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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)...

Yibo Yang
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  • 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 Jan 15 '21 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 $X - {x} = {3}$. – Yibo Yang Jan 15 '21 at 17:23
  • yes, you are right. I misunderstood your definition of atom. – UBM Jan 15 '21 at 18:35

1 Answers1

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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.

Michael Jesurum
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    Thanks, I think that makes sense. Would the problem have worked if we required the sigma algebra in question to be countably generated? – Yibo Yang Jan 15 '21 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. – Michael Jesurum Jan 15 '21 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. – Michael Jesurum Jan 15 '21 at 21:23
  • Could you elaborate on how the $B$ would be constructed in this case? – Yibo Yang Jan 15 '21 at 21:57