Prove if $E$ is a Lebesgue measurable set, there exists a continuous function $f$ differing from $\chi_{E}$ on a set of measure $< \epsilon$?

Question

I am reviewing my analysis notes, and I don't really understand the proof given by my professor. He first proved if $E$ is a Lebesgue measurable set and $\epsilon > 0$, then there is an open set $O$ such that $m(O) - m(E) < \epsilon$, which makes perfect sense. Then, using that fact, he proved that, given $\epsilon > 0$, $\exists$ a continuous function $f$ such that $m (\{ x \mid f(x) \neq \chi_{E}(x) \} ) < \epsilon$. Anyone know how to construct the function $f$?

Answer 1

Let $O \subseteq \mathbb{R}$ be an open set and define

$$O_n := \left\{x \in \mathbb{R}; d(x,\mathbb{R} \backslash O) \geq \frac{1}{n} \right\}.$$

It is not difficult to show that $O_n$ is closed and $O_n \uparrow O$. Now choose $n$ sufficiently large such that $m(O \backslash O_n)<\varepsilon$. It is not difficult to show that there exists a continuous functions $f_n$ such that $0 \leq f_n \leq 1$ and

$$f_n|_{O_n}=1 \qquad \qquad f_n|_{\mathbb{R} \backslash O} = 0;$$

this follows from Urysohn's Lemma. In particular, $$m(\{x; f(x) \neq 1_{O}(x)\})< \varepsilon.$$ Using that $m(O \backslash E)<\varepsilon$, the claim follows.