} \neq \emptyset$$, and since $$\mathcal{G}$$ is atomless we can find $$\emptyset \neq F \subsetneq E$$ with $$F \in \mathcal{G}$$, which also means $$F \in \mathcal{F}$$. Thus, $$E$$ contains a proper nonempty subset in $$\mathcal{F}$$, so is not an atom.

As noted in the comments below, the problem works out if the $$\sigma$$-algebra is countably generated. In this case, we can follow the idea in the question, making just one modification.

Suppose $$\mathcal{F}$$ is generated by some countable collection $$\mathcal{E}$$. We might as well assume $$\mathcal{E}$$ is an algebra, since the algebra generated by a countable collection is still countable. With this in hand, we can restrict our intersection defining B to sets $$S \in \mathcal{E}$$, and B should still be an atom.

• Thanks, I think that makes sense. Would the problem have worked if we required the sigma algebra in question to be countably generated? Commented Jan 15, 2021 at 19:17
• Yes, every countably generated sigma algebra has an atom, and in the construction above $\mathcal{F}$ is countably generated if and only if $\mathcal{G}$ is. Commented Jan 15, 2021 at 19:19
• I should add that your original idea goes through if the sigma algebra is countably generated, since the intersection defining your $B$ can be made to be countable in this case. Commented Jan 15, 2021 at 21:23
• Could you elaborate on how the $B$ would be constructed in this case? Commented Jan 15, 2021 at 21:57