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Cardinalic flaw of Riemann integral

  1. Aug 22, 2012 #1
    I have learnt that integral is the Riemann sum of infinite rectangle, that:
    Ʃ[itex]^{n=1}_{∞}[/itex]f(xi)Δxi = ∫[itex]^{b}_{a}[/itex]f(x)dx
    However, I think that (a,b) is the continuous interval, so the number of rectangle should be c instead of [itex]\aleph[/itex]0 (cardinality of natural number N).
    So I wonder whether there are some problem that this definition is not valid anymore.
  2. jcsd
  3. Aug 22, 2012 #2


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    How so? The oo you're using is the countable infinity. An uncountable sum will

    necessarily diverge , unless only countably-many are non-zero. Still, good


    Edit: after reading SteveL's comment, I guess I should be more precise:

    The limit in the sum you describe is a limit as you approach countable infinity;

    so you are selecting one point x_i* in each subinterval , and , as N-->oo (countable

    infinity) there is a bijection between the number of rectangles and the x_i* you choose.

    Since the x_i* are indexed by countable infinity, so are the rectangles.
    Last edited: Aug 23, 2012
  4. Aug 22, 2012 #3


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    Each Δxi is a continuum - there is no contradiction.
  5. Aug 22, 2012 #4
    I'm a little confused about this definition. Typically the Riemann integral is the limit of Riemann sums, each one of which is a finite sum over a partition of the interval. Each partition is a finite set of subintervals.

    There is no infinite sum such as you've notated. Is this a definition you saw in class or in a book?
  6. Aug 23, 2012 #5
    Thanks for explanation, I have understood.
    And I mean it's the limit of finite sum, but I am a bit lazy so I remove the limit part for convenience.
  7. Aug 24, 2012 #6


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    No, it isn't. It is a limit of Riemann sums, each of which involves a finite sum. That is not "the Riemann sum of infinite rectangles" which is not defined.
    It should be no suprise that your mistaken definition is not valid.
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