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A semiring $\Pi$ on a set $X$ is a non-empty family of subsets of $X$ with the following properties.

1) $P \cap Q \in \Pi$ whenever $P\in \Pi$ and $Q\in\Pi$.

2) $P - Q$ is a finite disjoint union of members of $\Pi$ whenever $P\in \Pi$ and $Q\in\Pi$.

Let $\lambda: \Pi\rightarrow [0, \infty]$ be a non-negative extended real-valued set function such that $\lambda(\emptyset) = 0$. Suppose $\lambda$ is additive, i.e. if $P, Q_1, Q_2$ are members of $\Pi$ and $P = Q_1\cup Q_2$ is a disjoint union, then $\lambda(P) = \lambda(Q_1) + \lambda(Q_2)$.

Then $\lambda$ is finitely additive? In other words, if $P, Q_1, \cdots, Q_n$ are members of $\Pi$ and $P = \bigcup_{i=1}^n Q_i$ is a disjoint union, then $\lambda(P) = \sum_{i=1}^n \lambda(Q_i)$?

I came across this problem when trying to solve this question: Explicit construction of Haar measure on a profinite group

I also found this is an exercise in Halmos's Measure Theory, Ch. II, sec. 7. It has some hints, but I was unable to solve it.

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JHW
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  • The definition of semiring as given in Halmos is not just that $P-Q$ is a finite disjoint union of members of $\Pi$, but that we can order these members of $\Pi$ as $D_1, D_2, \dots, D_n$ such that $Q \cup D_1 \cup D_2 \cup \dots \cup D_i \in \Pi$ for each $i=1,\dots,n$. – Misha Lavrov Dec 06 '17 at 04:07

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The answer is no:

Consider $X=\{a,b,c\}$ and $\Pi=\{\emptyset,\{a\},\{b\},\{c\},X\}$ with $$\lambda(\{a\})=\lambda(\{b\})=\lambda(\{c\})=\lambda(X)=1\,.$$ The cheat is that $\Pi$ doesn't contain any nontrivial triple of sets with $P=Q_1\cup^* Q_2$, so the 'additivity' (rather 2-additivity) holds vacuously.

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  • Great. So Halmos was wrong at his exercise. – JHW Aug 30 '15 at 01:03
  • The answer above is wrong. Semiring has a property that if $F-E=D_1 \cup D_2 \cup\dotsb\cup D_n $ then $D_i \cup D_{i+1} \in \prod$, so it is not a semiring. The proof of Halmos' exercise is that for each partition of $E$, for $E_i \subseteq E$ we can consider a disjoint sequence ${D_{ik} }$ which is proved to be a $\mu $-partition, and consider the intersection of all these $\mu $-partitions. – TheWildCat Dec 06 '17 at 02:54
  • It's not the answer that's wrong, but the question: the definition of a semiring is misstated. (I think your version of the definition is not quite right either, and that we want $E \cup D_1 \cup \dots \cup D_i \in \Pi$, instead. Halmos instead defines a chain $E = C_0 \subset C_1 \subset \dots \subset C_n=F$ such that $C_i \in \Pi$ and $D_i = C_i\setminus C_{i-1} \in \Pi$.) Either way, it's good that this mistake is pointed out. – Misha Lavrov Dec 06 '17 at 04:11