Almost-everywhere convergence under one integrable majorant gives $L^1$ convergence and convergence of integrals.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\colon X\to\CC\) for \(n\geq1\) and \(f\colon X\to\CC\) be measurable with \(f_n\to f\) almost everywhere. Suppose there is an integrable \(g\colon X\to[0,\infty)\) with \(\abs{f_n} \leq g\) almost everywhere for every \(n\). Then \(f\) is integrable and \[ \lim_{n\to\infty}\int_X \abs{f_n - f}\dmu = 0 ; \] consequently \[ \lim_{n\to\infty} \int_X f_n\dmu = \int_X f\dmu . \]
The functions \(f_n\) may be complex-valued and of arbitrary sign; the only sign condition is \(g\geq0\) on the dominating function.