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 1


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

  • $\begingroup$ 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? $\endgroup$ May 20, 2020 at 15:46
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    $\begingroup$ 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. $\endgroup$ May 20, 2020 at 16:06
  • $\begingroup$ 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/… $\endgroup$ May 20, 2020 at 17:18
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    $\begingroup$ 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. $\endgroup$ May 20, 2020 at 17:26
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    $\begingroup$ 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. $\endgroup$ May 20, 2020 at 17:32

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