f(x) = x , if x is rational = 0 , if x is irrational on the interval [0,1] i just wanted to check if my reasoning is right. take the equipartition of n equal subintervals with choices of t_r's as r/n for each subinterval. calculating the integral as limit of this sum (and sending the norm to 0) i got 1/2 as my value. now if f were to be R - integrable the value of the integral must be 1/2. but each subinterval for any partition would contain an irrational so the lower R sum would be 0, for all partitions of [0,1] this yields 0 as the value of the integral. the two values contradict.