Let \((X,\mcm,\mu)\) be a measure space, let \(f_n\colon X\to\CC\) for \(n\geq1\) be measurable, and let \(S(x) \coloneqq \sum_{n\geq1} \abs{f_n(x)}\in[0,\infty]\). Then \[ \int_X S\dmu = \sum_{n\geq1} \norm{f_n}_1 . \] In particular, if \(\sum_{n\geq1} \norm{f_n}_1<\infty\), then \(S\in L^1(X,\mu)\) and \(\sum_{n\geq1} \abs{f_n(x)} < \infty\) for \(\mu\)-almost every \(x\in X\).
The partial sums \(S_N \coloneqq \sum_{n=1}^N \abs{f_n}\) are nonnegative, measurable, and increase pointwise to \(S\). By the monotone convergence theorem, \[ \int_X S\dmu = \lim_{N\to\infty} \int_X S_N\dmu = \lim_{N\to\infty} \sum_{n=1}^N \int_X \abs{f_n}\dmu = \sum_{n\geq1} \norm{f_n}_1. \] Now assume \(\sum_{n\geq1} \norm{f_n}_1<\infty\), so \(\int_X S\dmu<\infty\). Let \(A\coloneqq\theset{x\in X\suchthat S(x)=\infty}\), which is measurable. For every \(t>0\) we have \(S\geq t\,\one_A\), so \(t\,\mu(A)\le\int_X S\dmu\); letting \(t\to\infty\) gives \(\mu(A)=0\). Therefore \(\sum_{n\geq1} \abs{f_n(x)} < \infty\) for almost every \(x\).