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Math Help - Deceptive Lebesgue measure theory problem

  1. #1
    Jeh
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    Deceptive Lebesgue measure theory problem

    Let f:[0,1]->R be a measurable function and E a subset of {x : f'(x) exists}. If m(E)=0, show that m(f(E))=0.
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    Re: Deceptive Lebesgue measure theory problem

    m(f(E)) = \int_{E} f'(x) dm \leq m(E) sup_{x\in E} f'(x)=0 for the lower bound take infimum of f'(x).
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    Jeh
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    Re: Deceptive Lebesgue measure theory problem

    Quote Originally Posted by InvisibleMan View Post
    m(f(E)) = \int_{E} f'(x) dm \leq m(E) sup_{x\in E} f'(x)=0 for the lower bound take infimum of f'(x).
    Where is the first equality coming from? For example, m(\sin([0,\pi]))=1, but \int_{[0,\pi]}\sin'\, dm=0.
    Last edited by Jeh; July 21st 2012 at 05:58 AM. Reason: bad code
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    Re: Deceptive Lebesgue measure theory problem

    Well if I am not mistaken:

     m(\sin ([0,\pi])) = m(\sin ([0,\frac{\pi}{2}])+m(\sin ([\frac{\pi}{2},\pi]))= m([0,1])+m([1,0])=m([0,1])-m([0,1])=0
    Last edited by InvisibleMan; July 21st 2012 at 07:44 PM.
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    Jeh
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    Re: Deceptive Lebesgue measure theory problem

    You're a bit confused. The image of \sin([0,\pi]) is [0,1], which has Lebesgue measure 1. Your reversal of the endpoints of the intervals makes no sense. We're just looking at the image of the function.
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    Re: Deceptive Lebesgue measure theory problem

    I guess by m you mean the Lebesgue measure function?

    Well, it is intuitively clear that a function cannot turn a set of Lebesgure-measure zero to a set which has Lebesgue-measure non-zero.

    Your example doesn't work because the sine function is differentiable on whole of [0,1].
    Think about it: Is the set that you chose where sin is differentiable really a set of measure zero? And think about what you integrate over.
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  7. #7
    Jeh
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    Re: Deceptive Lebesgue measure theory problem

    Quote Originally Posted by mastermind2007 View Post
    I guess by m you mean the Lebesgue measure function?

    Well, it is intuitively clear that a function cannot turn a set of Lebesgure-measure zero to a set which has Lebesgue-measure non-zero.

    Your example doesn't work because the sine function is differentiable on whole of [0,1].
    Think about it: Is the set that you chose where sin is differentiable really a set of measure zero? And think about what you integrate over.
    That's not true. The Cantor function on [0,1] takes a set of Lebesgue measure zero to a set of Lebesgue measure 1. It doesn't matter that \sin(x) is differentiable everywhere. E merely needs to be a subset of of places where it is differentiable. That's besides the point, anyway. I merely was questioning the validity of the first inequality posted by InvisibleMan.
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    Re: Deceptive Lebesgue measure theory problem

    Okay, you are right that the Cantor function does that.

    I mentioned the fact that sin is differentiable everywhere because the set you chose for the sin-function isn't a set of Lebesgue-measure zero (interval [0,pi]).

    As for the inequality that InvisibleMan posted:
    I think there might be some absolute value signs missing.
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