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)


\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.



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