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

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

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


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

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

    So, MY QUESTIONS ARE :
    1) What is this set $\displaystyle E_{m,n} = \bigcap_{i\geq n}$ { $\displaystyle x : |f_i (x)-f(x)|< \frac{1}{m}$ } 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 $\displaystyle n_m$ such that $\displaystyle m(A\backslash E_{m,n_m}) $ < $\displaystyle \frac{\varepsilon}{2^m} $ "

    3) Where does the properties m(A) < $\displaystyle \infty$ 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 $\displaystyle f_n$ converges pointwise to f as an union of such sets. We take $\displaystyle \frac 1m$ instead of $\displaystyle \delta\in\mathbb R_+^*$ in order to get a countable family.
    2) Since $\displaystyle \lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\displaystyle \delta=\varepsilon 2^{-m}$ to choose $\displaystyle n_m$ such that $\displaystyle m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $\displaystyle m(E_{m,k})\leq \varepsilon 2^{-m}$ for $\displaystyle k\geq n_m$, but it's not necessary here).
    3) We used the fact that $\displaystyle m(A)<\infty$ when we wrote $\displaystyle m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.
    Last edited by girdav; Nov 23rd 2011 at 12:06 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 $\displaystyle f_n$ converges pointwise to f as an union of such sets. We take $\displaystyle \frac 1m$ instead of $\displaystyle \delta\in\mathbb R_+^*$ in order to get a countable family.
    2) Since $\displaystyle \lim_{n\to\infty}m(E_{m,n})=0$ for any m, we use the definition of the limit to $\displaystyle \delta=\varepsilon 2^{-m}$ to choose $\displaystyle n_m$ such that $\displaystyle m(E_{m,n_m})\leq \varepsilon 2^{-m}$ (we can choose it to have the inequality $\displaystyle m(E_{m,k})\leq \varepsilon 2^{-m}$ for $\displaystyle k\geq n_m$, but it's not necessary here).
    3) We used the fact that m(A)<\infty when we wrote $\displaystyle m(A\setminus E_{m,n})=m(A)-m(E_{m,n})$.
    Can the set on which converges pointwise to f be written as $\displaystyle \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} ?
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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 $\displaystyle \bigcap_{m\geq1} \bigcup_{n\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$} AND
    $\displaystyle \bigcap_{m\geq1} \bigcap_{i\geq n}${$\displaystyle x: |f_i(x)-f(x)|\lneq \frac{1}{m}$}

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