Let \(E \subseteq {\mathbf{R}}^d\) be measurable of positive finite measure with \(f_k\to f\) almost everywhere on \(E\). Then for every \({\varepsilon}> 0\) there is a closed \(A_{\varepsilon}\subseteq E\) with \(\mu(E\setminus A_{\varepsilon}) < {\varepsilon}\) and \(f_k\to f\) uniformly on \(A_{\varepsilon}\).
If \(f\) is measurable and finite-valued on \(E\) with \(\mu(E) < \infty\) then for every \({\varepsilon}>0\) there exists a closed set \(F_{\varepsilon}\) with \begin{align*} F_{\varepsilon}\subset F && \mu(E - F_{\varepsilon}) \leq {\varepsilon} \end{align*} where \(f\) restricted to \(F_{\varepsilon}\) is continuous.
Note: this means that the separate function \(\tilde f \coloneqq{ \left.{{f}} \right|_{{F_{\varepsilon}}} }\) is continuous, not that the function \(f\) defined on all of \(E\) is continuous at points of \(F_{\varepsilon}\).
