Additive set function on a semiring of sets

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.

Solution 1:

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.