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Math Help - Uniform convergence in measures

  1. #1
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    Uniform convergence in measures

    Just stuck on a Measure theory question right now:

    Suppose  \{f_n\}\subset {L^1} and {f_n}\longrightarrow f uniformly. Show that if  \mu(X)<\infty , then f\in {L^1} and  \int f_n\longrightarrow \int f. Find an example that shows that the result is not necessarily true when  \mu(X)= \infty

    Little bit stumped on how to start. Initially I thought I could use the dominated convergence theorem, but not fully sure. We do have uniform convergence and the dominated convergence theorem claims if we can bound a function by g then the  \int f =lim_{n\rightarrow\infty} \int {f_n}.

    If not I'm thinking of using chebychev's inequality.

    Any help would be great.

    I also apologise for any problems with the LaTex, it's my first time using it.

    Thanks
    Last edited by ster; December 9th 2009 at 07:30 AM.
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  2. #2
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    Since f_n \rightarrow f uniformly we have that for all x\in X there exists an N such that m,n\geq N implies \vert f_n(x)\vert - \vert f_m(x) \vert \leq \vert f_m(x)-f_n(x) \vert \leq 1 now taking m=N we get \vert f_n(x) \vert \leq 1+ \vert f_N(x)\vert for all n\geq N and this last one is integrable since \int_{X} d\mu =\mu (X)<\infty so picking g(x)=\max \{ \vert f_1(x) \vert ,...,\vert f_{N-1}(x) \vert , 1+ \vert f_N(x) \vert \} we get our dominating function.

    When \mu (X)= \infty see here
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  3. #3
    GTO
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    uniform convergence

    looks like what u used cauchy sequence rather than uniformly convergence. can you still do what you did with uniform convergent sequence?
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  4. #4
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    |\int_Xf_n-f|\le\int_X|f_n-f|\le\mu(X)\cdot\sup|f_n-f|<\epsilon\mu(X).

    For a counter example consider f_n=\frac{1}{n}\chi_{[0,n]}. Then f_n\to 0 but \int_\mathbb{R}f_n=1 for all n.
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  5. #5
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    L1 space

    can you show me how to show that f is in L1, please?
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  6. #6
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    I think that the following version of the triangle inequality should help: |a|-|b|\le |a-b|
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