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Math Help - Lebesgue Differentiation Theorem

  1. #1
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    Post Lebesgue Differentiation Theorem

    The Lebesgue Differentiation Theorem states:

    If f: \mathbf{R}^{n} \rightarrow C is a summable function then for almost every x \in \mathbf{R}^{n} the following limit converges to 0:

    lim_{r \rightarrow 0}\text{ } (\frac{1}{m(B(x,r)})\int_{B(x,r)}|f(y) - f(x)|dy = 0

    Where m is the lebesgue measure and B(x,r) is an open ball of radius r.

    How does it follow that this Theorem is satisfied if f is only assumed to be continuous?

    Let me know if anything is unclear. Thanks.
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  2. #2
    Newbie Phantasma's Avatar
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    Re: Lebesgue Differentiation Theorem

    I believe it is required that f be Lebesgue integrable.

    See the proof here.
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  3. #3
    Super Member Rebesques's Avatar
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    Re: Lebesgue Differentiation Theorem

    Quote Originally Posted by Kramer View Post
    The Lebesgue Differentiation Theorem states:

    If f: \mathbf{R}^{n} \rightarrow C is a summable function then for almost every x \in \mathbf{R}^{n} the following limit converges to 0:

    lim_{r \rightarrow 0}\text{ } (\frac{1}{m(B(x,r)})\int_{B(x,r)}|f(y) - f(x)|dy = 0

    Where m is the lebesgue measure and B(x,r) is an open ball of radius r.

    How does it follow that this Theorem is satisfied if f is only assumed to be continuous?


    When f is (additionaly) continuous, then it is uniformly continuous over the closed balls \overline{B(x,r)}, r>0. Choose such a ball of radius r_0>0. Then, due to uniform continuity, for every \epsilon>0 there is  0<r_1<r_0 such that |f(x)-f(y)|<\epsilon , \forall y\in B(x,r_1).
    This implies

    \frac{1}{m(B(x,r_1))}\int_{B(x,r_1)}|f(y) - f(x)|dy \leq \epsilon

    from where the result follows.
    Last edited by Rebesques; August 8th 2013 at 12:14 PM.
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