# Lusin Theorem.

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• Dec 21st 2011, 04:45 AM
younhock
Lusin Theorem.
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?
• Dec 21st 2011, 05:15 AM
girdav
Re: Lusin Theorem.
What do we know about $\displaystyle A$?
• Dec 21st 2011, 05:27 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by girdav
What do we know about $\displaystyle A$?

A is a bounded measurable set.
• Dec 21st 2011, 05:46 AM
girdav
Re: Lusin Theorem.
On which measurable space are you working?
• Dec 21st 2011, 05:56 AM
vincisonfire
Re: Lusin Theorem.
I understand the concern if you look at $\displaystyle C_0=%20\bigcap_{n=1}^{\infty}%20C_n$ alone. It seems to me that you should not worry because of your construction. Doesn't the next line
Quote:

Originally Posted by younhock
$\displaystyle m(A\backslash C_0)$= $\displaystyle m(A\backslash \bigcap_{n=1}^{\infty} C_n )$=$\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}}$ =$\displaystyle \frac{\varepsilon}{2}$.

make you confident that $\displaystyle C_0$ is not empty, unless $\displaystyle A$ is.
• Dec 21st 2011, 06:19 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by girdav
On which measurable space are you working?

lebegue measure. Sorry.
• Dec 21st 2011, 06:22 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by vincisonfire
I understand the concern if you look at $\displaystyle C_0=%20\bigcap_{n=1}^{\infty}%20C_n$ alone. It seems to me that you should not worry because of your construction. Doesn't the next line

make you confident that $\displaystyle C_0$ is not empty, unless $\displaystyle A$ is.

But if $\displaystyle C_0$ is empty, then $\displaystyle C \subseteq C_0$ must be empty too. Then what is meant by f is continuous on $\displaystyle C$ which is an empty set?
• Dec 21st 2011, 06:36 AM
vincisonfire
Re: Lusin Theorem.
Well, what I was thinking is the following. You derived that $\displaystyle m(A \backslash C_0)<\epsilon$ for arbitrary $\displaystyle \epsilon$. If $\displaystyle C_0$ is empty then $\displaystyle m(A)<\epsilon$ for arbitrary $\displaystyle \epsilon$ or $\displaystyle m(A)=0$. In that case, doesn't the problem become trivial?
• Dec 21st 2011, 06:46 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by vincisonfire
Well, what I was thinking is the following. You derived that $\displaystyle m(A \backslash C_0)<\epsilon$ for arbitrary $\displaystyle \epsilon$. If $\displaystyle C_0$ is empty then $\displaystyle m(A)<\epsilon$ for arbitrary $\displaystyle \epsilon$ or $\displaystyle m(A)=0$. In that case, doesn't the problem become trivial?

It is not trivial to me because if $\displaystyle C_\varepsilon$ is empty. then the conclusion for that theorem will become $\displaystyle f$ is continous on $\displaystyle C_\varepsilon$ but here the $\displaystyle C_\varepsilon$ is empty. It is very unclear to me about " f is continuous on an empty set."
• Dec 21st 2011, 06:54 AM
vincisonfire
Re: Lusin Theorem.
$\displaystyle f$ is continuous if the preimage of an open set is an open set.
I think this condition is satisfied for $\displaystyle f:\phi\rightarrow\phi$.
• Dec 21st 2011, 07:05 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by vincisonfire
$\displaystyle f$ is continuous if the preimage of an open set is an open set.
I think this condition is satisfied for $\displaystyle f:\phi\rightarrow\phi$.

So do you mean that lusin theorem is true for any set A as long as f is measurable on lebesgue measurable set A and m(A) < $\displaystyle \infty$.
Because $\displaystyle \phi \subseteq A$ for any set A. and thus $\displaystyle \phi$ will be the $\displaystyle C_\varepsilon$ we are searching for.
• Dec 21st 2011, 07:09 AM
vincisonfire
Re: Lusin Theorem.
No because if $\displaystyle m(A)>0$ then you won't have $\displaystyle m(A\backslash\phi)<\epsilon$.
• Dec 21st 2011, 07:21 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by vincisonfire
No because if $\displaystyle m(A)>0$ then you won't have $\displaystyle m(A\backslash\phi)<\epsilon$.

I see. Thank you very much. I try to look back again.
• Dec 21st 2011, 07:48 AM
vincisonfire
Re: Lusin Theorem.
In conclusion, I think your proof doesn't require modification. However, one could want clarification for the case were $\displaystyle C_0 = \phi$. In this case, you can still write $\displaystyle m(A\backslash C_0)$= $\displaystyle m(A\backslash \bigcap_{n=1}^{\infty} C_n )$=$\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}}$ =$\displaystyle \frac{\varepsilon}{2}$. This means that $\displaystyle m(A)=0$. But in the case where $\displaystyle m(A)=0$, the empty set $\displaystyle \phi$ satisfies the conditions of the theorem.
• Dec 21st 2011, 08:05 AM
younhock
Re: Lusin Theorem.
Quote:

Originally Posted by vincisonfire
In conclusion, I think your proof doesn't require modification. However, one could want clarification for the case were $\displaystyle C_0 = \phi$. In this case, you can still write $\displaystyle m(A\backslash C_0)$= $\displaystyle m(A\backslash \bigcap_{n=1}^{\infty} C_n )$=$\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}}$ =$\displaystyle \frac{\varepsilon}{2}$. This means that $\displaystyle m(A)=0$. But in the case where $\displaystyle m(A)=0$, the empty set $\displaystyle \phi$ satisfies the conditions of the theorem.

I Think i get it d. Thanks very much ^.^
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