I have a doubt in the proof of lusin theorem. Before i ask the question, let me state down the proof first.

Lusin Theorem : If f is a measurable function on A where m(A)<$\displaystyle \infty$, then given any $\displaystyle \varepsilon > 0$, there exist a closed set $\displaystyle C_\varepsilon \subseteq A $ such that $\displaystyle m(A\backslash C_\varepsilon)<\varepsilon$ and f is continuous on $\displaystyle C_\varepsilon$.

PROOF :

First, let $\displaystyle f=\sum_{k=1}^{n}a_k\chi_{E_k}$ be a simple function on A. Given $\displaystyle \varepsilon>0$, for each k there exists a closed set $\displaystyle C_k \subset E_k$ such that

$\displaystyle m(E_k\backslash C_k)<\frac{\varepsilon}{n}$

Now C=$\displaystyle \bigcup_{k=1}^{n}C_k$ is closed, $\displaystyle m(A\backslash C)<\varepsilon$ and f is continuous on C since C is equal to the disjoint union of closed sets $\displaystyle C_k$ and f is constant on each $\displaystyle C_k$.

Now let f be any measurable function. Since $\displaystyle f=f^+-f^-$, we may assume f is non-negative. Since f is measurable, $\displaystyle f=\lim_{n\rightarrow\infty}f_n$ for a sequence {$\displaystyle f_n$} of simple functions.

By the first part of the proof, given $\displaystyle \varepsilon >0 $, there exists for each$\displaystyle n$ a closed set $\displaystyle C_n$ such that $\displaystyle m(A\backslash C_n) < \frac{\varepsilon}{2^{n+1}}$ and $\displaystyle f_n$ is continuous on $\displaystyle C_n$.

Let.$\displaystyle C_0= \bigcap_{n=1}^{\infty} C_n$

Then $\displaystyle C_0$ is closed and

$\displaystyle m(A\backslash C_0)$= $\displaystyle m(A\backslash \bigcap_{n=1}^{\infty} C_n ) $=$\displaystyle m(\bigcup_{n=1}^{\infty} (A\backslash C_n))$$\displaystyle \leq $$\displaystyle \sum_{n=1}^{\infty} m(A\backslash C_n) $<$\displaystyle \varepsilon \sum_{n=1}^{\infty} \frac{1}{2^{n+1}}$ =$\displaystyle \frac{\varepsilon}{2}$.

By egoroff theorem, there is a measurable subset $\displaystyle C\subseteq C_0 $ such that $\displaystyle m(C_0 \backslash C) < \frac{\varepsilon}{4}$ and $\displaystyle f_n \rightarrow f$ uniformly on C.

Now each $\displaystyle f_n$ is continuous on $\displaystyle C \subseteq C_0 $ so that $\displaystyle f$ , being the uniform limit of the $\displaystyle f_n$, is continuous on $\displaystyle C$. If $\displaystyle C $ is closed we are through. If not, take closed set $\displaystyle F \subseteq C$ such that $\displaystyle m(C\backslash F) < \frac{\varepsilon}{4}$.

MY QUESTION is : Can $\displaystyle C_0 $ be empty set ? Why?