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Thread: 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 $\displaystyle f_n \rightarrow f $ almost uniformly, then $\displaystyle f_n \rightarrow f $ a.e. and $\displaystyle f_n \rightarrow f $ in measure.

    Proof so far.

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

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

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

    Claim: $\displaystyle \mu (A)=0 $

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

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

    Why can I choose a B in A in that fashion
    I don't think you can, since by definition if you want $\displaystyle f_n$ to converge uniformly in $\displaystyle B^c$ you need $\displaystyle A \subseteq B$
    Last edited by Jose27; Nov 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 $\displaystyle B$ such that $\displaystyle \mu(B)<\mu(A)$ and $\displaystyle (f_n)_n$ converges uniformly to $\displaystyle f$ on $\displaystyle B^c$.

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