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Prove or disprove: If $\mu$ is an additive (i.e., 2-additive) set function on a semi-ring $S$, then $\mu$ is finitely additive (i.e., n-additive for every finite $n$) on $S$.

If $S$ is a ring, this is true by induction. For a semi-ring, the naive induction proof fails because if $A_1,\ldots,A_n$ are disjoint sets in $S$ and if $\bigcup_{i=1}^{n} A_i$ is in $S$, it does NOT generally follow that $\bigcup_{i=1}^{n-1} A_i$ is in $S$. Every measure theory book I've looked at works with finitely additive sets functions on semi-rings, never just additive set functions on semi-rings. So maybe additive doesn't imply finite additive on a semi-ring.

JasonJones
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Additive function on a semiring may not be finitely-additive. There is a simple example.

Consider a set $X = \{a,b,c\}$ and a semiring $S = \{X, \{a\}, \{b\}, \{c\}, \emptyset\}$ and a function $\mu:S \rightarrow \mathbb{R}$ that is defined in the following way: $\mu(X) = 1$, $\mu(A) = 0$ for all $A \in S$ s.t. $A \ne X$. $\mu$ is additive but not finitely-additive.

Matsmir
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