On finite measure spaces, uniform convergence forces convergence of integrals.Suggest a slogan
Let \((X,\mcm,\mu)\) be a measure space with \(\mu(X)<\infty\). If \(f_n,f\in L^1(X,\mu)\) and \(f_n\to f\) uniformly on \(X\), then \[ \abs{\int_X(f_n-f)\dmu} \leq \mu(X)\sup_{x\in X}\abs{f_n(x)-f(x)} \convergesto{n\to\infty} 0, \] so \(\int_X f_n\dmu\to\int_X f\dmu\).
\[ \abs{\int_X(f_n-f)\dmu} \leq \int_X\abs{f_n-f}\dmu \leq \mu(X)\sup_{x\in X}\abs{f_n(x)-f(x)}, \] and the right-hand side tends to \(0\) by uniform convergence.