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Is the characteristic function of the irrationals Riemann integrable on [a,b]?

  1. Feb 17, 2009 #1
    The characteristic function of the RATIONALS is a well-known example of a bounded function that is not Riemann integrable. But is the characteristic function of the IRRATIONALS (that is, the function that is 1 at every irrational number and 0 at every rational number) Riemann integrable on an arbitrary interval [a,b]? It seems like it would be, and that its integral would be equal to 1.

    But maybe I'm wrong. Anyone know for a fact?
     
    Last edited: Feb 17, 2009
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  3. Feb 17, 2009 #2

    HallsofIvy

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    No, the characteristic function of the irrationals on an integral is not Riemann integrable for the same reason the the characteristic function of the rationals isn't. Within any interval, no matter how small, there exist both rational and irrational functions. That means that there exist values of 1 and 0 in any interval. The "upper sum" between a and b is 1(b-a)= b-a while the lower sum is 0(b-a)= 0 for any partition. This is true for both functions.
     
  4. Feb 18, 2009 #3
    I buy that. Thanks very much for your help.
     
  5. Feb 18, 2009 #4
    The (powerful) Vitali theorem states that a bounded function [tex]f: D \subset \mathbb R \longrightarrow \mathbb R[/tex] defined on a bounded domain is Riemann integrable IF AND ONLY IF it has a set of point of discontinuity of measure zero.

    Now in your function you have that [tex][a,b][/tex] is the set of the points where the function is not continuous. So it has no measure zero and is not Riemann integrabel.. unless [tex]b=a[/tex].
     
  6. Feb 18, 2009 #5

    HallsofIvy

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    Indeed, both functions, f(x)= 1 if x is rational and 0 if x is irrational or g(x)= 0 if x is rational or 1 if x is irrational have all real numbers as points of discontinuity- which is NOT of measure 0!
     
    Last edited: Feb 19, 2009
  7. Feb 18, 2009 #6
    Keep in mind that the indicator function of the irrationals is one minus the indicator function of the rationals. That seems to be a pretty quick proof.
     
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