proving the failure of riemann integrability

• May 10th 2008, 07:28 PM
squarerootof2
proving the failure of riemann integrability
i want to prove that the function f:[0,1] -> R where [0,1] is the closed interval 0,1 and R is the set of reals, defined by f(x)=1 if x is rational and 0 otherwise, is not riemann integrable. i know that this is the dirischlet's function and that it is not integrable, but i'm not sure how to go about proving it. thanks for any help!
• May 10th 2008, 07:30 PM
ThePerfectHacker
Because any upper Darboux sum is 1 while any lower Darboux sum is 0.
This means the upper integral is 1 while the lower integral is 0.
Thus the function is not Riemann integrable.
• May 10th 2008, 07:43 PM
TheEmptySet
Quote:

Originally Posted by squarerootof2
i want to prove that the function f:[0,1] -> R where [0,1] is the closed interval 0,1 and R is the set of reals, defined by f(x)=1 if x is rational and 0 otherwise, is not riemann integrable. i know that this is the dirischlet's function and that it is not integrable, but i'm not sure how to go about proving it. thanks for any help!

Let P be any sequnce of partitions of [0,1] thats mesh (the length of each sub interval) goes to zero. The upper sum for the integral U(f,p) = 1 by the denisty of the rationals and the lower sum L(f,p)=0 by the density of irrationals

$|U(f,p)-L(f,p)|=1$

so the integral does not exist