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Thread: Measuring the reals

  1. #1
    MHF Contributor Swlabr's Avatar
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    Measuring the reals

    Let $\displaystyle (\mathbb{R}, \Sigma, \mu)$ be a measure space, $\displaystyle \mu$ some measure. There exists at least one Lebesgue integrable function. Then does it hold that,

    $\displaystyle \mu(\mathbb{R}) < \infty \Rightarrow \mu(\mathbb{R}) = 0$?
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  2. #2
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    What exactly are you asking? As for the last part you can just take $\displaystyle \mu$ to be the zero measure (not interesting but a measure none the less).
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by putnam120 View Post
    What exactly are you asking? As for the last part you can just take $\displaystyle \mu$ to be the zero measure (not interesting but a measure none the less).
    It's okay - I found an example (although I can't remember what it is...).

    Basically, my question was: does there exist a (positive) measure such that $\displaystyle \mu(\mathbb{R}) < \infty$ but which is not the zero measure.
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  4. #4
    Moo
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    Hello,

    Well just make $\displaystyle \mu$ a Dirac measure
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  5. #5
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    Let $\displaystyle \nu$ be any measure defined on $\displaystyle \mathbb{R}$ such that there exists $\displaystyle A\in \Sigma$ with $\displaystyle 0<\nu(A)<\infty$. Then the measure $\displaystyle \mu(E)=\nu(E\cap A)$ will be finite.
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