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Lebesgue Integration

  1. Aug 29, 2010 #1
    Hello all,

    I am wondering the implication between almost everywhere bounded function and Lebesgue integrable.

    In the theory of Lebesgue integration, if a non-negative function [tex]f[/tex] is bounded a.e., then it should be Lebesgue integrable, i.e. [tex]\int f d\mu < \infty[/tex] because we do not take into account the unboundedness of [tex]f[/tex] in a null set when approximate by sequence of simple function, am I correct? So this means a.e. boundedness implies Lebesgue integrable?

    And seems there is a counterexample on the reverse implication, http://planetmath.org/encyclopedia/AnIntegrableFunctionWhichDoesNotTendToZero.html [Broken], so that means Lebesgue integrable does not imply bounded a.e.

    So is this because in finding the Lebesgue integral, it is indeed an infinite series of products, which is [tex] \sum s_{n} \cdot \mu(A_{n})[/tex], so as long as the increase in [tex] s_{n} [/tex] is not faster than the decrease in [tex] \mu(A_{n}) [/tex], it is possible to have a finite value of this infinite sum? So in this way we may end up with a non-bounded a.e. function but Lebesgue integrable?

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 29, 2010 #2


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    One standard example of a non-Lebesgue integrable function is the characteristic function of a non-measurable set.

    While the Lebesgue integral (and the Riemann integral!) are limits of finite sums, neither is an infinite sum. (In the usual formulations, anyways)
  4. Aug 29, 2010 #3


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    This assumes that the total measure of the space is finite.
  5. Aug 29, 2010 #4
    Thanks so much.

    So, in conclusion, relationship between a.e. boundedness and Lebesgue integrable is not definitive in general, right?

    Regarding the characteristic function of a non-measurable set, it is then a non-measurable function, hence, its Lebesgue integral is not well-defined? Or is there any reference about this?

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