$L^1$ convergence forces norm convergence; with almost-everywhere convergence, norm convergence forces $L^1$ convergence.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space and let \(f, f_n\in L^1(\mu)\) for \(n\geq1\).
If \(\norm{f_n-f}_1\to0\), then \(\norm{f_n}_1\to\norm{f}_1\).
If \(f_n\to f\) \(\mu\)-almost everywhere and \(\norm{f_n}_1\to\norm{f}_1\), then \(\norm{f_n-f}_1\to0\) [@Fol13].
Without almost everywhere convergence, convergence of norms does not imply convergence in \(L^1\): for \(f\in L^1(\mu)\) with \(\norm{f}_1>0\) and \(f_n\coloneqq-f\), \(\norm{f_n}_1=\norm{f}_1\) for all \(n\) but \(\norm{f_n-f}_1=2\norm{f}_1\).