TITLE: How to integrate for a countable sum of measures?
QUESTION [5 upvotes]: Generalizing this question, when can we integrate in a similar way for a countable sum of measures?
$$\int f\,dm=\sum_{n \in \mathbb{N}} a_n \int f\,dm_n$$

REPLY [4 votes]: There is a more precise answer to the question, as a matter of fact a necessary and sufficient condition, as I pointed out a few months ago answering another question on the MathOverflow. Since
$$
m=\sum_{n\in\mathbb{N}}a_n m_n\triangleq\lim_{N\to\infty} \sum_{|n|\leq N}a_n m_n\tag{1}\label{1}
$$
the OP question is equivalent to asking when it is possible the passage to the limit under the integral symbol, i.e. when
$$
\int f\mathrm{d}m\triangleq\int f\mathrm{d}\left(\sum_{n\in\mathbb{N}}a_n m_n\right)=\lim_{N\to\infty} \sum_{|n|\leq N}a_n\int f \mathrm{d}m_n\triangleq\sum_{n\in\mathbb{N}}a_n \int f \mathrm{d}m_n\; ?\tag{2}\label{2}
$$
The answer requires the two following definitions
Definition 1. Let $(E,\mathcal{E})$ be a measure space and $\phi:\mathcal{E}\to\overline{\mathbb{R}}$ a numerical set function:
$\phi$ is called exhaustive if
$$
\lim_n\phi(A_n)=0
$$
for all families $\{A_n\}$ of pairwise disjoint sets in $\mathcal{E}$.
Definition 2. Let $(E,\mathcal{E})$ be a measure space and $H$ a set (and thus possibly a family) of numerical set functions defined on $\mathcal{E}$: $H$ is called uniformly exhaustive if the numerical set function
$$
A\mapsto\sup_{\phi\in H} \vert\phi(A)\vert\;\text{ is exhaustive.}
$$
By Cafiero's theorem (on the passage to the limit under the integral), formula \eqref{2} holds if and only if the limit \eqref{1} exist pointwise and
$$
\bigg(f\cdot\sum_{|n|\leq N}a_n m_n\bigg)_{N\geq 1}=\bigg(\sum_{|n|\leq N}f\cdot a_n m_n\bigg)_{N\geq 1}=\bigg(\sum_{|n|\leq N} a_n\, f\cdot m_n\bigg)_{N\geq 1}\text{ is uniformly exaustive.}
$$
A minimal bibliography on the theorem is available in my MathOverflow post cited above