a function $=x(1-x)$ on rationals, and $=0$ on irrationals is not Riemann integrable on $[0,1]$

Question

Let $f(x)=x(1-x), x\in [0,1]\cap \Bbb Q$, and $=0,\ x\in [0,1]\backslash \Bbb Q$. Can we show that $f(x)$ in not Riemann integrable on $[0,1]$.

If the inteval is $[a,b]$ with $0<a<b<1$, then it is easy to see that the sup of Riemann sum $>$ the inf of it. However, what about the case $[0,1]$?

Answer 1

The endpoints do not matter: just choose partitions $P_n=\left \{ 0,1/n,2/n,\cdots, (n-1)/n,1 \right \}$ of $[0,1]$ and note that

$U(P_n,f)\ge \sum^{n}_{k=1}\frac{k}{n}(1-\frac{k}{n})\frac{1}{n}=\frac{n(n+1)}{n^{2}}-\frac{n(n+1)(2n+1)}{6n^{3}}\to 1-\frac{1}{3}=\frac{2}{3}$ as $n\to \infty$,

whereas

$L(P_n,f)=0$ for all integers $n.$

Or better yet, note that $f$ is discontinuous on an uncountable set, so can not be Riemann integrable.

Answer 2

The end points will not have any effect on the existence or not existence of the Riemann integral. In this case the integral does not exist with or without the end points because as you mentioned the upper sum is different from the lower sum on any partition of $[0,1]$