I want to ask if the indicator function of rational numbers is Riemann-integragle, cause I read that a function is Riemann-integrable if the set of discontinuities is at most of Lebesgue measure $0$ on a compact set, And $\mathbb{Q}$ has zero Lebesgue measure. However I also read arguments about the non integrability using Darboux sums, so it is not clear for me if it is integrable or not.
Thanks.
This function is not continuous at any point, not just at the rationals. Let's call the function $D$. Given any $x\in\mathbb{R}$ you can pick a sequence $x_n$ of rational numbers and a sequence $y_n$ of irrational numbers in your interval such that both converge to $x$. However, $D(x_n)\to 1$ and $D(y_n)\to 0$, so $D$ is not continuous at the point $x$. So the set of discontinuities of $D$ in an interval $[a,b]$ has full measure $b-a$, and hence $D$ is not Riemann integrable. So there are no contradictions here.
That being said, this function is Lebesgue integrable. (it is even a simple function)
Yes, the rationals have measure $0$, but the indicator function of the rationals is actually discontinuous everywhere.
