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Integrals and continuum vs discrete

  1. Jun 4, 2010 #1
    One the things which always bugged me about the definition of the integral is that limit of a Reimann sums consists of a countably infinite number of terms, yet it is supposed to be giving the under area under a curve which varies in a continuous manner. Has anyone else thought seriously about this?

    I suspect it is closely related to the fact some functions on a finite interval can be represented by a fourier series.

    [tex]f(x) =\frac{a_0}{2}+\sum_{n=1}^\infty\left[a_n cos(nx)+ b_n sin(nx)\right][/tex]

    Here we see that a function f consisting of a continuum of points [tex]f(x)[/tex] is represented entirely by a countably infinite set of numbers [tex]\{a_0,a_1,...,b_1,b_2,...\}[/tex].

    I'm not able to pose a coherent question about this. There is something interesting here but I can't get my head around putting it into words.

    I'm interested in thoughts anyone cares to share on continuum vs countable. Is there a relation between integrable and able to be represented by a Fourier series?
  2. jcsd
  3. Jun 4, 2010 #2
  4. Jun 4, 2010 #3
    Thanks, VeeEight.
  5. Jun 10, 2010 #4
    Here's my best guess about this. The rationals are a countable dense set in [tex] \mathbb{R} [/tex]. So give any function on [tex] \mathbb{R} [/tex] you can approximate it arbitrarily close working strictly from a countable subset of your domain. This along with the fact that we can use a net to take the limit as the partitions over rational intervals become arbitrarily fine means that we can approximate continuous functions at all of its domain (using only rational points) closely enough to get a well defined integral. Also, VeeEight, I feel like the Lebesgue integral might actually have the same issue that the Riemann integral has considering that it is defined similarly as the supremum of simple functions that are composed as countable infinite sums.
    Last edited: Jun 10, 2010
  6. Jun 10, 2010 #5


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    Does it help you to observe that such a continuous function is itself completely determined by its values on the rationals, which are countable?
  7. Jun 12, 2010 #6
    What sorts of functions are completely determined by their values on the rationals?
  8. Jun 12, 2010 #7
    See the last line under properties: http://en.wikipedia.org/wiki/Dense_set#Properties
    This is due to the fact that continuous functions behave nicely with regards to limits, so you can extend a continuous function on the rationals to the entire real line using limits of sequences of rationals that converge to an irrational.
  9. Jun 12, 2010 #8
    Thanks. Very interesting.
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