Exercise 1.4.21 (Finitely additive measure on a finite Boolean algebra)

Let $\mu$ be a finitely additive measure on $B$, and let $E\in B$ be arbirtary. For each atom $A_i$ denote $c_i:=\mu (A_i)$. We argue that $c_1,\dots ,c_k$ are exactly the coefficients given in the exercise.

(i)

By definition, we can represent $E$ using the atoms $E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ for some $I\subseteq \{1,\dots ,k\}$. Since the atoms are disjoint we can write $I=\{1\le i\le k:A_i\subseteq E\}$ (if we had some $i$ in the latter set but not $i\in I$ this would imply a non-empty intersection of $A_i$ with at least one element of $A_j,j\in I$). In other words, $E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{1\le i\le k:A_i\subseteq E}{A}_i$. This a finite union of the disjoint sets, applying the finite additivity of $\mu$ we see $$\mu (E)=\mu \left(\bigcup _{1\le i\le k:A_i\subseteq E}{A}_i\right)=\sum _{1\le i\le k:A_i\subseteq E}\mu (A_i)=\sum _{1\le i\le k:A_i\subseteq E}{c}_i.$$

(ii)

Now suppose that we have two such collections $c_1,\dots ,c_k$ and $c_1^{\prime },\dots ,c_k^{\prime }$. We then must have by the previous part for any $1\le j\le k$: $$\mu (A_j)=\sum _{1\le i\le k:A_i\subseteq A_j}{c}_i=c_j\mu (A_j)=\sum _{1\le i\le k:A_i\subseteq A_j}{c}_i^{\prime }=c_j^{\prime }.$$

Thus, for all $1\le j\le k$ we have $c_j=c_j^{\prime }$.