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Robustness of the Riemann integral

  1. Jan 28, 2005 #1
    Could someone explain me what robustness is(in ur words), and how it works in proofs. All i kno is that basically u have two functions and u jiggle them alot until u make them integrable if its not or destroy their integrability if they are integrable. Geometric explanation would really help here but the main question is what is its purpose, it doesn't really make any difference in the end, right.
  2. jcsd
  3. Jan 29, 2005 #2


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    I have absolutely no idea at what you are trying to get...
  4. Jan 29, 2005 #3


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    When it is said that the Riemann integral is not "robust" enough for "theoretical" purposes, it roughly means there are lots of functions we would have liked to be "integrable" but which aren't in the Riemann sense of integrability.
    The above is perhaps more appriopriately concerned with "limitedness" rather than "robustness".

    The following example, however, is more to the issue of "robustness":
    Consider the unit interval, and make a bijection between the rationals in the unit interval and N, i.e, [tex]x_{n}[/tex]

    Now, consider the functions:
    [tex]f_{n}(x_{m})=1, m<=n[/tex]
    with [tex]f_{n}=0[/tex] otherwise.

    The limiting function f of this sequence, is clearly 1 on the rationals, and 0 at the irrationals.

    All [tex]f_{n}[/tex] are Riemann integrable, yet f is not!
    We would have liked that the limiting function were integrable as well, but the Riemann integral isn't robust enough to allow the integrability of f.
    Last edited: Jan 29, 2005
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