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Math Help - convergence in measure and convergence a.e

  1. #1
    GTO
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    convergence in measure and convergence a.e

    suppose we have a space ([0,1],B,\lambda ), where \lambda is Lebesgue measure. and let f_n(x)=e^{-nx} be a sequence of functions: [0,1]\rightarrow [0,1]

    Show f_n \rightarrow 0 in measure and almost eveywhere.

    To show the convergence in measure, for given \epsilon >0, let E=\{x:f_n>\epsilon\} . Then e^{-x})^n > \epsilon \}=" alt="E=\{x:f_n>\epsilon \}=\{xe^{-x})^n > \epsilon \}=" />  \{x:e^{-x}>\epsilon ^{1/n}\}=\{x<-Ln(\epsilon ^{1/n})\}=[0,-Ln(\epsilon)/n) . So \lambda ( [0,-Ln(\epsilon)/n)) \rightarrow 0 as n \rightarrow 0. So f_n \rightarrow 0 in measure.

    here is my question. In this question, i dont really see the difference between the convergence in measure and convergence almost everywhere. Can i prove the convergence almost everywhere in the same way i did above? or i have to show the different proof?
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  2. #2
    Moo
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    Hello,

    No you just have to say that \lim_{n\to\infty} f_n(x)=0 ~,~ \forall x\in (0,1], the limit is 1 if x=0. So that's almost everywhere in [0,1] (you just have to exclude 0, and since we're dealing with the Lebesgue measure, {0} is of measure 0)
    That's a tad simpler
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