For Lebesgue measure \(m\) on \(\RR^d\), Egorov’s theorem upgrades almost-everywhere convergence to uniform convergence, and Lusin’s theorem upgrades measurability to continuity, on a closed set whose complement has arbitrarily small measure.
Egorov: almost uniform convergence
Egorov's theorem (Tag FT-OGS76)
Let \(m\) be Lebesgue measure on \(\RR^n\), let \(E\subseteq \RR^n\) be Lebesgue measurable with \(m(E) < \infty\), and let \(f_n\colon E\to\RR\) for \(n\geq1\) and \(f\colon E\to\RR\) be measurable with \(f_n \to f\) almost everywhere on \(E\). Then for every \(\varepsilon > 0\) there exists a closed set \(F\subseteq E\) such that \(m(E\setminus F)<\varepsilon\) and \(f_n\to f\) uniformly on \(F\).
Almost everywhere convergence implies convergence in measure on a set of finite measure (Tag FR-GG457)
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.
(Egorov's theorem)
Let \(E\subseteq\RR^d\) be measurable with \(m(E)<\infty\) and let \(f_k\to f\) almost everywhere on \(E\); removing a null set, assume \(f_k(x)\to f(x)\) for every \(x\in E\). For integers \(n\geq1\) and \(k\geq0\), let \[ E_k^n\coloneqq\theset{x\in E \st \abs{f_j(x)-f(x)}<1/n \text{ for all } j>k}. \] For fixed \(n\), \(E_k^n\subseteq E_{k+1}^n\) and \(\bigcup_k E_k^n = E\), so by continuity of measure from below and \(m(E)<\infty\) there is \(k_n\) with \(m(E\setminus E_{k_n}^n)<2^{-n}\). By construction, \(\abs{f_j(x)-f(x)}<1/n\) whenever \(j>k_n\) and \(x\in E_{k_n}^n\).
Given \(\varepsilon>0\), choose \(N\) with \(\sum_{n\geq N}2^{-n}<\varepsilon/2\) and let \(\tilde A_\varepsilon\coloneqq\bigcap_{n\geq N}E_{k_n}^n\). Then \[ m(E\setminus\tilde A_\varepsilon)\leq\sum_{n\geq N}m(E\setminus E_{k_n}^n)<\varepsilon/2. \] For \(\delta>0\), choose \(n\geq N\) with \(1/n<\delta\); every \(x\in\tilde A_\varepsilon\) lies in \(E_{k_n}^n\), so \(\abs{f_j(x)-f(x)}<\delta\) for all \(j>k_n\). Hence \(f_k\to f\) uniformly on \(\tilde A_\varepsilon\). Finally, by inner regularity of Lebesgue measure choose a closed \(A_\varepsilon\subseteq\tilde A_\varepsilon\) with \(m(\tilde A_\varepsilon\setminus A_\varepsilon)<\varepsilon/2\); then \(m(E\setminus A_\varepsilon)<\varepsilon\) and \(f_k\to f\) uniformly on \(A_\varepsilon\).
Lusin: almost continuity
Lusin's theorem (Tag T-CGFCU)
Let \(E\subseteq\RR^n\) be Lebesgue measurable with \(m(E)<\infty\), and let \(f\colon E\to\RR\) be measurable. For every \(\varepsilon>0\) there exists a closed set \(F_\varepsilon\subseteq\RR^n\) with \[ F_\varepsilon \subseteq E \quad\text{and}\quad m(E \setminus F_\varepsilon) \leq \varepsilon \] such that the restriction \(\ro{f}{F_\varepsilon}\colon F_\varepsilon\to\RR\) is continuous.
Continuity of the restriction \(\ro{f}{F_\varepsilon}\) does not make \(f\) continuous at points of \(F_\varepsilon\). Let \(E=[0,1]\) and \(f=\chi_{\QQ\cap[0,1]}\), which is discontinuous at every point of \([0,1]\). Enumerate \(\QQ\cap[0,1]=\theset{q_1,q_2,\ldots}\) and put \(F_\varepsilon\coloneqq[0,1]\setminus\bigcup_{k\geq1}(q_k-\varepsilon2^{-k-1},q_k+\varepsilon2^{-k-1})\). Then \(F_\varepsilon\) is closed, \(F_\varepsilon\subseteq E\), \(m(E\setminus F_\varepsilon)\leq\varepsilon\), and \(\ro{f}{F_\varepsilon}=0\) is continuous.
Lusin's theorem (Tag FT-XIVPL)
Let \(m\) be Lebesgue measure on \(\RR^n\), let \(E\subseteq\RR^n\) be Lebesgue measurable with \(m(E) < \infty\), and let \(f\colon E\to\RR\) be measurable. Then for every \(\varepsilon>0\) there exists a closed set \(F\subseteq E\) such that \(m(E\setminus F) < \varepsilon\) and the restriction \(f|_F\) is continuous.
(Lusin's theorem)
Let \(E\subseteq\RR^d\) be measurable with \(m(E)<\infty\) and \(f\) measurable and finite almost everywhere on \(E\), and let \(\varepsilon>0\). Choose step functions \(f_n\) with \(f_n\to f\) almost everywhere on \(E\). Each step function is continuous outside the boundaries of finitely many rectangles, so there are sets \(E_n\) with \(m(E_n)<2^{-n}\) such that \(f_n\) is continuous on \(E\setminus E_n\). By Egorov’s theorem there is \(A_{\varepsilon/3}\subseteq E\) with \(m(E\setminus A_{\varepsilon/3})\leq\varepsilon/3\) on which \(f_n\to f\) uniformly. Choose \(N\) with \(\sum_{n\geq N}2^{-n}<\varepsilon/3\) and let \[ F'\coloneqq A_{\varepsilon/3}\setminus\bigcup_{n\geq N}E_n. \] For \(n\geq N\), \(f_n\) is continuous on \(F'\), so the uniform limit \(f\) is continuous on \(F'\). By inner regularity choose a closed \(F_\varepsilon\subseteq F'\) with \(m(F'\setminus F_\varepsilon)<\varepsilon/3\). Then \(f\) is continuous on \(F_\varepsilon\) and \(m(E\setminus F_\varepsilon)<\varepsilon\).