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Math Help - Egoroff's Theorems Proof

  1. #1
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    Egoroff's Theorems Proof

    There are some part of this proof I couldn't understand. First, let me state the theorem and its proof which quoted from an analysis book.

    EGOROFF's THEOREM :
     m(A)< \infty where A\subset$\mathbb{R}$ . Let  f_n and  f be functions defined on A such that each  f_n \rightarrow f almost everywhere on A. Then given any  \epsilon > 0 there exists a measurable subset A_\epsilon \subset A such that  m(A\backslash A_\epsilon)<\epsilon and f_n \rightarrow f uniformly on A_\epsilon .

    PROOF :
    Given that  \varepsilon >0 and positive integers m and n, define the set
    E_{m,n} = \bigcap_{i\geq n} { x : |f_i (x)-f(x)|< \frac{1}{m} } where  i\in $\mathbb{N}$ .
    If U is the subset on which  f_n \rightarrow f , then for any m ,  U\subset \bigcup_{n\geq 1}E_{m,n} \subset A . And Hence we get  m(\bigcup_{n\geq 1} E_{m,n})=m(A) .
    But,  E_{m,n} \subset E_{m,n+1} for all m and n, so lim_{n\rightarrow \infty} m(A\backslash E_{m,n}) =  lim_{n\rightarrow\infty}[m(A)-m(E_{m,n})] = m(A)-m(\bigcup_{n\geq1}E_{m,n}) = 0

    Thus for each m, there exists an integer n_m such that  m(A\backslash E_{m,n_m}) <  \frac{\varepsilon}{2^m} .


    Now let A_\epsilon =  \bigcap_{m\geq1}E_{m,n_m} . Then A_\epsilon is measurable and m(A\backslash A_\epsilon)=  m(\bigcup_{m\geq1}(A\backslash E_{m,n_m})) \leq \sum_{m\geq1}m(A-E_{m,n_m})\lneq \sum_{m\geq1}\frac{\epsilon}{2^m}= \varepsilon.

    It is easy to show that f_n \rightarrow f uniformly on A.

    So, MY QUESTIONS ARE :
    1) What is this set E_{m,n} = \bigcap_{i\geq n} { x : |f_i (x)-f(x)|< \frac{1}{m} } defined at the beginning of this proof and why it is defined like this?

    2) How can we conclude that " for each m, there exists an integer n_m such that  m(A\backslash E_{m,n_m}) <  \frac{\varepsilon}{2^m} "

    3) Where does the properties m(A) < \infty used in the proof?
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  2. #2
    Super Member girdav's Avatar
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    Re: Egoroff's Theorems Proof

    1) We write the set on which f_n converges pointwise to f as an union of such sets. We take \frac 1m instead of \delta\in\mathbb R_+^* in order to get a countable family.
    2) Since \lim_{n\to\infty}m(E_{m,n})=0 for any m, we use the definition of the limit to \delta=\varepsilon 2^{-m} to choose n_m such that m(E_{m,n_m})\leq \varepsilon 2^{-m} (we can choose it to have the inequality m(E_{m,k})\leq \varepsilon 2^{-m} for k\geq n_m, but it's not necessary here).
    3) We used the fact that m(A)<\infty when we wrote m(A\setminus E_{m,n})=m(A)-m(E_{m,n}).
    Last edited by girdav; November 23rd 2011 at 01:06 PM.
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  3. #3
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    Re: Egoroff's Theorems Proof

    Quote Originally Posted by girdav View Post
    1) We write the set on which f_n converges pointwise to f as an union of such sets. We take \frac 1m instead of \delta\in\mathbb R_+^* in order to get a countable family.
    2) Since \lim_{n\to\infty}m(E_{m,n})=0 for any m, we use the definition of the limit to \delta=\varepsilon 2^{-m} to choose n_m such that m(E_{m,n_m})\leq \varepsilon 2^{-m} (we can choose it to have the inequality m(E_{m,k})\leq \varepsilon 2^{-m} for k\geq n_m, but it's not necessary here).
    3) We used the fact that m(A)<\infty when we wrote m(A\setminus E_{m,n})=m(A)-m(E_{m,n}).
    Can the set on which converges pointwise to f be written as \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}{ x: |f_i(x)-f(x)|\lneq \frac{1}{m}} ?
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  4. #4
    Super Member girdav's Avatar
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    Re: Egoroff's Theorems Proof

    Yes, you can see that using the definition of the limit.
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  5. #5
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    Re: Egoroff's Theorems Proof

    Quote Originally Posted by girdav View Post
    Yes, you can see that using the definition of the limit.
    Then what is difference between \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}{ x: |f_i(x)-f(x)|\lneq \frac{1}{m}} AND
    \bigcap_{m\geq1}  \bigcap_{i\geq n}{ x: |f_i(x)-f(x)|\lneq \frac{1}{m}}

    both also refer to pointwise convergence? or one of them is refer to uniform convergence?
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