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Math Help - Absolute continuity (measures): finiteness

  1. #1
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    Absolute continuity (measures): finiteness

    Hi there

    I don't know how to solve this exercise:
    """"
    Let \mu, \lambda be two measures on the borel set with \lambda << \mu (this means that \lambda is absolute continuous related to \mu). Prove the following statement or the contrary:

    If \mu is finite then \lambda is finite.
    """"

    I'd guess that the statement is true. What's clear is that according to Radon-Nikodym there exists a measurable f such that:

     \lambda(A)=\int_A f d\mu and the assumption is  \mu(\mathbb{R})<\infty

    So now: \lambda(\mathbb{R})=\int_{\mathbb{R}} f d\mu=...

    How can I complete this? Is there an upperbound for f I can take (a real number c)? If yes what theorem says that there is one? Then this proof was easy.

    \int_{\mathbb{R}} f d\mu \le \int_{\mathbb{R}} c d\mu = c*\mu(\mathbb{R})

    But I can't find a theorem or a corollary that says that. How can I show this?

    Regards
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  2. #2
    Super Member girdav's Avatar
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    Re: Absolute continuity (measures): finiteness

    Are you working with non-negative measures?
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  3. #3
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    Re: Absolute continuity (measures): finiteness

    yes regular measures non-negative ones.

    what do you think? is the statement true?
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  4. #4
    Super Member girdav's Avatar
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    Re: Absolute continuity (measures): finiteness

    Let \nu(A):=\int_Ae^{-x^2}d\lambda(x), where \lambda is the Lebesgue measure. \nu is a finite measure, and if \nu(A)=0, then \lambda(A\cap [n,n+1])=0 for all integer n.
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    Re: Absolute continuity (measures): finiteness

    Thank you. But if I define \lambda to be the Lebesgue measure restricted to \mathbb{R}_{\ge 0}.

    \nu(A)=\int_A e^{-x^2} d \lambda and \nu(\mathbb{R})=\frac{\sqrt{\pi}}{2} so this is finite.


    But how can I prove that if \nu(A)=0 then  \lambda(A)=0 ??

    Since \nu is zero for singletons and the emptyset only ( e^{-x^2} is a non-negative even function on \mathbb{R}_{\ge 0}) it follows because of the properties of the Lebesgue measure?

    Is this enough? What do you think?

    Regards.
    Last edited by huberscher; October 31st 2012 at 08:06 AM.
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    Super Member girdav's Avatar
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    Re: Absolute continuity (measures): finiteness

    What are the properties of Lebesgue measure you use?

    If \nu(A)=0, then 0=\nu_{A\cap [n,n+1]}=\int_{A\cap (n,n+1)}e^{-x^2}dx\geq \exp(-(n+1)^2)\lambda(A\cap [n,n+1]).
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  7. #7
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    Re: Absolute continuity (measures): finiteness

    In your example I don't understand the whole relation now because \lambda(A) \neq \lambda(A\cap[n,n+1]) ??
    Hasn't the \nu to be made this way:
    \nu(A)=\int_A e^{-x^2}*d\lambda(A \cap [n,n+1])
    So this n can be fixed?


    Concerning my example:

    \nu(A)=0 \Leftrightarrow A=\emptyset \lor A=\{x\}  with x \in \mathbb{R} (this is because f is an even non-negative function on R)

    For the Lebesgue measure \lambda restricted to \mathbb{R}_{\ge 0} it follows that
    \lambda(\emptyset)=0 \land \lambda(\{x\})=0 \forall x \in \mathbb{R}

    So my example works as well, does it?
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  8. #8
    Super Member girdav's Avatar
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    Re: Absolute continuity (measures): finiteness

    Not only countable of finite subsets of the real line have 0 measure. For example, Cantor set is uncountable, but has zero measure.

    In my argument, for a fixed n I show that \lambda(A\cap [n,n+1]) is 0. This proves, by \sigma-additivity, that \lambda(A)=0.
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    Re: Absolute continuity (measures): finiteness

    What is the difference between absolute continuity and differential continuity?






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