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Math Help - lebesgue measure and differentiability

  1. #1
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    lebesgue measure and differentiability

    Hi there

    I've got the following two exercises to solve:

    -
    Consider the measure space \([0,\infty\),B,\lambda\) where B denotes the Borel-sigma-algebra and \lambda the Lebesgue-measure.
    (1)
    Let \mu << \lambda and  \frac{d\mu}{d\lambda} continuous. Show that
    f(x)=\mu([0,x]) is differentiable and that
    f'(x)=\frac{d\mu}{d\lambda}(x)

    (2)
    If  \frac{d\mu}{d\lambda} is \lambda- almost everywhere continuous only is then
    f'(x)=\frac{d\mu}{d\lambda}(x) \lambda-almost everywhere as well?
    Why? Prove your answer!
    -

    (1)
    According to the theorem of Radon-Nikodym for g(x)=\frac{d\mu}{d\lambda}(x) then

    f(x)= \int_{[0,x]} g d\lambda(x)

    Obviously the derivative of this is g but how can one show this in a proper way?

    f'(x)= \int_{[0,x]} g \frac{d\lambda(x)}{d\lambda(x)} =\int_{[0,x]} g 1 ????

    I also tried it this way:

    f'(x)=\frac{d\mu}{d\lambda}(x)-\frac{d\mu}{d\lambda}(0) but what then is x and 0? Definitely not a one-point set... cause then it was all zero.



    (2)
    I'd say that this is right because where

    f'(x)  \neq \frac{d\mu}{d\lambda} is a Lebesgue-zero set but how to show?



    Could please someone help me to solve this exercise? Any hint would be appreciated.

    Regards
    Huberscher
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  2. #2
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    Re: lebesgue measure and differentiability

    I think you're trying to make an easy problem hard. Here's a solution:
    lebesgue measure and differentiability-radonnikodym.png
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  3. #3
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    Re: lebesgue measure and differentiability

    Thank you very much.

    Concerning 2)
    A kind of "prove" or argument hmmm

    assuming g is continuous at x_0 and x_0 is contained in a non-zero-set of \lambda then this proof works also, ok.

    But when x satisfies |x-x_0|<\delta and x is on the left of x_0...doesn't that mean that at x the function g is continuous from the right since the \delta is arbitraty ? That's right yeah?
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  4. #4
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    Re: lebesgue measure and differentiability

    Hi there

    Well I understand the proof but I still don't understand why this is also valid for a.e. continuity. Can someone please explain me that?

    Cause then I have

    \int_{[x_0,x_0+h]}|g(x)-g(x_0) | dx is 0 because the set [x_0,x_0+h] is a zero set but then I still have this \frac{1}{h} which goes to infinity when h goes to 0.

    Is there anyone who can answer me that please?

    Regards
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