Additivity
0.1 Additivity
Definition 1 (Finite additivity) Let \(\mathcal{S}\) be a \(\sigma\)-algebra on a set \(S\). A function \(\mu: \mathcal{S}\to [0, \infty]\), with values in the extended non-negative reals, is finitely additive [1] if, for every finite collection of pairwise disjoint sets \(A_1, \ldots, A_n\) in \(\mathcal{S}\), the value of their union is the sum of their values:
\[\mu(A_1 \cup \cdots \cup A_n) = \sum_{i=1}^n\mu(A_i)\]
Example 1 (Counting elements is finitely additive) For the die rolls \(D = {\mathopen{}\left\{1, 2, 3, 4, 5, 6\right\}\mathclose{}}\) (sets of die rolls), let \(\mu(A) \stackrel{\text{def}}{=}\mathopen{}\left|A\right|\mathclose{}\), the number of elements of \(A\) (cardinality), for each set \(A\) in the \(\sigma\)-algebra of all subsets of \(D\) (Example 1 in Sigma-Algebras). The sets \({\mathopen{}\left\{1, 2\right\}\mathclose{}}\) and \({\mathopen{}\left\{5, 6\right\}\mathclose{}}\) are pairwise disjoint (low, middle and high die rolls), and:
\[ \begin{aligned} \mu({\mathopen{}\left\{1, 2\right\}\mathclose{}} \cup {\mathopen{}\left\{5, 6\right\}\mathclose{}}) &= \mu({\mathopen{}\left\{1, 2, 5, 6\right\}\mathclose{}}) && \text{(take the union)} \\ &= 4 && \text{(count the elements)} \\ &= 2 + 2 && \text{(write 4 as a sum)} \\ &= \mu({\mathopen{}\left\{1, 2\right\}\mathclose{}}) + \mu({\mathopen{}\left\{5, 6\right\}\mathclose{}}) && \text{(count each set's elements)} \end{aligned} \]
The same holds for any pairwise disjoint sets, because the sizes of disjoint sets add, so \(\mu\) is finitely additive.
Example 2 (Squaring the count is not finitely additive) On the same \(\sigma\)-algebra, let \(\nu(A) \stackrel{\text{def}}{=}\mathopen{}\left|A\right|\mathclose{}^2\). The sets \({\mathopen{}\left\{1\right\}\mathclose{}}\) and \({\mathopen{}\left\{2\right\}\mathclose{}}\) are pairwise disjoint, but
\[ \begin{aligned} \nu({\mathopen{}\left\{1\right\}\mathclose{}} \cup {\mathopen{}\left\{2\right\}\mathclose{}}) &= \nu({\mathopen{}\left\{1, 2\right\}\mathclose{}}) && \text{(take the union)} \\ &= 2^2 \\ &= 4, && \text{(count, then square)} \\ \nu({\mathopen{}\left\{1\right\}\mathclose{}}) + \nu({\mathopen{}\left\{2\right\}\mathclose{}}) &= 1^2 + 1^2 \\ &= 2, && \text{(count each set, then square)} \end{aligned} \]
and \(4 \ne 2\), so \(\nu\) is not finitely additive.
Definition 2 (Countable additivity (\(\sigma\)-additivity)) Let \(\mathcal{S}\) be a \(\sigma\)-algebra on a set \(S\). A function \(\mu: \mathcal{S}\to [0, \infty]\) is countably additive [1] (also called \(\sigma\)-additive) if, for every sequence of pairwise disjoint sets \(A_1, A_2, \ldots\) in \(\mathcal{S}\), the value of their union is the sum of their values:
\[\mu\mathopen{}\left(\bigcup_{i=1}^{\infty} A_i\right)\mathclose{} = \sum_{i=1}^{\infty} \mu(A_i)\]
Example 3 (Counting elements is countably additive) For the counting function \(\mu(A) \stackrel{\text{def}}{=}\mathopen{}\left|A\right|\mathclose{}\) of Example 1, take any sequence of pairwise disjoint sets \(A_1, A_2, \ldots\). No two of them share an element, and \(D\) has only six elements, so at most six of the \(A_i\) contain any elements; the rest are \(\emptyset\), with \(\mu(\emptyset) = 0\). The infinite sum therefore has at most six nonzero terms, and finite additivity (Example 1) shows that those terms add up to \(\mu\) of the union. So \(\mu\) is countably additive.
