Let $\mu_n$ be measures and $\mu=\sum_{n=1}^\infty \mu_n$. Show for measurable, positive $f$: $\int_Xf\ d\mu = \sum\int_X f\ d\mu_n$ Let $(X,\mathscr{S})$ be a measurable space, $\mu_n$ be measures and $\mu=\sum_{n=1}^\infty \mu_n$. I want to show for measurable $f:X\rightarrow[0,\infty]$: 
$$\int_Xf\ d\mu = \sum_{n=1}^\infty\int_X f\ d\mu_n$$
holds. The exercise gives two hints: use the construction of the Lebesgue-Integral and Beppo-Levi.
I have fully expanded both the left and right-hand side using the construction of the Lebesgue-Integral (supremum of step-functions). Now it looks like I need to swap an infinite sum and the supremum of a sum, which I don't think I'm allowed to do.
Following the second hint, I noticed that if we define $m_k=\sum_{n=1}^k\mu_n$, we have an increasing sequence of measurable functions. However, Beppo-Levi is about integrating a series of functions, not about integrating with respect to a series of functions, and I just can't see how BL could be useful.
Is it possible to somehow switch the integration to integrating the measures themselves? Or is there a different approach?
 A: The Bepo-Levi theorem really states that if $(f_n)_{n\in \mathbb{N}}$ is an increasing family of positive simple functions converging to $f,$ then
$$
\int f\textrm{d}\nu=\lim_{n\to\infty} \int f_n\textrm{d}\nu=\sup_{n\in \mathbb{N}} \int f_n\textrm{d}\nu
$$
for any measure $\nu$, where the last inequality follows directly from the fact the sequence of integrals will also be increasing. Hence, in your setup, you have
$$
\int f \textrm{d}\mu=\sup_{n\in \mathbb{N}} \int f_n\textrm{d}\mu=\sup_{n\in \mathbb{N}} \sum_{k=1}^{\infty} \int f_n\textrm{d}\mu_k,
$$
where the last equality holds by definition of the infinite sum of measures, since $f_n$ is simple. Now, note that since the series is positive, we have
$$
\sup_{n\in \mathbb{N}} \sum_{k=1}^{\infty} \int f_n\textrm{d}\mu_k=\sup_{n\in \mathbb{N}}\sup_{K\in \mathbb{N}} \sum_{k=1}^K \int f_n\textrm{d}\mu_k,
$$
and it's a general fact that if $A$ and $B$ are any sets and $g:A\times B\to \mathbb{R}$ is any function, then
$$
\sup_{a\in A}\sup_{b\in B} g(a,b)=\sup_{b\in B}\sup_{a\in A} g(a,b)
$$
Indeed, this holds because clearly we have $\sup_{a'\in A} g(a',b)\geq g(a,b)$ for any $a$ and $b$ and hence,
$$\sup_{b\in B}\sup_{a'\in A} g(a',b)\geq \sup_{b\in B} g(a,b)$$
Now, the left-hand side here is just a number, so using the characterising property of the supremum, you get 
$$
\sup_{b\in B}\sup_{a'\in A} g(a',b)\geq  \sup_{a\in A}\sup_{b\in B} g(a,b)
$$
This argument is, of course, completely symmetric and so, we have equality.
Thus, returning to the original problem, we now have
$$
\int f\textrm{d}\mu=\sup_{n\in\mathbb{N}}\sup_{K\in \mathbb{N}} \sum_{k=1}^K \sum_{k=1}^K\int f_n\textrm{d}\mu_k=\sup_{K\in \mathbb{N}}\sup_{n\in\mathbb{N}} \sum_{k=1}^K \int f_n\textrm{d}\mu_k=\sup_{K\in \mathbb{N}}\sum_{k=1}^K \int f\textrm{d}\mu_k=\sum_{k=1}^{\infty} \int f\textrm{d}\mu_k
$$
