Let \(a<b\), let \(m\) be Lebesgue measure on \(\RR\), and let \(f_n\colon\RR\to\RR\) for \(n\geq1\) be Lebesgue measurable functions that vanish outside \([a,b]\). Suppose there is \(M\geq0\) with \(\abs{f_n(x)}\leq M\) for all \(n\) and all \(x\), and that \(f_n\to f\) pointwise for a function \(f\colon\RR\to\RR\). Then \[ \lim_{n\to\infty}\int f_n\,dm=\int f\,dm. \]
As a pointwise limit of measurable functions, \(f\) is measurable; it vanishes outside \([a,b]\) and satisfies \(\abs{f}\leq M\), so \(f_n\) and \(f\) are integrable. Let \(\varepsilon>0\). By Egorov's theorem applied on \([a,b]\), there is a closed set \(B\subseteq[a,b]\) with \(m(A)<\varepsilon\) for \(A\coloneqq[a,b]\setminus B\), such that \(f_n\to f\) uniformly on \(B\). Then \[ \abs{\int_B (f_n-f)\,dm}\leq(b-a)\sup_{x\in B}\abs{f_n(x)-f(x)}\longrightarrow0, \] and, since \(\abs{f_n-f}\leq 2M\), \[ \abs{\int_A (f_n-f)\,dm}\leq 2M\,m(A)<2M\varepsilon. \] Since \(f_n-f\) vanishes outside \(A\cup B\), \[ \limsup_{n\to\infty}\abs{\int f_n\,dm-\int f\,dm}\leq 0+2M\varepsilon. \] As \(\varepsilon>0\) was arbitrary, \(\int f_n\,dm\to\int f\,dm\).
The theorem is a special case of the dominated convergence theorem, with dominating function \(M\chi_{[a,b]}\).