# Lusin's Theorem - Proof Verification (Folland Chap 2, Exercise 44)

Conisder Lusin's Theorem (Folland Chap 2 Ex 44):

If $$f: [a,b] \rightarrow \mathbb{C}$$ is Lebesgue measurable and $$\epsilon > 0$$, there is a compact subset $$E \in [a,b]$$ s.t. $$\mu(E^C) < \epsilon$$ and $$f$$ restricted to $$E$$ is continuous.

To prove this, I basically:

1. Defined $$E_k = \{x | x \in [a,b], |f(x)| and $$f_k' = f \chi_{E_k}$$. Then $$f_k' \rightarrow f$$ p.w.
2. Approximated each $$f_k'$$ is a continuous function $$f_k$$ (one can show each $$f_k'$$ is $$L^1$$)
3. $$f_k \rightarrow f$$ p.w. so in particular, $$f_k \rightarrow f$$ a.e
4. By Egoroff's Theorem $$\exists F \subset [a,b]$$ s.t. $$\mu(F) < \epsilon$$ and $$f_k \rightarrow f$$ unifomly on $$F^c$$. Ergo $$f$$ is continuous on $$F^C$$

To complete, I think I can get a compact set by finding one inside $$F^C$$ because $$F^C$$ is Borel. But I really didn't use compactness in an essential way to obtain the conclusion. Am I missing something?

• The definition of $E_k$ does not seem to be correct. Since $f(x)$ is a complex number, $f(x)<k$ doesn't make sense. – Edmundo Martins Oct 3 '18 at 14:07
• It should be magnitude of $f(x)$ – yoshi Oct 3 '18 at 14:13
• See this link people.math.gatech.edu/~heil/6337/spring11/lusin.pdf – xpaul Oct 3 '18 at 15:51

Let $$f: [a,b]\to \mathbb{C}$$ be Lebesgue measurable and $$\epsilon > 0$$. By theorem 2.26 we can build a sequence of continuous functions $$\{g_n\}$$ such that
$$g_n\to f \ \ \text{in} \ \ L^1$$
Then by Corollary 2.32 there is a sub-sequence $$\{g_{n_j}\}$$ of $$\{g_n\}$$ such that $$g_{n_j}\to f$$ almost everywhere. Now by Egoroff's theorem for any $$\epsilon > 0$$ there exists a set $$F\subset [a,b]$$ with $$\mu(F) < \epsilon/2$$ such that $$g_{n_j}\to f$$ uniformly on $$F^{c}$$.
Now by theorem 1.18, since $$\mu([a,b]) < \infty$$, there is $$E$$ compact subset of $$[a,b]$$ such that $$E\subset F^c$$ and
\begin{align*} \mu(E^c) &= \mu(F) + \mu(E^c\setminus F)\\ &= \mu(F) + \mu(E^c\cap F^c)\\ &= \mu(F) + \mu(F^c\setminus E)\\ &= \mu(F) + (\mu(F^c) - \mu(E))\\ &\leq \epsilon/2 + \epsilon/2\\ &= \epsilon \end{align*} Note since $$E\subset F^c$$ and $$g_{n_j}\to f$$ uniformly on $$F^c$$, we have that $$g_{n_j}\to f$$ uniformly on $$E$$. Since, for all $$j$$, $$g_{n_j}$$ is continuous, we have that $$f$$ is continuous on $$E$$, that is, $$f|E$$ is continuous.