Series of Measures is Measure

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of measures on $\left({X, \Sigma}\right)$.

Let $\sequence {a_n}_{n \mathop \in \N} \subseteq \R_{\ge 0}$ be a sequence of positive real numbers.

Then the series of measures $\mu: \Sigma \to \overline \R$, defined by:


 * $\ds \map \mu E := \sum_{n \mathop \in \N} a_n \map {\mu_n} E$

is also a measure on $\struct {X, \Sigma}$.

Proof
Let us verify the conditions for a measure in turn.

Let $E \in \Sigma$.

Then:


 * $\forall n \in \N: \map {\mu_n} E \ge 0$

and so:


 * $a_n \map {\mu_n} E \ge 0$

Therefore, by Series of Positive Real Numbers has Positive Limit:


 * $\ds \map \mu E = \sum_{n \mathop \in \N} a_n \map {\mu_n} E \ge 0$

For every $n \in \N$, also $\map {\mu_n} \O = 0$.

Therefore, it immediately follows that:


 * $\ds \map \mu \O = \sum_{n \mathop \in \N} a_n \map {\mu_n} \O = \sum_{n \mathop \in \N} 0 = 0$

Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.

Then:

Therefore, having verified all three axioms, $\mu$ is a measure.