A Question on Luzin's Theorem

From: Li Yi (l-o-w-a-i_at_163.com)
Date: 08/23/04 

Date: Mon, 23 Aug 2004 14:30:02 +0000 (UTC)

THEOREM Let $f(x)$ be a measurable function (in Lebesgue's sense) on
$E\subset \mathbb{R}$ and almost everywhere bounded over $E$. Then
$\forall \delta>0$, $\exists F \subset E$, that is closed, and
(i) $\mu (E \subset F)<\delta$
(ii) $f(x)$ is continous on $F$.

PROOF First consider that $f$ is measurable simple function.

\[f(x) = \sum_{i=1}^p c_i \chi_{E_i} (x), \quad x \in E =
\bigcup_{i=1}^p E_i, \quad E_i \cap E_j = \emptyset (i\neq j)\]

Now, $\forall \delta>0$ and every $E_i$, we can find closed set
$F_i\subset E_i$ such that \[m(E_i\setminus F_i) <
\frac{\delta}{p},\quad i=1,2,\dots,p\]

because $f(x)=c_i$ when $x\in F_i$, $f(x)$ is continous on $F_i$. And
$F_1,\dots,F_p$ are disjoint, we know that $f$ is continous on
\[F=\bigcup_{i=1}^p F_i.\]
It is easy to see that $F$ is closed and
\[\mu(E\setminus F) = \sum_{i=1}^p \mu(E_i\setminus F_i) < \sum_{i=1}^p
\frac{\delta}{p} = \delta\]

Now we come back to the general case.

Since [tex]g(x) = \frac{f(x)}{1+|f(x)|}[/tex] ([tex]f(x) =
\frac{g(x)}{1-|g(x)|}[/tex])
we can suppose $f(x)$ is bounded without loss of generality. There
exists measurable simple functions $\{\phi_k(x)\}$ converges to $f(x)$
uniformly on $E$ (See also Theorem 1.17, Real and Complex Analysis,
Rudin). For any given $\delta > 0$ and each $\phi_k(x)$ we can find
closed set $F_k\subset E$ such that
\[\mu(E\setminus F_k) < \frac{\delta}{2^k},\]
and $\phi_k(x)$ is continuous on $F_k$. Let
\[F = \bigcap_{k=1}^\infty F_k,\]
we have $F\subset E$ and \[\mu(E\setminus F) \leq \sum_{k=1}^\infty
m(E\setminus F_k) < \delta\]

Since every $\phi_k(x)$ is continuous on $F$ and from the uniformly
convergence, we know that $f(x)$ is continuous on $F$.

Q.E.D

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My question is: Where is Egorov's Theorem used in the proof above?