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Math Help - Converges almost uniformly implies a.e. and in measure.

  1. #1
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    Converges almost uniformly implies a.e. and in measure.

    If f_n \rightarrow f almost uniformly, then f_n \rightarrow f a.e. and f_n \rightarrow f in measure.

    Proof so far.

    Now, since f_n \rightarrow f almost uniformly, I know that for any  \epsilon > 0 , there exists E \subset X such that  \mu (E)< \epsilon and f_n \rightarrow f uniformly on E^c , that is,  | f_n(x)-f(x)| \rightarrow 0 \ \ \ \ \forall x \in E^c .

    First, I want to show that f_n \rightarrow f a.e.

    Let  A = \{ x : f_n(x) \ not \ \rightarrow f \}

    Claim:  \mu (A)=0

    Suppose to the contrary that  \mu (A) > 0 , then there exists B \subset A such that  \mu (B) < \frac { \mu (A) }{2} and that  f_n \rightarrow f uniformly on A \backslash B , and  \mu (A \backslash B ) > \frac { \mu (A) }{2} > 0

    But since fn do not converge to f on A , that is a contradication.

    My question is: Why can I choose a B in A in that fashion and to get fn converges uniformly to A for all x in A \backslash B ?

    Thank you.
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  2. #2
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    As it is, take E_{\epsilon } the set that makes f_n converge uniformly on E_{\epsilon }^c then \mu (A) \leq \mu (E_{\epsilon}) < \epsilon thus proving \mu (A) =0.

    Why can I choose a B in A in that fashion
    I don't think you can, since by definition if you want f_n to converge uniformly in B^c you need A \subseteq B
    Last edited by Jose27; November 11th 2009 at 08:03 PM.
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  3. #3
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    Indeed you can't write it this way. You could say: There exists B such that \mu(B)<\mu(A) and (f_n)_n converges uniformly to f on B^c.

    Then you have A\subset B, because uniform convergence on B^c implies convergence at all points of B^c (this sentence tells in fact that B^c \subset A^c). And this can't be because of what you wrote: the measure of B is less than that of A.
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