# Proof of Lusin's Theorem with the characteristic function

Let $$m$$ be the Lebesgue measure and the set $$E$$ be Lebesgue-measurable, and $$m(E)<\infty$$. Prove that for any $$\epsilon>0$$ there is a compactly supported continuous function $$g:\mathbb{R} \to \mathbb{R}$$ such thath $$m(\{ x:\chi_E(x) \neq g(x)\})<\epsilon$$

My attempt: We can define $$E=\bigcup_{k=0}^n E_k$$ where each $$E_k$$ has finite measure and they are disjoint and $$n<\infty$$.

By regularity of the Lebesgue measure, we know that $$m(E) = \sup \{m(K):K \text{ compact and }K\subset E\}$$

and for each $$1\leq k \leq n$$ then $$\forall \epsilon/n > 0$$, we have compact sets $$F_k \subset E_k$$, such that $$m(E_k) < m(F_k) + \epsilon/n$$, therefore $$m(E_k-F_k) < \epsilon/n$$

since each $$E_k$$ is disjoint

$$m \bigg(\bigcup_{k=1}^n (E_k - F_k)\bigg) = \sum_{k=1}^n(E_k-F_k) < \frac{\epsilon}{n}n = \epsilon$$

Define the function $$g$$

$$g = \begin{cases} 1 & x \in (a_k, b_k) \\ \frac{x+\delta_k-a_k}{\delta_k} & x\in[a_k - \delta_k, a_k] \\ \frac{-x + b_k+\delta_k}{\delta_k} & x \in [b_k, b_k + \delta_k] \\ 0 & \text{otherwise} \end{cases}$$

For each interval $$E_k = [a_k, b_k]$$ and arbitrary $$\delta_k$$ for each interval as well.

• This does not work. $g$ need not be continuous. – Kavi Rama Murthy Feb 22 at 7:14
• I need to prove it for a continuous one though. It is part of the problem prompt. I read that I can argue it is continuous on the restriction to the union of the $F_k$ sets – The Bosco Feb 22 at 7:15
• The characteristic function of a set $A$ is continuous iff $A$ is either the empty set or the whole space. So your approach doesn't work. – Kavi Rama Murthy Feb 22 at 7:17
• Oh, the approach does not work. Could you give me a hint please? – The Bosco Feb 22 at 7:18
• A continuous function $\mathbb R\to\mathbb R$ cannot take just two values. – MPW Feb 22 at 7:44