# Applications of Lebesgue integration to elementary analysis problems

In elementary analysis one might have come across the problem of showing

${\displaystyle \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}a_{nk}=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_{nk}}$

for $a_{nk} \geq 0$. It turns out (Real and Complex Analysis, Rudin) that there is a more advanced proof using Lebesgue integration with respect to a counting measure!

Specifically, we can consider the counting measure $\mu : P(\mathbb{Z}^+) \rightarrow [0,\infty]$ where $P(\mathbb{Z}^+)$ is the power set of $\mathbb{Z}^+$, so that $\mu(E)$ is the cardinality of $E$ if $E$ is finite, and otherwise it is $\infty$ (e.g. $\mu(\{1,5,8\})=3$ while $\mu(\{1,3,5,\ldots\})=\infty$). Indeed the domain is $P(\mathbb{Z}^+)$ so that in fact all functions $f : \mathbb{Z}^+ \rightarrow [0,\infty]$ are measurable.

It turns out that integration of such a function $f$ with respect to the counting measure is

$\displaystyle \int_{\mathbb{Z}^+} f d\mu = \sum_{k=1}^{\infty} f(k)$

That is, integrals with respect to the counting measure are just sums (if you are interested, see this document for a proof of this). A consequence (Real and Complex Analysis, Rudin) of Lebesgue’s Monotone Convergence Theorem is: If $f_n : \mathbb{Z}^+ \rightarrow [0,\infty]$ (are measurable) and

$\displaystyle f(x) = \sum_{n=1}^\infty f_n(x)$

then

$\displaystyle \int_{\mathbb{Z}^+} f d\mu = \sum_{n=1}^\infty \int_{\mathbb{Z}^+} f_n d\mu$

From what was noted earlier the integrals can be replaced with sums:

$\displaystyle \sum_{k=1}^\infty \sum_{n=1}^\infty f_n(k) = \sum_{k=1}^\infty f(k) = \sum_{n=1}^\infty \sum_{k=1}^\infty f_n(k)$

Consequently simply choosing $f_n(k) = a_{nk}$, which is measurable because as noted earlier all functions are measurable with respect to the counting measure, does the trick! This completes the proof.