Let $\mathcal{S}$ be a semiring on a set $X$. This means that $\mathcal{S}$ contains $\emptyset$, is closed under intersections, and every relative complement of two sets in $\mathcal{S}$ is a disjoint union of finitely many sets in $\mathcal{S}$.
Let $\mu:\mathcal{S} \to [0,\infty]$ such that $\mu(\emptyset)=0$.
Some terminology:
Saying $\mu$ is finitely additive means that if $A_1,\ldots,A_n$ belong to $\mathcal{S}$ and are disjoint, and if $\bigcup_{i=1}^n A_i$ belongs to $\mathcal{S}$, then $\mu(\bigcup_{i=1}^n A_i) = \sum_{i=1}^{n} \mu(A_i)$.
Saying $\mu$ is countably additive means that if the $A_1,A_2,\ldots$ belong to $\mathcal{S}$ and are disjoint, and if $\bigcup_{i=1}^{\infty} A_i$ belongs to $\mathcal{S}$, then $\mu(\bigcup_{i=1}^n A_i) = \sum_{i=1}^{\infty} \mu(A_i)$. If $\mu$ is countably additive, it is called a premeasure.
Saying $\mu$ is continuous at $\emptyset$ means that if $A_1,A_2,\ldots$ is a decreasing sequence of sets in $\mathcal{S}$ such that $\bigcap_{i=1}^{\infty} A_i = \emptyset$, then $\lim_{i \to \infty} \mu(A_i) = 0$.
Question
Prove or disprove the following statement.
(Q): If $\mu(A)$ is finite for all $A$ in $\mathcal{S}$, if $\mu$ is finitely additive, and if $\mu$ is continuous at $\emptyset$, then $\mu$ is countably additive.
Remark
If $\mathcal{S}$ is a ring (so that it is closed under relative complements), rather than a semi-ring, then (Q) is true and the proof is fairly easy. Ineed, it can found in many textbooks, such as those by Bauer, by Cohn, by Klenke, and by Yeh).
Related
How do i prove that "countably monotone + Finitely Additive" implies "Premeasure" on a semiring?
Best Answer
This is, perhaps somewhat surprisingly, false.
Let $\mathcal{S}$ be the semiring of rational intervals with rational endpoints. More precisely, a member of $\mathcal{S}$ is of the form $$ I\cap \mathbb{Q} $$ where $I$ is some interval with rational endpoints.
Let $$ \mu(I\cap\mathbb{Q})=m(I) $$ where $m$ is the Lebesgue measure (i.e. length).
This function is finitely additive, continuous at $\emptyset$, but not countably additive. Indeed, $$ \mu(\mathbb{Q}\cap [0,1])=1\neq 0=\sum_{q\in \mathbb{Q}\cap[0,1]}\mu(\{q\}) $$