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Riemann Integrable

  1. Feb 6, 2006 #1
    Is it possible to show by induction that f:[a,b]->R, a bounded function, is Riemann integrable if f has a countable number of discontinuities? I'm told this is usually done with Lebesgue integrals, but I don't see why an inductive proof of this using Riemann/Darboux integrals can't work.
     
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  3. Feb 6, 2006 #2

    matt grime

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    What are you inducting on? It can't be the unmber of discontinuities. showing something for a finite number of things in no way says anything about the infinite case.
     
  4. Feb 6, 2006 #3

    NateTG

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    No.

    Consider, for example, the characteristic function of the irrationals.
    f(x) is 1 if x is irrational, and 0 if x is rational.

    This has a countable number of discontinuities -- one at every rational number -- but is most certainly not Riemann integrable as the lower sum will always be 0, and the upper sum will always be 1.
     
    Last edited: Feb 6, 2006
  5. Feb 6, 2006 #4

    matt grime

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    A 'rational' number of discontinuities? No, it has an uncountable number of discontinuities. It is discontinuous at every point at of the interval [0,1]
     
  6. Feb 6, 2006 #5
    Well, f is integrable if it has 1 discontinuity, and whenever it has n discontinuities and is integrable, it implies that having n+1 discontinuities is integrable.
     
  7. Feb 6, 2006 #6

    matt grime

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    and that tells you nothing (as given) about countably (but not finitely) many discontinuities. (Proof... well, according to this logic, the set of natural numbers is finite by 'induction')
     
  8. Feb 6, 2006 #7
    What's wrong with it?
     
  9. Feb 7, 2006 #8

    matt grime

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    Because countable includes infinite, and induction on finite things tells you nothing about infinite things. as i said according to your logic all countable sets (ie the set of natural numbers) are finite. proof: a set with n elements is finite, adding one more it remains finite, hence all countable sets are finite, but the natural numbers are a countable set that are not finite. (induction does not apply.)
     
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