TITLE: Countable sum of measures is a measure
QUESTION [10 upvotes]: Prove that if $\mu_1, \mu_2, \dots$ are measures on a measurable space and $a_1, a_2, \dots \in [0,\infty)$, then $\sum_{n=1}^\infty a_n\mu_n$ is also a measure. 
I need some help justifying the third equality in the final line of my proof. My idea was to break this into finite and infinite cases and use facts about absolute convergence, but I'm not sure of the details.
My solution:
First, we let $(X, \mu_n, \mathcal A)$ be a measure space for all $n\in \mathbb N$ and define $\nu_n = a_n\mu_n$.  Since $\nu_n(\emptyset) = a_n\mu_n(\emptyset) = a_n\cdot0=0$ and $$\nu_n(\cup_{i=1}^\infty A_i) = a_n\mu_n(\cup_{i=1}^\infty A_i) = a_n\sum_{i=1}^\infty \mu_n(A_i) = \sum_{i=1}^\infty a_n\mu_n(A_i) = \sum_{i=1}^\infty \nu_n(A_i),$$
we know that $\nu_n$ is a measure for all $n\in \mathbb N$. So we are reduced to the case that a countable sum of measures is again a measure. 
Let $\mu = \sum_{n=1}^\infty \nu_n$. Since $\nu_n(\emptyset) = 0$ for all $n \in \mathbb N$, then $\mu(\emptyset) = \sum_{n=1}^\infty \nu_n(\emptyset) = 0$. So, we show that $\mu$ is countably additive. That is, if $A_i\in \mathcal A$ for all $i \in \mathbb N$ are pairwise disjoint, we show $\mu(\cup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \mu(A_i)$. Then,
\begin{align*}
\mu(\cup_{i=1}^\infty A_i) &= \sum_{n=1}^\infty \nu_n(\cup_{i=1}^\infty A_i) =\sum_{n=1}^\infty\sum_{i=1}^\infty \nu_n(A_i) = \sum_{i=1}^\infty\sum_{n=1}^\infty \nu_n(A_i) = \sum_{i=1}^\infty \mu(A_i). \\
\end{align*}

REPLY [7 votes]: Not absolute convergence, the sum can be infinite. But all terms are non-negative, hence we can change the order of summation. Fitting to the theme, we can appeal to the monotone convergence theorem for the counting measure on $\mathbb{N}$ for the justification.