Remark 1 (The sum in countable additivity always has a value). By sums of non-negative terms, the right-hand side of the equation in Definition 2, \(\sum_{i=1}^{\infty} \mu(A_i)\), always has a value, because each \(\mu(A_i)\) is in \([0, \infty]\). For example, if \(\mu(A_i) = 1/2^i\) for each \(i\), the partial sums are \(s_n = 1 - 1/2^n\), and the sum is \(1\); if \(\mu(A_i) = 1\) for each \(i\), the partial sums are \(s_n = n\), and the sum is \(\infty\).
Lemma 1 (Countable additivity and the empty set) If \(\mu\) is a countably additive function on a \(\sigma\)-algebra \(\mathcal{S}\), then \(\mu(\emptyset) = 0\) or \(\mu(\emptyset) = \infty\).
Proof. The set \(\emptyset = S \setminus S\) is in \(\mathcal{S}\), as the complement of \(S\), and the sequence \(\emptyset, \emptyset, \ldots\) is pairwise disjoint with union \(\emptyset\), so:
\[ \begin{aligned} \mu(\emptyset) &= \mu\mathopen{}\left(\bigcup_{i=1}^{\infty} \emptyset\right)\mathclose{} && \text{(the union of copies of } \emptyset \text{ is } \emptyset \text{)} \\ &= \sum_{i=1}^{\infty} \mu(\emptyset) && \text{(countable additivity)} \end{aligned} \]
If \(\mu(\emptyset) = c\) for a finite \(c > 0\), the right-hand side is \(c + c + \cdots = \infty \neq c\), a contradiction. So \(\mu(\emptyset)\) is 0 or \(\infty\).
Example 4 (A countably additive function with \(\mu(\emptyset) = 0\)) The counting function \(\mu(A) \stackrel{\text{def}}{=}\mathopen{}\left|A\right|\mathclose{}\) of Example 3 is countably additive, and:
\[ \begin{aligned} \mu(\emptyset) &= \mathopen{}\left|\emptyset\right|\mathclose{} && \text{(definition of } \mu\text{)} \\ &= 0 && \text{(} \emptyset \text{ has no elements)} \end{aligned} \]
Example 5 (A countably additive function with \(\mu(\emptyset) = \infty\)) For the die rolls \(D = {\mathopen{}\left\{1, 2, 3, 4, 5, 6\right\}\mathclose{}}\) (sets of die rolls), let \(\mu(A) \stackrel{\text{def}}{=}\infty\) for every set \(A\) in the \(\sigma\)-algebra of all subsets of \(D\) (Example 1 in Sigma-Algebras), including \(A = \emptyset\). For any sequence of pairwise disjoint sets \(A_1, A_2, \ldots\), the left-hand side of the countable additivity equation is:
\[ \begin{aligned} \mu\mathopen{}\left(\bigcup_{i=1}^{\infty} A_i\right)\mathclose{} &= \infty && \text{(definition of } \mu\text{)} \end{aligned} \]
and the right-hand side, the limit of its partial sums (sums of non-negative terms), is:
\[ \begin{aligned} \sum_{i=1}^{\infty} \mu(A_i) &= \infty + \infty + \cdots && \text{(definition of } \mu\text{)} \\ &= \infty && \text{(every partial sum is } \infty \text{)} \end{aligned} \]
The two sides agree, so \(\mu\) is countably additive, and \(\mu(\emptyset) = \infty\).
Theorem 1 (Countable additivity implies finite additivity) If \(\mu\) is a countably additive function on a \(\sigma\)-algebra \(\mathcal{S}\), and \(\mu(\emptyset) = 0\), then \(\mu\) is finitely additive.
