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Math Help - dominated convergence theorem for convergence in measure

  1. #1
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    dominated convergence theorem for convergence in measure

    hi,
    Ive just read this on wikipedia:
    "(X, M, μ) - measure space. If μ is σ-finite, Lebesgue's dominated convergence theorem also holds if almost everywhere convergence is replaced by convergence in measure."
    How could i go about proving this?

    I know that if a sequence f_n --> f in measure, then there is a subsequence which converges to f a.e. I can apply the DCT on this subsequence, but how would i show the it works for the whole sequence? Also, how would i use the fact that μ is σ-finite?
    Thank you
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  2. #2
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    Re: dominated convergence theorem for convergence in measure

    I think you have the right idea. The point is that every subsequence has in turn a subsequence that converges to f a.e.
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  3. #3
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    Re: dominated convergence theorem for convergence in measure

    Sorry, I didn't read your question carefully enough.

    The measure \mu is \sigma-finite iff every subsequence has in turn a subsequence that converges a.e.
    Last edited by Juju; October 1st 2014 at 12:21 PM.
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