Lusin's theorem

Let $E \subseteq \mathbb{R}^n$ be measurable with $\lambda_n(E) < \infty$ and let $f : E \to \mathbb{R}$ be measurable. Then for every $\varepsilon > 0$ there is a closed set $K \subseteq E$ with

$\lambda_n(E \setminus K) < \varepsilon \qquad \text{and} \qquad f|_{K} \text{ continuous}.$

$K$ may be taken compact when $E$ is bounded, and $f|_K$ extends to a continuous function on all of $\mathbb{R}^n$ by Tietze. The assertion is about the restriction $f|_K$ and not about continuity of $f$ at the points of $K$: the Dirichlet function $\mathbf{1}_{\mathbb{Q}}$ is nowhere continuous on $\mathbb{R}$, yet its restriction to the closed set $\mathbb{R} \setminus U$, where $U$ is an open set of measure below $\varepsilon$ containing $\mathbb{Q}$, is identically zero and so continuous.

Not proved in this library. It is recorded with citations and used in no proof here.

What would prove it. Regularity of Lebesgue measure, which supplies closed sets from inside and open sets from outside, applied to the preimages of a countable base of intervals, plus Egorov's theorem (Egorov's theorem ‡) in the version where a measurable function is an almost everywhere limit of simple functions. Both ingredients are measure-theoretic (Lebesgue measure and the Lebesgue integral ‡).

Which page it serves. The continuity page and the uniform convergence page, where the question "how badly can a function fail to be continuous" is answered only for specific examples (Dirichlet, Thomae, Volterra). Lusin's theorem is the general answer: measurability is exactly continuity after deleting a set of arbitrarily small measure, which is Littlewood's second principle. It also belongs beside the Riesz-Markov-Kakutani theorem (Riesz-Markov-Kakutani representation theorem ‡), since both express that continuous functions are dense in the measurable world.