# R integrable ?

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.