# 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?

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 $$\int f\textrm{d}\mu=\sup_{n\in\mathbb{N}}\sup_{K\in \mathbb{N}} \sum_{k=1}^K \sum_{k=1}^K\int f_n\textrm{d}\mu_k=\sup_{K\in \mathbb{N}}\sup_{n\in\mathbb{N}} \sum_{k=1}^K \int f_n\textrm{d}\mu_k=\sup_{K\in \mathbb{N}}\sum_{k=1}^K \int f\textrm{d}\mu_k=\sum_{k=1}^{\infty} \int f\textrm{d}\mu_k$$
• By definition, $\int \sum_{n=1}^N c_n 1_{A_n}\textrm{d}\mu=\sum_{n=1}^N c_n \mu(A_n)\sum_{n=1}^N c_n \sum_{k=1}^{\infty} \mu_k(A_n)=\sum_{k=1}^{\infty} \sum_{n=1}^N c_n \mu(A_n)$. Hence, the result is automatic for simple functions. May 20, 2020 at 16:06
• I believe you can save some work by noting that, as $f_n$ is increasing, you can just swap the supremum in. See math.stackexchange.com/questions/3424451/… May 20, 2020 at 17:18
• I mean... that's more or less the same as the argument about swapping the order of the supremums of $g$. Anyway, the technical part of the argument is that you need to swap a supremum and an infinite sum at some point and for this purpose, positivity of the terms is absolutely vital. May 20, 2020 at 17:26