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Proving that a measurable function is integrable

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data[/b]

    Let [itex]f:ℝ\rightarrowℝ[/itex] be measureable and [itex]A_{k}=\left\{x\inℝ:2^{k-1}<\left|f(x)\right|≤2^{k}\right\}[/itex], [itex]k\in[/itex] [itex]\mathbb{Z}[/itex].

    Show that [itex]f[/itex] is integrable only if [itex]\sum^{∞}_{k=-∞}2^{k}m(A_{k}) < ∞ [/itex].

    2. Relevant equations
    By the definition [itex]f[/itex] is integrable in ℝ if and only if [itex]f[/itex] is measurable and [itex]∫_{ℝ}\left|f\right|<∞[/itex].

    Now we know that [itex]f[/itex] is measureable, thus we should show that [itex]∫_{E}\left|f\right|≤ \sum^{∞}_{k=-∞}2^{k}m(A_{k}) < ∞ [/itex].


    3. The attempt at a solution

    Let [itex]f = \sum^{∞}_{k=-∞}f_{k}[/itex], we get [itex] \left|f \right| = \left| \sum^{∞}_{k=-∞} f_{k} \right| \Rightarrow \int_{ℝ}\left|f\right| = \int_{ℝ}\left|\sum^{∞}_{k=-∞}f_{k}\right| [/itex]

    (?)We also notice that [itex]\int_{ℝ}\left|\sum^{∞}_{k=-∞}f_{k}\right| = \sum^{∞}_{k=-∞}\int_{ℝ}\left|f_{k}\right| = 2 \sum^{∞}_{k=0}\int_{ℝ}\left|f_{k}\right| [/itex] (Beppo Levi's lemma?) (?)

    Let [itex]I_{k} = [2^{k-1}, 2^{k}][/itex]. The length of this interval is [itex]\ell(I_{k}) = 2^{k}-2^{k-1}=2^{k}(1-2^{-1}) = (1/2)2^{k}[/itex]

    We get [itex]m(A_{k})≤\ell(I_{k}) \Rightarrow \sum^{∞}_{k=-∞}m(A_{k})≤\sum^{∞}_{k=-∞}\ell(I_{k}) = \frac{1}{2}\sum^{∞}_{k=-∞}2^{k}[/itex].

    And thereby [itex]2\sum^{∞}_{k=-∞}m(A_{k}) = \sum^{∞}_{k=-∞}2m(A_{k}) ≤ \sum^{∞}_{k=-∞}2^{k}[/itex].

    Does my inference make any sense?
     
  2. jcsd
  3. Jul 9, 2012 #2

    jbunniii

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    What is [itex]f_k[/itex]? You didn't define it.

    Also, your assertion that

    [tex]m(A_{k})≤\ell(I_{k})[/tex]

    is nonsense. [itex]A_k[/itex] is a subset of the domain of the function, whereas [itex]I_k[/itex] is a subset of the range. You certainly do not have [itex]A_k \subset I_k[/itex], which I suspect is the unstated assertion leading you to claim that [itex]m(A_{k})≤\ell(I_{k})[/itex]. E.g. if I put [itex]f(x) = 2^{k}[/itex] for all x, then [itex]A_k = \mathbb{R}[/itex] and thus [itex]m(A_k) = \infty[/itex].
     
    Last edited: Jul 9, 2012
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