Let \(m\) be Lebesgue measure on \(\RR^d\), let \(E \subseteq \RR^d\) be Lebesgue measurable with \(m(E) < \infty\), and let \(f_k\colon E\to\RR\) for \(k\geq1\) and \(f\colon E\to\RR\) be measurable. If \(f_k \to f\) almost everywhere on \(E\), then \(f_k\to f\) in measure on \(E\).
Fix \(\varepsilon > 0\) and \(\delta > 0\). For each \(N\geq1\), define the measurable set \[ E_N \coloneqq \theset{x \in E \suchthat \abs{f_k(x) - f(x)} < \varepsilon \text{ for all } k \ge N}. \] The sets \(E_N\) increase with \(N\), and every \(x\in E\) at which \(f_k(x) \to f(x)\) lies in some \(E_N\), so \(E\setminus\bigcup_N E_N\) is a null set. By continuity of measure from below, \(m(E_N)\to m\qty{\bigcup_N E_N} = m(E)\); since \(m(E)<\infty\), it follows that \(m(E \setminus E_N) = m(E)-m(E_N) \to 0\) as \(N \to \infty\). Choose \(N\) with \(m(E \setminus E_N) < \delta\). For every \(k \ge N\) and every \(x \in E_N\) we have \(\abs{f_k(x) - f(x)} < \varepsilon\), so \[ m\qty{\theset{x \in E \suchthat \abs{f_k(x) - f(x)} \ge \varepsilon}} \le m(E \setminus E_N) < \delta \] for all \(k \ge N\). Since \(\delta>0\) was arbitrary, \(m\qty{\theset{x \in E \suchthat \abs{f_k(x) - f(x)} > \varepsilon}}\to0\) as \(k\to\infty\) for every \(\varepsilon>0\), which is convergence in measure.