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Lusin's Theorem asserts that a measurable function f is nearly continuous in the sense that for all $\epsilon>0$ there is a set S of measure less than $\epsilon$ such that f is continuous on the complement of S.

I am looking for an example of such a function which is not continuous on the complement of any zero set. For instance, something like the characteristic function of the rationals won't do, since it is continuous on the complement of the rationals, which is a zero set.

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1 Answer 1

The characteristic function of a fat Cantor set gives an example (or any closed set with positive measure and empty interior). If $C$ is such a set, and $A$ is any null set, then $\mathbb{R}\setminus A$ contains an element $x$ of $C$. Every neighborhood of $x$ intersects $\mathbb{R}\setminus C$ in a set of positive measure (because nonempty open sets have positive measure), so for all $\delta>0$, $(x-\delta,x+\delta)\setminus A$ contains elements of $\mathbb R\setminus C$. This implies that the restriction of $\chi_C$ to $\mathbb{R}\setminus A$ is discontinuous at $x$.

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