Riemann Integrability of f(x) = x on [0,1]

[sihag]

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.

 

[HallsofIvy]

Yes, if that function were integrable, then you would get two contradictory values. Conclusion: that function is not (Riemann) integrable.

(It is Lebesque integrable: since the rational numbers have measure 0, it Lebesque integral is 0. If the function were f(x)= x if is is irrational, 0 if x is rational, then its Lebesque integral would be 1/2.)