Proof. Let \(A_1, \ldots, A_n\) be pairwise disjoint sets in \(\mathcal{S}\), and extend them to a sequence by setting
\[ \begin{aligned} A_{n+1} &= A_{n+2} \\ &= \cdots \\ &= \emptyset. \end{aligned} \]
The extended sequence is still pairwise disjoint, since \(\emptyset\) shares no element with any set, and its union is \(A_1 \cup \cdots \cup A_n\). So:
\[ \begin{aligned} \mu(A_1 \cup \cdots \cup A_n) &= \mu\mathopen{}\left(\bigcup_{i=1}^{\infty} A_i\right)\mathclose{} && \text{(} A_i = \emptyset \text{ for } i > n \text{)} \\ &= \sum_{i=1}^{\infty} \mu(A_i) && \text{(countable additivity)} \\ &= \sum_{i=1}^n\mu(A_i) + \sum_{i=n+1}^{\infty} \mu(\emptyset) && \text{(} A_i = \emptyset \text{ for } i > n \text{)} \\ &= \sum_{i=1}^n\mu(A_i) && \text{(} \mu(\emptyset) = 0 \text{)} \end{aligned} \]
Example 6 (Finitely additive but not countably additive) Let \(S = {\mathopen{}\left\{0, 1, 2, \ldots\right\}\mathclose{}}\), with the \(\sigma\)-algebra of all subsets of \(S\) (Theorem 2 in Sigma-Algebras), and define:
\[ \mu(A) \stackrel{\text{def}}{=}\begin{cases} 0 & \text{if } A \text{ is finite} \\ \infty & \text{if } A \text{ is infinite} \end{cases} \]
\(\mu\) is finitely additive. Let \(A_1, \ldots, A_n\) be pairwise disjoint subsets of \(S\). If every \(A_i\) is finite, then so is their union, and:
\[ \begin{aligned} \mu(A_1 \cup \cdots \cup A_n) &= 0 && \text{(a finite union of finite sets is finite)} \\ &= \sum_{i=1}^n0 && \text{(a sum of zeros is 0)} \\ &= \sum_{i=1}^n\mu(A_i) && \text{(each } A_i \text{ is finite)} \end{aligned} \]
If some \(A_k\) is infinite, then the union, which contains \(A_k\), is infinite too, and:
\[ \begin{aligned} \mu(A_1 \cup \cdots \cup A_n) &= \infty && \text{(the union is infinite)} \\ &= \mu(A_k) + \sum_{i \in \mathopen{}\left\{1, \ldots, n\right\}\mathclose{} \setminus \mathopen{}\left\{k\right\}\mathclose{}} \mu(A_i) && \text{(} \mu(A_k) = \infty \text{, and } \infty + x = \infty \text{)} \\ &= \sum_{i=1}^n\mu(A_i) && \text{(regroup the terms)} \end{aligned} \]
\(\mu\) is not countably additive. The single-element sets \({\mathopen{}\left\{0\right\}\mathclose{}}, {\mathopen{}\left\{1\right\}\mathclose{}}, {\mathopen{}\left\{2\right\}\mathclose{}}, \ldots\) are pairwise disjoint, and their union is \(S\), which is infinite. So:
\[ \begin{aligned} \mu\mathopen{}\left(\bigcup_{k=0}^{\infty} {\mathopen{}\left\{k\right\}\mathclose{}}\right)\mathclose{} &= \mu(S) && \text{(the union is } S \text{)} \\ &= \infty && \text{(} S \text{ is infinite)} \end{aligned} \]
but:
\[ \begin{aligned} \sum_{k=0}^{\infty} \mu({\mathopen{}\left\{k\right\}\mathclose{}}) &= 0 + 0 + \cdots && \text{(each } {\mathopen{}\left\{k\right\}\mathclose{}} \text{ is finite)} \\ &= 0 && \text{(every partial sum is 0)} \end{aligned} \]
Remark 3 (Finite additivity does not imply countable additivity). Example 6 shows that the converse of Theorem 1 fails, even when \(\mu(\emptyset) = 0\): a finitely additive function need not be countably additive [1]. The function \(\mu\) in Example 6 has \(\mu(\emptyset) = 0\), because \(\emptyset\) is finite.