Let \((\Omega,\mcb)\) be a measurable space with a Borel \(\sigma\dash\)algebra and \(\mu_n: \mcb \to [0, \infty]\) be a \(\sigma\dash\)additive measure for each \(n\). Show that the following map is again a \(\sigma\dash\)additive measure on \(\mcb\): \[ \mu(B) \da \sum_{n\geq 1} \mu_n(B) .\]
Solution
1. \(\mu(\emptyset) = 0\) and \(\mu\) takes values in \([0,\infty]\).
Each \(\mu_n(\emptyset) = 0\), and a series of terms in \([0,\infty]\) has a sum in \([0,\infty]\).
2. For pairwise disjoint \(E_1, E_2, \ldots \in \mcb\), \(\mu\qty{\bigcup_{k\geq1} E_k} = \sum_{k\geq1}\mu(E_k)\).
By \(\sigma\)-additivity of each \(\mu_n\), \[ \mu\qty{\bigcup_{k\geq 1} E_k} = \sum_{n\geq 1} \mu_n\qty{\bigcup_{k\geq 1} E_k} = \sum_{n\geq 1} \sum_{k\geq 1} \mu_n(E_k) = \sum_{k\geq 1}\sum_{n\geq 1} \mu_n(E_k) = \sum_{k\geq 1} \mu(E_k). \] The third equality exchanges two sums of nonnegative terms, which is Tonelli’s theorem for counting measure on \(\NN \times \NN\).
3. Q.E.D.
Steps 1 and 2 are the axioms of a measure.