Lusin Theorem.

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• Dec 21st 2011, 05:45 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)< $\infty$, then given any $\varepsilon > 0$, there exist a closed set $C_\varepsilon \subseteq A$ such that $m(A\backslash C_\varepsilon)<\varepsilon$ and f is continuous on $C_\varepsilon$.

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

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

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

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

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

Then $C_0$ is closed and

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

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

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

MY QUESTION is : Can $C_0$ be empty set ? Why?
• Dec 21st 2011, 06:15 AM
girdav
Re: Lusin Theorem.
What do we know about $A$?
• Dec 21st 2011, 06:27 AM
younhock
Re: Lusin Theorem.
Quote:

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

A is a bounded measurable set.
• Dec 21st 2011, 06:46 AM
girdav
Re: Lusin Theorem.
On which measurable space are you working?
• Dec 21st 2011, 06:56 AM
vincisonfire
Re: Lusin Theorem.
I understand the concern if you look at $C_0= \bigcap_{n=1}^{\infty} C_n$ alone. It seems to me that you should not worry because of your construction. Doesn't the next line
Quote:

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

make you confident that $C_0$ is not empty, unless $A$ is.
• Dec 21st 2011, 07:19 AM
younhock
Re: Lusin Theorem.
Quote:

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

lebegue measure. Sorry.
• Dec 21st 2011, 07:22 AM
younhock
Re: Lusin Theorem.
Quote:

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

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

But if $C_0$ is empty, then $C \subseteq C_0$ must be empty too. Then what is meant by f is continuous on $C$ which is an empty set?
• Dec 21st 2011, 07:36 AM
vincisonfire
Re: Lusin Theorem.
Well, what I was thinking is the following. You derived that $m(A \backslash C_0)<\epsilon$ for arbitrary $\epsilon$. If $C_0$ is empty then $m(A)<\epsilon$ for arbitrary $\epsilon$ or $m(A)=0$. In that case, doesn't the problem become trivial?
• Dec 21st 2011, 07:46 AM
younhock
Re: Lusin Theorem.
Quote:

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

It is not trivial to me because if $C_\varepsilon$ is empty. then the conclusion for that theorem will become $f$ is continous on $C_\varepsilon$ but here the $C_\varepsilon$ is empty. It is very unclear to me about " f is continuous on an empty set."
• Dec 21st 2011, 07:54 AM
vincisonfire
Re: Lusin Theorem.
$f$ is continuous if the preimage of an open set is an open set.
I think this condition is satisfied for $f:\phi\rightarrow\phi$.
• Dec 21st 2011, 08:05 AM
younhock
Re: Lusin Theorem.
Quote:

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

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) < $\infty$.
Because $\phi \subseteq A$ for any set A. and thus $\phi$ will be the $C_\varepsilon$ we are searching for.
• Dec 21st 2011, 08:09 AM
vincisonfire
Re: Lusin Theorem.
No because if $m(A)>0$ then you won't have $m(A\backslash\phi)<\epsilon$.
• Dec 21st 2011, 08:21 AM
younhock
Re: Lusin Theorem.
Quote:

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

I see. Thank you very much. I try to look back again.
• Dec 21st 2011, 08:48 AM
vincisonfire
Re: Lusin Theorem.
In conclusion, I think your proof doesn't require modification. However, one could want clarification for the case were $C_0 = \phi$. In this case, you can still write $m(A\backslash C_0)$= $m(A\backslash \bigcap_{n=1}^{\infty} C_n )$= $m(\bigcup_{n=1}^{\infty} (A\backslash C_n))$ $\leq$ $\sum_{n=1}^{\infty} m(A\backslash C_n)$< $\varepsilon \sum_{n=1}^{\infty} \frac{1}{2^{n+1}}$ = $\frac{\varepsilon}{2}$. This means that $m(A)=0$. But in the case where $m(A)=0$, the empty set $\phi$ satisfies the conditions of the theorem.
• Dec 21st 2011, 09:05 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 $C_0 = \phi$. In this case, you can still write $m(A\backslash C_0)$= $m(A\backslash \bigcap_{n=1}^{\infty} C_n )$= $m(\bigcup_{n=1}^{\infty} (A\backslash C_n))$ $\leq$ $\sum_{n=1}^{\infty} m(A\backslash C_n)$< $\varepsilon \sum_{n=1}^{\infty} \frac{1}{2^{n+1}}$ = $\frac{\varepsilon}{2}$. This means that $m(A)=0$. But in the case where $m(A)=0$, the empty set $\phi$ satisfies the conditions of the theorem.

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