If $\abs{f_n}\le g_n$, $f_n\to f$ and $g_n\to g$ almost everywhere, and $\int g_n\to\int g<\infty$, then $\int f_n\to\int f$.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space, and let \(f_n\colon X\to\CC\) and \(g_n\colon X\to[0,\infty)\) for \(n\geq1\), \(f\colon X\to\CC\), and \(g\colon X\to[0,\infty)\) be measurable. Suppose that
\(f_n\to f\) almost everywhere;
\(g_n\in L^1(X,\mu)\) and \(\abs{f_n}\leq g_n\) almost everywhere, for every \(n\);
\(g_n\to g\) almost everywhere, \(g\in L^1(X,\mu)\), and \(\lim_{n\to\infty}\int_X g_n\dmu=\int_X g\dmu\).
Then \(f\in L^1(X,\mu)\) and \[ \lim_{n\to\infty}\int_X f_n\dmu=\int_X f\dmu. \]
The dominated convergence theorem is the case \(g_n=g\) for all \(n\): the single dominating function \(g\) is replaced by dominating functions \(g_n\) that converge to \(g\) almost everywhere and in integral.