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Suppose $A$ is a Borel measurable subset of [0,1], $m$ is Lebesgue measure, and $\varepsilon\in (0,1)$. Prove that there exists a continuous function $f: [0,1]\to \mathbb{R}$ such that $0\le f\le 1$ and $$ m(\{x:f(x)\ne\chi_A(x)\})<\varepsilon. $$ Here $\chi_A(x)$ is the indicator function on $A$.

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  • $\begingroup$ Hint: Apply Dynkin's $\pi-\lambda$ theorem. $\endgroup$ – pre-kidney Mar 8 '17 at 5:28
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If $A$ is Borel measurable, then it is certainly Lebesgue measurable. The Lebesgue measure is regular, meaning that for any $\epsilon > 0$, you can find a closed set $F$ and an open set $U$ with $F \subset A \subset U$ such that $ m(U \backslash F) < \epsilon.$

By Urysohn's lemma, there exists a continuous function $f:[0,1]\to \mathbb R$ with $0 \leq f \leq 1$ such that $f(x) = 0$ for $x \notin U$ and $f(x) = 1$ for $x \in F$. Thus the only place where $f \neq \chi_A$ is the region $U \backslash F$, whose measure is less than $\epsilon$ by assumption.

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