Lusin's theorem

Remarks

One proof approximates $f$ by simple functions and uses that each measurable set is nearly closed. For $f = \mathbf{1}_B$, choose closed $F_1 \subseteq B$ and $F_2 \subseteq A \setminus B$ with small leftover measure; $f$ is constant on each of the disjoint closed sets $F_1$ and $F_2$, so it is continuous on $F_1 \cup F_2$. For general $f$, take a sequence of simple functions that converges to $f$ pointwise, and use Egorov's theorem to make the convergence uniform on a large compact set. Example: $f = \mathbf{1}_{\Q}$ on $A = [0, 1]$ is discontinuous at every point, but its restriction to a compact set $K$ of irrational numbers with $\lambda(K) > 1 - \varepsilon$ is the constant $0$. The conclusion is about $f|_K$, not about $f$: the points of $K$ need not be points of continuity of $f$. By the Tietze extension theorem, there is also a continuous $g \colon \R^n \to \R$ with $g = f$ on $K$. The same result holds for every Radon measure on a locally compact Hausdorff space, for functions on sets of finite measure.

Going deeper

For a student

This is the second of Littlewood's three principles. It explains why continuous functions with compact support are dense in $L^p$ spaces for $1 \le p < \infty$, together with the density of simple functions. It is a basic tool for passing from statements about continuous functions to statements about measurable ones. In the other direction, a function with this property for every $\varepsilon$ is measurable, so the property characterises measurability on sets of finite measure.