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Thread: Lebesgue Differentiation Theorem

  1. #1
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    Post Lebesgue Differentiation Theorem

    The Lebesgue Differentiation Theorem states:

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

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

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

    How does it follow that this Theorem is satisfied if $\displaystyle f$ 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 $\displaystyle f$ 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 $\displaystyle f: \mathbf{R}^{n} \rightarrow C$ is a summable function then for almost every $\displaystyle x \in \mathbf{R}^{n}$ the following limit converges to $\displaystyle 0$:

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

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

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


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

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

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