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Math Help - convergence in measure

  1. #1
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    convergence in measure

    f_n \rightarrow f in L^2. Show that f_n \rightarrow f in measure.
    any hint would be appreciated.
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  2. #2
    Moo
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    Hello,

    \forall \epsilon>0,\int (f_n-f)^2 ~d\mu \geq \int_{\{|f_n-f|>\epsilon\}} (f_n-f)^2 ~d\mu>\epsilon^2 \mu(|f_n-f|>\epsilon)\geq 0

    and since the LHS goes to 0 (convergence in Lē), then the RHS goes to 0 too, and hence the convergence in measure.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    \forall \epsilon>0,\int (f_n-f)^2 ~d\mu \geq \int_{\{|f_n-f|>\epsilon\}} (f_n-f)^2 ~d\mu>\epsilon^2 \mu(|f_n-f|>\epsilon)\geq 0

    and since the LHS goes to 0 (convergence in Lē), then the RHS goes to 0 too, and hence the convergence in measure.
    RHS goes to 0 because \epsilon goes to 0, is it correct? how do i know that \mu (|f-f_n >|\epsilon) goes to 0?
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  4. #4
    Moo
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    I'm rephrasing it

    \forall \epsilon>0,{\color{red}\int (f_n-f)^2 ~d\mu} \geq \int_{\{|f_n-f|>\epsilon\}} (f_n-f)^2 ~d\mu>{\color{blue}\epsilon^2 \mu(|f_n-f|>\epsilon)}\geq 0<br />

    Since f_n converges to f in Lē, then by definition of the convergence in Lē, the red part goes to 0 as n goes to infinity
    Thus by the sandwich theorem, the blue part goes to 0 as n goes to infinity.
    When it is said "for all epsilon", it's like you choose any epsilon and you fix it. So it's like a constant !

    Thus \lim_{n\to\infty} {\color{blue} \mu(|f_n-f|>\epsilon)}=0, which means, by definition, that f_n converges to f in measure.


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  5. #5
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    Quote Originally Posted by Moo View Post
    I'm rephrasing it

    \forall \epsilon>0,{\color{red}\int (f_n-f)^2 ~d\mu} \geq \int_{\{|f_n-f|>\epsilon\}} (f_n-f)^2 ~d\mu>{\color{blue}\epsilon^2 \mu(|f_n-f|>\epsilon)}\geq 0<br />

    Since f_n converges to f in Lē, then by definition of the convergence in Lē, the red part goes to 0 as n goes to infinity
    Thus by the sandwich theorem, the blue part goes to 0 as n goes to infinity.
    When it is said "for all epsilon", it's like you choose any epsilon and you fix it. So it's like a constant !

    Thus \lim_{n\to\infty} {\color{blue} \mu(|f_n-f|>\epsilon)}=0, which means, by definition, that f_n converges to f in measure.


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