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Riemann-integrable functions

  1. Oct 8, 2008 #1
    Hello,

    can you provide me an example where the limits of a Riemann-integrable functiosn (or even continuous function may fail to be Riemann-integrable?

    Thanks
     
  2. jcsd
  3. Oct 8, 2008 #2

    CompuChip

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    You mean: is there an example of a converging series [itex]\{ f_n | n \in \mathbb Z \} [/itex] of Riemann integrable functions of which the limit
    [tex]f \binop{:=} \lim_{n \to \infty} f_n[/tex]
    is not Riemann integrable?
     
  4. Oct 8, 2008 #3
    Here's a hint:

    Do you know any of the "popular" functions that fail to be Riemann integrable? Why not try to construct a sequence of Riemann integrable functions that converge (pointwise) to this function.
     
  5. Oct 27, 2008 #4
    I was referring to page 322 in Rudin.
     
  6. Oct 27, 2008 #5
    So if you're given a fxn

    f:[-2,3]->R defined by f(x) = 0 if x is rational and 4 is rational, then if is not RI, that is int(f(x),-2,3) DNE.

    How does Lebesgue theory make it integrable?
     
  7. Oct 27, 2008 #6

    HallsofIvy

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    Did you mean "f(x)= 4 if x is not rational"?

    Lebesque theory measures sets differently from Riemann theory, in a way that gives "measure" to a much larger collection of sets. In particular, in Lebesque measure any countable set (such as the rational numbers) has measure 0. Since the measure of the interval [-2, 3] has measure 3-(-2)= 5, just as in Riemann theory, and the disjoint union of the rational and irrational numbers give all of the interval, the measure of the set of irrational numbers in [-2, 3] is 5. If a function is constant on a measurable set, its integral over that set is that constant times the measure of the set. The integral of "f(x)= 0 if x is rational and f(x)= 4 if x is irrational" has integral 0(0)+ 4(5)= 20.

    Of course, that has nothing to do with the original question.
     
  8. Oct 28, 2008 #7

    CompuChip

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    Personally I like that in measure theory, statements like
    "If f is constantly equal to c almost everywhere on [a, b], then the integral of f over that interval is equal to (b - a) * c"
    can be rigorously defined (once I'd be more clear about the measure). In particular, the statement "almost everywhere" has a well-defined meaning which usually corresponds to ones intuition (although admittedly, intuitively Q may seem larger than it is :smile:)
     
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