Signed measures on algebras (fields) and their boundedness properties

Question

I asked this question here on math.StackEchange, but it might be too technical so I re-post it here.

Let $X$ be a compact Hausdorff second countable topological space. Let $\mathcal{B}$ a countable basis of the topology of $X$, which we can assume to be closed under finite unions and intersections, and let $\mathcal{A}$ be the ring induced by $\mathcal{B}$. We assume that $X\in\mathcal{B}$, so that $\mathcal{A}$ is an algebra. Let $\mu$ be a real measure on $\mathcal{A}$ (in the notation of Rao and Rao "Theory of Charges", 1983), namely $\mu$ is a countably additive set function on $\mathcal{A}$ with values in $(-\infty,\infty)$ and $\mu(\emptyset)=0$.

By Theorem 2.5.3 in (Rao and Rao "Theory of Charges", 1983) we know that $\mu$ has a Jordan decomposition composed by two positive measures $\mu^+$ and $\mu^-$ on $\mathcal{A}$. Denote by $|\mu|:=\mu^++\mu^-$, which is a positive measure on $\mathcal{A}$. From the same theorem we know that $|\mu|$ is bounded if and only if $\mu$ is bounded.

Assume that $|\mu|(A_n)\to0$ for every sequence of sets $(A_n)_{n\in\mathbb{N}}$ such that $A_n\in\mathcal{A}$ (for every $n\in\mathbb{N}$) and $A_n\downarrow\emptyset$.

Question: Can we conclude that $|\mu|$ is bounded? If not, is there a counterexample?

Answer 1

I already answered on MSE. Here is the answer, copied here as well:

Here is a counterexample, if I did not misunderstand the notations:

Let $X = \mathbb{N} \cup \{\infty\}$ be the one-point compactification of the naturals. Let $\mathcal{B}$ consists of finite subsets of $\mathbb{N}$ as well as their complements, which is indeed a countable basis of topology for $X$. $\mathcal{B}$ is already closed under finite unions, finite intersections, and taking complements, so $\mathcal{A} = \mathcal{B}$. Let $\mu$ be defined by $\mu(A) = |A|$ if $A$ is finite and $\mu(A) = -|X \setminus A|$ if $A$ is co-finite. Then $\mu$ is a real measure, in the sense you wrote down. (We note that there is no countable collection of disjoint subsets of $\mathcal{A}$ whose union is in $\mathcal{A}$ unless all but finitely many of them are empty, since all finite sets in $\mathcal{A}$ do not contain $\infty$ but all infinite sets in $\mathcal{A}$ are co-finite and contain $\infty$. So, countable additivity just reduces to finite additivity, which is easy to verify.)

If I’m not mistaken, $\mu^+$ is simply the counting measure and $\mu^-$ is $0$ on finite sets and $\infty$ on co-finite sets, so $|\mu|$ is the counting measure. Regardless, $\mu$ and $|\mu|$ are certainly not bounded. But if $A_n \downarrow \varnothing$, since all infinite sets in $\mathcal{A}$ contain $\infty$, we must have $A_n$ are eventually finite, and thus eventually the empty set, so you do have $|\mu|(A_n) \to 0$.