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Math Help - Lusin Theorem.

  1. #1
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    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?
    Last edited by younhock; December 21st 2011 at 06:43 AM.
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  2. #2
    Super Member girdav's Avatar
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    Re: Lusin Theorem.

    What do we know about A?
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  3. #3
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    Re: Lusin Theorem.

    Quote Originally Posted by girdav View Post
    What do we know about A?
    A is a bounded measurable set.
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  4. #4
    Super Member girdav's Avatar
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    Re: Lusin Theorem.

    On which measurable space are you working?
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  5. #5
    Senior Member vincisonfire's Avatar
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    Re: Lusin Theorem.

    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
    Quote Originally Posted by younhock View Post
    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.
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  6. #6
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    Re: Lusin Theorem.

    Quote Originally Posted by girdav View Post
    On which measurable space are you working?
    lebegue measure. Sorry.
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  7. #7
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    Re: Lusin Theorem.

    Quote Originally Posted by vincisonfire View Post
    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?
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  8. #8
    Senior Member vincisonfire's Avatar
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    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?
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  9. #9
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    Re: Lusin Theorem.

    Quote Originally Posted by vincisonfire View Post
    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."
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  10. #10
    Senior Member vincisonfire's Avatar
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    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.
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  11. #11
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    Re: Lusin Theorem.

    Quote Originally Posted by vincisonfire View Post
    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.
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  12. #12
    Senior Member vincisonfire's Avatar
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    Re: Lusin Theorem.

    No because if m(A)>0 then you won't have m(A\backslash\phi)<\epsilon.
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  13. #13
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    Re: Lusin Theorem.

    Quote Originally Posted by vincisonfire View Post
    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.
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  14. #14
    Senior Member vincisonfire's Avatar
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    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.
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  15. #15
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    Re: Lusin Theorem.

    Quote Originally Posted by vincisonfire View Post
    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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