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Bounded Lebesgue integrals

  1. Nov 1, 2011 #1
    Say f is a non-negative, integrable function over a measurable set E. Suppose
    [tex]
    \int_{E_k} f\; dm \leq \epsilon
    [/tex]
    for each positive integer [itex]k[/itex], where
    [tex]
    E_k = E \cap [-k,k]
    [/tex]
    Then, why is it true that
    [tex]
    \int_E f\; dm \leq \epsilon \quad ?
    [/tex]

    I know that
    [tex]
    \bigcup_k E_k = E
    [/tex]
    and intuitively it seems reasonable, but I don't know how to prove it.
     
    Last edited: Nov 1, 2011
  2. jcsd
  3. Nov 1, 2011 #2
    Proceed by indirect method : if the integral over E is strictly > e ,then the integral over E_k should also exceed e for sufficiently large k.
     
  4. Nov 1, 2011 #3
    Why is that so? Are you using some property of integrable functions I'm not seeing?
     
  5. Nov 1, 2011 #4
    I think you could also prove it this way. Given [itex] \epsilon > 0 [/itex], for each [itex]k \in \mathbb{Z}^+[/itex] [itex] \int_{E_k} f \; dm \leq \epsilon/2^k[/itex]. Then
    [tex]
    \int_E f \; dm = \int_{\bigcup_k E_k} f \; dm \leq \sum_{k=1}^\infty \int_{E_k} f \; dm \leq \sum_{k=1}^\infty \frac{\epsilon}{2^k} = \epsilon \frac{1}{2} \frac{1}{1 - \frac{1}{2}} = \epsilon \; .
    [/tex]

    The only question I have is if it's legal to have an epsilon that depends on k, but I think it is since I think your given information held for all [itex] \epsilon > 0 [/itex] and all positive integers k.
     
  6. Nov 1, 2011 #5
    Actually, the epsilon is a fixed number. That is why it's stumbling me. Sorry I wasn't clear.
     
  7. Nov 1, 2011 #6
    define [itex]f_k(x)=\textbf{1}_{E_k}(x)f(x)[/itex] where [itex]\textbf{1}[/itex] is the characteristic function.
    From the monotone convergence theorem you have
    [tex]\epsilon \ge \lim_{k\to\infty} \int_{E_k} f \; dm = \lim_{k\to\infty} \int_E f_k \; dm = \int_E f \; dm[/tex]
     
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