$f(x) = \begin{cases} 1 & x\in\Bbb Q \\[2ex] 0 & x\notin\Bbb Q \end{cases}$
Is this function Riemann integrable on $[0,1]$?
Since rational and irrational numbers are dense on $[0,1]$, no matter what partition I choose, there will always be rational and irrational numbers in every small interval. So the upper sum and lower of will always differ by $1$.
However, I know rational numbers in $[0,1]$ are countable, so I can index them from 1 to infinity. For each rational number $q$ in $[0,1]$, I can cover it by $[q-\frac\epsilon{2^i},q+\frac\epsilon{2^i}]$. So all rational numbers in $[0,1]$ can be covered by a set of measure $\epsilon$. On this set, the upper sum is $1\times\epsilon=\epsilon$. Out of this set, the upper sum is 0. So the upper sum and lower sum differ by any arbitrary $\epsilon$. Thus, the function is integrable.
One of the above arguments must be wrong. Please let me know which one is wrong and why. Any help is appreciated.