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Thread: Borel measure

  1. #1
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    Borel measure

    Hi,

    I need to prove that their exists a Borel measure v on $\displaystyle \mathbb{R}^+ $ which is absolutely continuous with respect to Lebesgue measure and which satisfies $\displaystyle v([a,b])=b^2-a^2 $ with $\displaystyle a< b$

    I know that the usual Lebesgue measure would satisfy $\displaystyle v([a,b])=b-a $, but I'm terrible at coming up with examples. If I could find one that satisfies the requirement, I'm sure I could prove it is absolutely continuous. Any hints?

    Everything I was trying put me in $\displaystyle \mathbb{R}^2 $, and I need to be in $\displaystyle \mathbb{R}$. Thanks.


    *Note I had a typo and edited that the Lebesque measure satisfies $\displaystyle v([a,b])=b-a $
    Last edited by Benmath; Aug 10th 2010 at 10:43 AM.
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  2. #2
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    Quote Originally Posted by Benmath View Post
    Hi,

    I need to prove that their exists a Borel measure v on $\displaystyle \mathbb{R}^+ $ which is absolutely continuous with respect to Lebesgue measure and which satisfies $\displaystyle v([a,b])=b^2-a^2 $ with $\displaystyle a< b$

    I know that the usual Lebesgue measure would satisfy $\displaystyle v([a,b])=b-a $, but I'm terrible at coming up with examples. If I could find one that satisfies the requirement, I'm sure I could prove it is absolutely continuous. Any hints? [/tex]
    It looks as though the measure v should be defined by $\displaystyle v(E) = \int_E2x\,dx$ for each Borel set E, where dx means integral with respect to Lebesgue measure. Is that enough of a hint?
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  3. #3
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    Thank you Opalg! That is incredibly obvious! I am usually terrible with the very obvious things. Again, thanks!
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