How to integrate for a countable sum of measures?

Question

Generalizing this question, when can we integrate in a similar way for a countable sum of measures?

$$\int f\,dm=\sum_{n \in \mathbb{N}} a_n \int f\,dm_n$$

Answer 1

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

Answer 2

It works as long as the $a_n$ are non-negativ. Let us denote with $\mathcal{A_n}$ the $\sigma$-Algebra of $m_n$-measurable sets. Then $$\mathcal{A}=\bigcap_n \mathcal{A_n}$$ is another $\sigma$-Algebra on which we need to work to define $m$: For $A\in \mathcal{A}$ we set $$m(A):=\sum_{n}a_nm_n(A).$$ Now $m$ becomes a measure because if you take pairwise disjoint and countable many $A_k\in\mathcal{A}$ you get $$m(\cup_k A_k)= \sum_{n}a_nm_n(\cup A_k) = \sum_n a_n\sum_k m_n(A_k) = \sum_k\sum_n a_n m_n(A_k)=\sum_k m(A_k),$$ since $A_k\in \mathcal{A_n}$ and by $a_n\geq0$ the corresponding sums converge absolutely.

Now $m$ is well defined and we can proceed to the integrals. First for simple $f(x)=\sum_{k=1}^N b_k 1_{B_k}(x)$, $B_k\in\mathcal{A}$ and $b_k\geq 0$ $$\int fdm= \sum_{k=1}^Nb_km(B_k) = \sum_{k=1}^N b_k \sum_n a_nm_n(B_k) = \sum_na_n\sum_{k=1}^Nb_km_n(B_k)\\ =\sum_n a_n\int fdm_n.$$ Now approximate a $\mathcal{A}$-measurable $f\geq0$ pointwise by a non-decreasing sequence of simple functions $f_k$. Beppo-Levi's theorem yields $$\int fd_m = \lim_{k\rightarrow\infty}\int f_kdm = \lim_{k\rightarrow\infty}\sum_{n}a_n\int f_kdm_n=\sum_n a_n\lim_{k\rightarrow\infty}\int f_kdm_n = \sum_na_n\int fdm_n.$$ Please note that the exchange of $\sum$ and $\lim$ is also due to Beppo-Levi applied to the counting measure and the fact that $k\mapsto a_n\int f_kdm_n$ is non-decreasing in $k$