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Lebesgue integration

  1. Sep 23, 2010 #1
    Hello all,

    Here is my question:

    Suppose a measureable space [tex] (S,\mathcal{S},\mu) [/tex] with [tex] \mu(S) < \infty [/tex] and [tex] f : S \mapsto [0,\infty) [/tex], this is not yet sufficient to ensure [tex] \int_{S} f d \mu < \infty [/tex].

    Am I correct?
     
  2. jcsd
  3. Sep 23, 2010 #2

    Office_Shredder

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    S=(0,1) and f=1/x
     
  4. Sep 23, 2010 #3

    LCKurtz

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    Yes. Consider the measure defined on R by

    [tex]\mu(E) =\int_E \frac 1 {1+x^2}\ dx[/tex]

    for Lebesgue measurable E. Let f(x) = 1/x2. Then

    [tex]\int_R \frac 1 {x^2}\cdot \frac 1 {1+x^2}\ dx \ge \int_{-1}^1 \frac 1 {x^2}
    \cdot \frac 1 2\ dx =\infty[/tex]
     
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