A measurable function on a set of finite measure has a continuous restriction to a closed subset whose complement has arbitrarily small measure.Suggest a slogan
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.
State Lusin's theorem.