# Series of Measures is Measure

## Theorem

Let $\left({X, \Sigma}\right)$ be a measurable space.

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

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

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

$\displaystyle \mu \left({E}\right) := \sum_{n \mathop \in \N} a_n \mu_n \left({E}\right)$

is also a measure on $\left({X, \Sigma}\right)$.

## Proof

Let us verify the conditions for a measure in turn.

Let $E \in \Sigma$.

Then for all $n \in \N$, $\mu_n \left({E}\right) \ge 0$ and so $a_n \mu_n \left({E}\right) \ge 0$.

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

$\displaystyle \mu \left({E}\right) = \sum_{n \mathop \in \N} a_n \mu_n \left({E}\right) \ge 0$

For every $n \in \N$, also $\mu_n \left({\varnothing}\right) = 0$.

Therefore, it immediately follows that:

$\displaystyle \mu \left({\varnothing}\right) = \sum_{n \mathop \in \N} a_n \mu_n \left({\varnothing}\right) = \sum_{n \mathop \in \N} 0 = 0$

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

Then:

 $\displaystyle \mu \left({\bigcup_{m \mathop \in \N} E_m}\right)$ $=$ $\displaystyle \sum_{n \mathop \in \N} a_n \mu_n \left({\bigcup_{m \mathop \in \N} E_m}\right)$ Definition of $\mu$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop \in \N} a_n \left({\sum_{m \mathop \in \N} \mu_n \left({E_m}\right)}\right)$ The $\mu_n$ are measures $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop \in \N} \sum_{m \mathop \in \N} a_n \mu_n \left({E_m}\right)$ $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop \in \N} \sum_{n \mathop \in \N} a_n \mu_n \left({E_m}\right)$ Double Series of Positive Real Numbers $\displaystyle$ $=$ $\displaystyle \sum_{m \mathop \in \N} \mu \left({E_m}\right)$ Definition of $\mu$

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

$\blacksquare$