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

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  • #1
AxiomOfChoice
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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?
 
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  • #2
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.
 
  • #3
I buy that. Thanks very much for your help.
 
  • #4
AxiomOfChoice said:
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?

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].
 
  • #5
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!
 
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  • #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.
 

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

1. What is the characteristic function of the irrationals?

The characteristic function of the irrationals is a mathematical function that maps all real numbers to either 0 or 1, depending on whether the number is irrational or not. It is denoted by χI and can also be defined as the indicator function of the set of irrational numbers.

2. What does it mean for a function to be Riemann integrable?

A function is Riemann integrable if it can be approximated by a sequence of sums of the form ∑i=1nf(xi)(xi-xi-1), where n is the number of subintervals of a partition and xi are the points within the subintervals. In simpler terms, a function is Riemann integrable if it can be integrated using the Riemann integral.

3. Is the characteristic function of the irrationals Riemann integrable?

No, the characteristic function of the irrationals is not Riemann integrable on any interval [a,b] where a and b are real numbers. This is because the set of irrational numbers is uncountable and therefore cannot be approximated by a sequence of sums.

4. Why is the characteristic function of the irrationals not Riemann integrable?

The characteristic function of the irrationals is not Riemann integrable because it is a discontinuous function. This means that it has a jump discontinuity at every rational number, which makes it impossible to approximate using a sequence of sums.

5. Can the characteristic function of the irrationals be integrated using other methods?

Yes, the characteristic function of the irrationals can be integrated using the Lebesgue integral. Unlike the Riemann integral, the Lebesgue integral can handle discontinuous functions and is not limited by the countability of a set. However, the result of this integration will still be 0, as the measure of the set of irrational numbers is 0.

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