Lusin Theorem.

**younhock** · December 21st 2011, 04:45 AM

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?

**girdav** · December 21st 2011, 05:15 AM

What do we know about $A$ ?

**younhock** · December 21st 2011, 05:27 AM

Originally Posted by girdav

What do we know about $A$ ?

A is a bounded measurable set.

**girdav** · December 21st 2011, 05:46 AM

On which measurable space are you working?

**vincisonfire** · December 21st 2011, 05:56 AM

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

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.

**younhock** · December 21st 2011, 06:19 AM

Originally Posted by girdav

On which measurable space are you working?

lebegue measure. Sorry.

**younhock** · December 21st 2011, 06:22 AM

Originally Posted by vincisonfire

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

make you confident that $C_0$ 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?

**vincisonfire** · December 21st 2011, 06:36 AM

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?

**younhock** · December 21st 2011, 06:46 AM

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."

**vincisonfire** · December 21st 2011, 06:54 AM

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

**younhock** · December 21st 2011, 07:05 AM

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.

**vincisonfire** · December 21st 2011, 07:09 AM

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

**younhock** · December 21st 2011, 07:21 AM

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.

**vincisonfire** · December 21st 2011, 07:48 AM

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.

**younhock** · December 21st 2011, 08:05 AM

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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