I am facing the following problem:
Is the given function $f$ Riemann integrable on the interval $[1, 2]$?
$$ f(x) = \begin{cases} x & \forall x \in \mathbb{Q}, \\\\ -x & \forall x \in \mathbb{R} \backslash \mathbb{Q} \end{cases} $$
I am not sure how to prove this. I assume that the continuity is no the way how to prove this. From the density of the ratinal numbers in rational numbers I assume that this function is not Riemann integrable as we cannot find the Upper Riemann Sum nor the Lower Riemann Sum.
Therefore it violates the criteria of the Riemann Integrability:
Function $f$ is Riemann Integrable on the interval $[a,b]$ if and only if $$ \forall \varepsilon \gt 0 \ \exists D: 0 \le S(D,f) - s(D,f) \lt \varepsilon $$
Are my assumptions correct? If not, any idea how to solve this one?
Consider any division $D$ of the interval $[1,2].$ Then $$s(D,f)=s(D,-x)\le s(D,-1)=-1$$ $$S(D,f)= S(D,x)\ge S(D,1)=1$$ Therefore $S(D,f)-s(D,f)\ge 2.$ The function $f$ is thus not Riemann integrable.