- #1

- 29

- 0

= 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.