Interchange of summation of measures and integral I want to show: If $(\mu_n)_{n\in\mathbb{N}}$ is a sequence of measures on $X$ and $\mu=\sum_{i=1}^\infty \mu_n$ and $f\colon X\longrightarrow \mathbb{R}$ a measurable function. Then $f$ is $\mu$ integrable if and only if 
$\sum_{n=1}^\infty \int_X |f|\mathrm{d}\mu_n < \infty $. It then follows that $\int_X f\mathrm{d}\mu =\sum_{n=1}^\infty \int_X f\mathrm{d}\mu_n$.
Firstly, I stated that there is a sequence of step functions $(S_k)_{k\in\mathbb{N}}$ which convergence p.w. to $f$.
I begin with $\int_X f \sum_{n=1}^\infty\mu_n = \sum_{i=1}^\infty c_i \sum_{n=1}^\infty\mu_n(X_i)$ by definition of the integral of a step function. Now i know that the sum of measure converges, which also means that it converges absolutely and I can interchange the double sum.
It then follows that $\sum_{i=1}^\infty \sum_{n=1}^\infty c_i\mu_n(X_i)=\sum_{n=1}^\infty \sum_{i=1}^\infty c_i \mu_n(X_i)=\sum_{n=1}^\infty \int_x f\mathrm{d}\mu_n$
I have my doubts that this is correct. 
 A: $\mu=\sum_{n=1}^{\infty}\mu_n$ states that $\mu(A)=\sum_{n=1}^{\infty}\mu_n(A)$ for every measurable set $A$.
Based on that it can be shown that: $$\int sd\mu=\sum_{n=1}^{\infty}\int sd\mu_n$$ for any nonnegative measurable function $s$ that has finite image.
Now let $g$ be a nonnegative measurable function. If $s$ is a function as described above with $s\leq g$ then $\int sd\mu=\sum_{n=1}^{\infty}\int sd\mu_n\leq\sum_{n=1}^{\infty}\int gd\mu_n$ and on base of the definition of $\int gd\mu$ we conclude that: $$\int gd\mu\leq\sum_{n=1}^{\infty}\int gd\mu_n$$
Now we will prove that for every $m$ we have $\int gd\mu\geq\sum_{n=1}^m\int gd\mu_n$ and consequently: $$\int gd\mu\geq\sum_{n=1}^{\infty}\int gd\mu_n$$
If $\sum_{n=1}^m\int gd\mu_n=\infty$ then $\int gd\mu_k=\infty$ for some index $k\in\{1,\dots,m\}$ and consequently $\int gd\mu=\infty$ since $\mu\geq\mu_k$.
If $\sum_{n=1}^m\int gd\mu_n<\infty$ then for $\epsilon>0$ we can find a function $s$ as described above with $\sum_{n=1}^m\int sd\mu_n\geq\sum_{n=1}^m\int gd\mu_n-\epsilon$ and consequently $\int gd\mu\geq\sum_{n=1}^m\int gd\mu_n-\epsilon$. 
Proved is now that: $$\int gd\mu=\sum_{n=1}^{\infty}\int gd\mu_n$$ for nonnegative measurable functions $g$.
This can be applied on $g:=|f|$ and gives the result you want: $$f\text{ is }\mu\text{-integrable iff }\sum_{n=1}^{\infty}\int |f|d\mu_n<\infty$$

edit to clarify.
Let $s$ be a nonnegative measurable function $s$ that has finite image. That means that we can write: $$s=\sum_{i=1}^ma_i1_{B_i}$$ where $a_i\geq0$ and the $B_i$ are measurable disjoint and covering sets.
Then: $$\begin{aligned}\int sd\mu & =\sum_{i=1}^{m}a_{n}\mu\left(B_{i}\right)\\
 & =\sum_{i=1}^{m}a_{n}\sum_{n=1}^{\infty}\mu_{n}\left(B_{i}\right)\\
 & =\sum_{n=1}^{\infty}\sum_{i=1}^{m}a_{n}\mu_{n}\left(B_{i}\right)\\
 & =\sum_{n=1}^{\infty}\int sd\mu_{n}
\end{aligned}
$$
