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Prove that an Lebesgue integral exists iff the other exists

  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Let g: ℝ → ℝ be Lebesgue measurable, let μ(A) be a measure of Lebesgue measurable sets defined by
    μ(A) = ∫A f dx, where f: ℝ → ℝ is non-negative, with finite integral on compact intervals.

    Prove that
    g(x)dμ(x)
    exists if and only if
    g(x)f(x) dx
    exists, in which case the two are equal

    2. Relevant equations
    N/A


    3. The attempt at a solution
    I'm reviewing the definitions/construction of Lebesgue integrals but that is not leading to anything fruitful. I have that the first integral exists iff the integral of |g| exists, which exists iff the sup of all integrals of simple, bounded, measurable, finitely supported functions less than |g| is finite ... and then I am stuck.

    I suppose that seeing the equation
    ∫|g|dμ = sup Σai μ(Ai) = sup ΣaiAi f dμ
    (aiχAi being some simple function approximating |g|) causes me to "get" the reason for the equality -- if we imagine the Ai partitioning ℝ and getting smaller and smaller (small enough to encompass some single point x) we see that the a_i ≈ g(x) and that the integral ≈ f(x)dx. But actually showing the truth of this rigorously is an entirely different matter. Please advise me! Thanks.

     
  2. jcsd
  3. Aug 6, 2015 #2

    micromass

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    The standard method of proof for this is sometimes called "bootstrapping". The idea is that you show the following for measurable functions.

    1) The theorem holds for ##g = \chi_A## an indicator function.
    2) If the theorem holds for positive functions ##g_i##, then it holds for finite linear combinations ##\sum \alpha_i g_i## with ##\alpha_i\geq 0##.
    3) If the theorem holds for positive functions ##g_n## with ##n\in \mathbb{N}## such that they increase pointswise to a limit function ##g##, then the theorem holds for ##g##.
    4) If the theorem holds for all positive functions, then it holds for all functions.
     
  4. Aug 7, 2015 #3
    Okay, I think I am starting to make progress here. A few questions:

    (1) We can assume that the a_i are positive because, if not, we're assuming the simple function is positive and so could just replace the simple function with another one with positive coefficients, right?
    (2) Does "exist" in this case include the possibility that the integral is infinite? Or does "integral of f exist" = "f is Lebesgue integrable"? I ask because Wikipedia tells me I have to assume that μ(Ai) < +∞ if ai ≠ 0. But if we have all positive coefficients then we don't have an issue where we get a ∞ - ∞ term, and we're allowed to approach infinity with our sum. Do I need Wikipedia's requirement, or not?
     
  5. Aug 7, 2015 #4

    micromass

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    Indeed, the positive requirements are there to rule out ##\infty-\infty## situations. In those cases, the integral is allowed to be infinite. It is only in the last step of my outline that you assume the integral to be finite.
     
  6. Aug 7, 2015 #5
    I proved that the integrals exist and are the same for positive, measurable functions. So if g is a general measurable function, then we have
    g = g+ - g-.

    Take the integral over ℝ. Each of the two integrands (of ∫ g+dμ and ∫g-dμ ) represents a positive, measurable function, so we can replace their integrals w/r to dm(x) with their integrals w/r to f(x)dx, then combine. We have that the two integrals are equivalent, thus are defined when and only when the other is defined, namely, when one of the two sub-integrals is finite.

    Sound correct?
     
  7. Aug 7, 2015 #6

    micromass

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    Sounds good.
     
  8. Aug 7, 2015 #7
    Thank you!
     
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