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.