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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.
Yes, the rationals have measure $0$, but the indicator function of the rationals is actually discontinuous everywhere.
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)
