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Pointwise convergence of Riemann integrable functions

  1. Sep 9, 2014 #1

    Normally in order to change the order of limit and integration in rimann integration, you need uniform convergence.

    But lets say that you are not able to prove uniform convergence, but only pointwise convergence. And lets say you are able to prove that the functions are also bounded. Could you then use something like the dominated convergence theorem for lebesgue measurable functions? The problem is offcourse that it holds for integrals defined another way.

    But lets say you have a sequence of riemann integrable functions, that converge pointwise to a function, and all the functions are bounded, can we then change the order of limit and integration?
  2. jcsd
  3. Sep 9, 2014 #2


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    Most of the time. To use bounded convergece as dominant convergence the domain of integration has to be finite.

    Simple counterexample for infinite domain.

    f_n(x) = 1 for n<x<n+1, =0 otherwise.
    functions converge to 0, but integral remains 1.
  4. Sep 9, 2014 #3
    Yeah, but what if we assume that it is finite domain? I mean technically the riemann integral is only defined on a bounded and closed interval?, but you can take indefinite integrals, but let's assume you don't do that in this case.
    Last edited: Sep 9, 2014
  5. Sep 10, 2014 #4


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    Not true. Riemann sums can be defined over unbounded closed intervals of ##\overline{\mathbb{R}}##.
  6. Sep 10, 2014 #5
    Yes there is an analog of dominated convergence for Riemann integrable functions but it is decidedly less useful. Suppose ## f_n: [a,b] \to \mathbb{R} ## and for some M we have that ##|f_n(x)| < M ## for each n. If ##f_n \to f ## pointwise and ##f## is riemann integrable, then you may interchange limits and integrals.
  7. Sep 10, 2014 #6


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    This is a simple consequence of the corresponding theorem for Lebesgue integrals, since a Riemann integrable function is also Lebesgue integrable, and its Riemann integral is the same as its Lebesgue integral.

    It may happen, however, that the limit function of a sequence of Riemann integrable functions on a closed interval exists and is bounded, but not Riemann integrable, so that the Riemann integral of the limit function does not exist.

    As an example, let q1, q2, q3,... be an enumeration of the rational numbers in [0,1].
    For each n, let fn:[0,1]→[0,1] be defined fn(x)=1 if x=qi for some i≤n, and fn(x)=0 otherwise. Each fn is Riemann integrable on [0,1] with Riemann integral 0.
    But fn → f, pointwise, where f:[0,1]→[0,1] is defined as f(x)=1 if x is rational and f(x)=0 otherwise. f is bounded but not Riemann integrable on [0,1].
    Last edited: Sep 10, 2014
  8. Sep 10, 2014 #7

  9. Sep 10, 2014 #8
    You are certainly free to define Riemann sums/integrals in whatever way pleases you, but the standard definitions of those terms do not apply to unbounded intervals in any meaningful way.

    You can definitely talk about improper Riemann integrals, which are typically defined in terms of limits of Riemann integrals and not in terms of Riemann sums, or Lebesgue integrals, which look kinda like Riemann integrals (but not really) when applied to compact intervals, both of which are perfectly well-defined on all intervals. And there are other way to define integrals that are "Riemann-like" and apply to functions defined on unbounded intervals. But the standard definitions define Riemann integrals as limits of Riemann sums, which are defined using finite partitions of compact intervals.
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