Exercise 1.7.5 (Elementary measure is a pre-measure)

First of all, by Exercise 1.4.1 (Elementary algebra) the set of all elementary and co-elementary sets $\overline{E[R^d]}$ forms a Boolean algebra.

Furthermore, we verify the pre-measure axioms of the elementary measure $m_E$. Let ${(F_n)}_{n\in N}$ be a set of disjoint sets in $\overline{E[R^d]}$. We have two cases:

(i)

There is at least one co-elementary set $F_j$ lurking inside this sequence. Then the theorem assertion holds automatically by the monotonicity property of the elementary measure: $$m_E\left(\bigcup _{n\in N}{F}_n\right)\ge m_E(F_j)=\infty \sum _{n\in N}{m}_E(F_n)\ge m_E(F_j)=\infty .$$

(ii)

All of the $F_n$ are elementary. Recall from Lemma 1.2.6 that the elementary measure of any elementary set coincides with the Lebesgue outer measure of an elementary set. Thus, $m_E$ must be countably subadditive for ${(F_n)}_{n\in N}$ because Lebesgue outer measure is (Exercise 1.2.3), while it is also finitely additive by remark on page 6.