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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?

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    $\begingroup$ You have a typo in the definition of "continuous at $\emptyset$". $\endgroup$ – WhatsUp Dec 30 '19 at 3:14
  • $\begingroup$ @WhatsUp Thanks. Fixed. $\endgroup$ – JasonJones Dec 30 '19 at 3:15
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    $\begingroup$ Not fixed... should be $\bigcap$ instead of $\bigcup$. $\endgroup$ – WhatsUp Dec 30 '19 at 3:16
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    $\begingroup$ @WhatsUp You can fix obvious typos yourself, as I have. $\endgroup$ – Kevin Arlin Dec 30 '19 at 3:27
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    $\begingroup$ @KevinCarlson Ah yes... still not used to this privilege (: $\endgroup$ – WhatsUp Dec 30 '19 at 3:30
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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, $$ \mu(\mathbb{Q}\cap [0,1])=1\neq 0=\sum_{q\in \mathbb{Q}\cap[0,1]}\mu(\{q\}) $$

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  • $\begingroup$ Could you please give an example of a sequence on which it is not countably additive? $\endgroup$ – JasonJones Dec 30 '19 at 4:04
  • $\begingroup$ @JasonJones Done $\endgroup$ – Reveillark Dec 30 '19 at 4:10
  • $\begingroup$ Notably, this function is continuous from below as well. So replacing continuity at $\emptyset$ by continuity from below does not save the statement. Wonderful. $\endgroup$ – JasonJones Dec 30 '19 at 4:17
  • $\begingroup$ @JasonJones and continuous from above too :) $\endgroup$ – Reveillark Dec 30 '19 at 4:19
  • $\begingroup$ Sorry for this. But I just had a doubt. How is $\{q\}$ a set of the form $\{q \in \mathbb{Q}: a \leq q < b\}$? $\endgroup$ – JasonJones Dec 30 '19 at 4:21

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