Exercise 6, Chapter 2, Rudin (Real and Complex Analysis): Construct a totally disconnected compact set $K \subset \mathbb R^1$ such that $m(K) > 0$
($K$ is to have no connected subset consisting of more than one point). If $v$ is lower semicontinuous and $v \le \chi_K$, show that actually $v \le 0$. Hence $\chi_K$ cannot be approximated from below by lower semicontinuous functions, in the sense of the Vitali–Caratheodory
theorem.
The first badly disconnected set that comes to mind is the Cantor set $F = \bigcap_{n\ge 1} E_n$, but $m(F) = 0$ so that doesn't work. Is there some way we can modify the Cantor set in its inductive construction so as to construct an appropriate totally disconnected compact set $K \subset [0,1]$ with positive measure?
About lower semincontinuity: $f$ is lower semicontinuous if $$\{x: f(x) > \alpha\}$$ is open for every $\alpha\in\mathbb R$. Since whatever $K$ we construct is compact, it is also closed. We know that characteristic functions of closed sets are upper semincontinuous, i.e. $\{x: f(x) > \alpha\}$ is open for every $\alpha\in\mathbb R$. Does this help in any way?
Lastly, what would this have to do with the Vitali-Caratheodory theorem?
Thanks for the help.
Best Answer
Consider any fat Cantor set $K$. There several ways to construct them. The Wikipedia article quoted here shows one way.
The important thing is to notice that $K$ has empty interior. If $v$ is a lower semicontinuous function and $v\leq \mathbb{1}_K$, then $\{v>0\}\subset K$ and so, being that $\{v>0\}$ is open, it must be the case that $\{v>0\}=\emptyset$; hence $v\leq 0$. Since $K$ is compact, $u=\mathbb{1}_{K}$ is upper semicontinuous. So, for any other upper semicontinuous function $\widetilde{u}\geq\mathbb{1}_K$ $$\int(\widetilde{u}-v)\geq \int(u-v)\geq m(K)$$ where $m$ is Lebesgue's measure on $\mathbb{R}$.
What the problem tells you is that for $f\in\mathcal{L}_1$, and $\varepsilon>0$, it is not always possible to find lower and upper semicontinuous function $v$ and $u$ such that $v\leq f\leq u$ and $\int(u-v)<\varepsilon$. (Notice that in Vitali-Caratheodory's theorem, the roles of the functions $v$ and $u$ are reversed, that is $u\leq f\leq v$ and $\int(v-u)<\varepsilon$).