Summable $L^1$ norms force absolute almost-everywhere convergence and convergence in $L^1$.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\in L^1(X,\mu)\) for \(n\geq1\) satisfy \[ \sum_{n=1}^\infty \norm{f_n}_1 < \infty. \] Then for almost every \(x\in X\) the series \(\sum_{n=1}^\infty f_n(x)\) converges absolutely, its sum \(f\), defined almost everywhere, lies in \(L^1(X,\mu)\), and \[ \norm{f-\sum_{n=1}^N f_n}_1 \longrightarrow 0 \quad\text{as } N\to\infty. \]