Let \((X,\mcm,\mu)\) be a measure space and let \(f_n\colon X\to\CC\) for \(n\geq1\) be measurable functions with \[ \sum_{n=1}^\infty \int_X \abs{f_n} \dmu < \infty . \] Then \(\sum_{n}f_n(x)\) converges absolutely for almost every \(x\), and \[ \sum_{n=1}^\infty \int_X f_n \dmu = \int_X \sum_{n=1}^\infty f_n \dmu. \] The method views the sum as an integral against counting measure on \(\NN\) and applies the Fubini–Tonelli theorem on \(X\times\NN\).
1. Encode the series as an integral over \(\NN\).
1.1. Let \(\nu\) be counting measure on \(\NN\), so that \(\int_{\NN} g \, d\nu = \sum_{n=1}^\infty g(n)\) for every \(g\colon\NN\to[0,\infty]\) and every \(\nu\)-integrable \(g\colon\NN\to\CC\).
1.2. Define \(F \colon X \times \NN \to \CC\) by \(F(x, n) \coloneqq f_n(x)\).It is measurable for the product \(\sigma\)-algebra, since \(F^{-1}(B)=\bigcup_n f_n^{-1}(B)\times\theset{n}\) for every Borel set \(B\subseteq\CC\).
1.3. The two sides of the identity are the two iterated integrals of \(F\):\[ \sum_{n=1}^\infty \int_X f_n \dmu = \int_{\NN} \int_X F(x, n) \, d\mu(x) \, d\nu(n), \qquad \int_X \sum_{n=1}^\infty f_n \dmu = \int_X \int_{\NN} F(x, n) \, d\nu(n) \, d\mu(x). \]
2. Reduce to a \(\sigma\)-finite measure space.
2.1. For each \(n\) and \(k\geq1\), \(\mu\qty{\theset{\abs{f_n}>1/k}}\le k\int_X\abs{f_n}\dmu<\infty\), so \(A\coloneqq\bigcup_n\theset{x\in X\suchthat f_n(x)\neq0}\) is a countable union of sets of finite measure.
2.2. Every \(f_n\) vanishes outside \(A\), so both sides of the identity are unchanged when \(X\) is replaced by \(A\) and \(\mu\) by its restriction to \(A\), which is \(\sigma\)-finite.Counting measure on \(\NN\) is \(\sigma\)-finite as well.
3. Verify the hypothesis of Fubini’s theorem.
3.1. Since \(\abs{F}\) is nonnegative and measurable on \(A \times \NN\), Tonelli’s theorem gives\[ \int_{A \times \NN} \abs{F} \, d(\mu \times \nu) = \int_{\NN} \int_A \abs{F} \dmu \, d\nu = \sum_{n=1}^\infty \int_X \abs{f_n} \dmu < \infty. \]
3.2. Hence \(F \in L^1(A \times \NN, \mu \times \nu)\).
4. Apply Fubini’s theorem.
4.1. Because \(F\) is integrable on the product space, Fubini’s theorem shows that \(\int_{\NN}\abs{F(x,n)}\,d\nu(n)=\sum_n\abs{f_n(x)}\) is finite for almost every \(x\) and that the iterated integrals agree:\[ \int_{\NN} \int_A F \dmu \, d\nu = \int_A \int_{\NN} F \, d\nu \dmu. \]
4.2. Substituting the definitions of \(F\) and \(\nu\) into step 1 yields\[ \sum_{n=1}^\infty \int_X f_n \dmu = \int_X \sum_{n=1}^\infty f_n \dmu. \]
The same argument, with counting measure on \(\NN\times\NN\), applies to a doubly indexed family \((f_{m,n})_{m,n\geq1}\) of measurable functions: the hypothesis \(\sum_{m,n} \int_X \abs{f_{m,n}}\dmu < \infty\) justifies interchanging the double sum with the integral.