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Lebesgue integration over sets of measure zero

  1. Sep 18, 2011 #1
    Is it true in general that if [itex]f[/itex] is Lebesgue integrable in a measure space [itex](X,\mathcal M,\mu)[/itex] with [itex]\mu[/itex] a positive measure, [itex]\mu(X) = 1[/itex], and [itex]E \in \mathcal M[/itex] satisfies [itex]\mu(E) = 0[/itex], then

    [tex]
    \int_E f d\mu = 0
    [/tex]

    This is one of those things I "knew" to be true yesterday, and the day before, and the day before...but now I can't show it! I need to be able to bound that integral, somehow, by [itex]\mu(E)[/itex], but how? Using Holder's inequality? But don't I need to know that [itex]f\in L^2[/itex] or [itex]f\in L^\infty[/itex] to do that? Do I know either of those? I don't think so...
     
  2. jcsd
  3. Sep 18, 2011 #2

    micromass

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    It's fairly easy, but it's tricky. Basically, we put

    [tex]\int_E fd\mu\leq \int \|f\|_\infty d\mu=\|f\|_\infty \int d\mu=\|f\|_\infty \mu(E)[/tex]

    Note that [itex]\|f\|_\infty=+\infty[/itex] can happen here, that's the tricky part. The thing isd that [itex]0.(+\infty)[/itex] is defined as 0 in measure theory.
     
  4. Sep 18, 2011 #3
    Ah yes! [itex]0 \cdot \infty = 0[/itex]! I had forgotten about that. Thanks micromass.
     
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