# 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?

## 1 Answer

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 $$\int f\textrm{d}\mu=\sup_{n\in\mathbb{N}}\sup_{K\in \mathbb{N}} \sum_{k=1}^K \sum_{k=1}^K\int f_n\textrm{d}\mu_k=\sup_{K\in \mathbb{N}}\sup_{n\in\mathbb{N}} \sum_{k=1}^K \int f_n\textrm{d}\mu_k=\sup_{K\in \mathbb{N}}\sum_{k=1}^K \int f\textrm{d}\mu_k=\sum_{k=1}^{\infty} \int f\textrm{d}\mu_k$$

• Thank you for the very comprehensive answer! Could you please add a bit to the statement "by the definition of the infinite sum of measures"? Why can you pull the sum for the measures out of the integral? May 20, 2020 at 15:46
• 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
• Anyway, I feel like it's worth stressing that this part of the argument holds in the highest possible generality: Supremums over free indices just always commute. May 20, 2020 at 17:32