Riemann sum of a zero-one function

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Find the Riemann sum $S^*$ for $$V(x) = \begin{cases} 1 & \text{at all fractions $x = p/q$} \\ 0 & \text{at all other points} \end{cases}$$ when $\Delta x = 1/n$ and each $x_k^*$ is the midpoint. This $S^*$ is well-behaved but still $V(x)$ is not Riemann integrable.

We have $S^*=\Delta x\sum_{k=1}^nV(x_k^*)$. Since $\Delta x = 1/n$, $\Delta x$ is a fraction and each midpoint $x_k^*$ is a fraction. Therefore, $V(x_k^*) = 1$ at each midpoint $x_k^*$, and $S^*$ becomes $1/n \times n = 1$. Why is $V(x)$ not Riemann integrable?

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$V$ is not Riemann integrable because all the Darboux lower sums vanish. As you proved that the upper Darboux sums are all equal to $1$, $ V$ is not Riemann integrable.