0
$\begingroup$

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 $E_k = \{x | x \in [a,b], |f(x)|<k\}$ 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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Edmundo Martins Oct 3 '18 at 14:07
  • $\begingroup$ It should be magnitude of $f(x)$ $\endgroup$ – yoshi Oct 3 '18 at 14:13
  • $\begingroup$ See this link people.math.gatech.edu/~heil/6337/spring11/lusin.pdf $\endgroup$ – xpaul Oct 3 '18 at 15:51
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.