Thread: 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!

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.

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



so the integral does not exist