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.

  • $\begingroup$ This does not work. $g$ need not be continuous. $\endgroup$ – Kavi Rama Murthy Feb 22 at 7:14
  • $\begingroup$ 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 $\endgroup$ – The Bosco Feb 22 at 7:15
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Feb 22 at 7:17
  • $\begingroup$ Oh, the approach does not work. Could you give me a hint please? $\endgroup$ – The Bosco Feb 22 at 7:18
  • $\begingroup$ A continuous function $\mathbb R\to\mathbb R$ cannot take just two values. $\endgroup$ – MPW Feb 22 at 7:44

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