Interchange of Integral and Discrete Sum

Theorem

Given a sequence of integrable functions \(g_{k}\) defined on a measurable space \(X\) where \(g_{k} \geq 0\) on \(X\) and \(\sum_{k = 1}^{\infty} g_{k} < \infty\) then:

\[ \sum_{k = 1}^{\infty} \int_{X} g_{k} \,\mathrm{d}\mu = \int_{X} \sum_{k = 1}^{\infty} g_{k} \,\mathrm{d}\mu.\]

This theorem is an application of the Beppo Levi monotone convergence theorem as is clear from the proof.

Proof

Let \(f_{n} = \sum_{k = 1}^{n} g_{k}\). We will apply the Beppo Levi monotone convergence theorem to \(f\), by first checking that the three criteria are satisfied.

First, \(f_{n}\) is Lebesgue integrable as it is a finite sum of Lebesgue integrable functions. Secondly, the fact that \(g_{k} \geq 0\) implies that as additional terms are added to the sum, the value of the sum is increasing, and thus \(f_{1} \leq f_{2} \leq f_{3} \leq \dots\). Finally, the fact that the sum is convergent: \(\sum_{k = 1}^{\infty} g_{k} < \infty\) implies that \(f_{n}\) has a finite supremum.

Now, from the monotone convergence theorem:

\[\begin{align*} \lim_{n \to \infty} \int_{X} f_{n} \,\mathrm{d}\mu &= \int_{X} \lim_{n \to \infty} f_{n} \,\mathrm{d}\mu \\ \lim_{n \to \infty} \int_{X} \sum_{k = 1}^{n} g_{k} \,\mathrm{d}\mu &= \int_{X} \lim_{n \to \infty} \sum_{k = 1}^{n} g_{k} \,\mathrm{d}\mu \\ \lim_{n \to \infty} \sum_{k = 1}^{n} \int_{X} g_{k} \,\mathrm{d}\mu &= \int_{X} \lim_{n \to \infty} \sum_{k = 1}^{n} g_{k} \,\mathrm{d}\mu. \\ \sum_{k = 1}^{\infty} \int_{X} g_{k} \,\mathrm{d}\mu &= \int_{X} \sum_{k = 1}^{\infty} g_{k} \,\mathrm{d}\mu. \\ \end{align*}\]