Why this function is not Riemann integrable on [0, 1] ? \begin{equation}
f\left(x\right)=\begin{cases}
x & x\in\mathbb{Q}\\
0 & x\notin\mathbb{Q}
\end{cases}
\end{equation}
We can calculate the upper integral and lower integral, $\overline{\int}f\left(x\right)dx=0.5$, $\underline{\int}f\left(x\right)dx=0$, therefore, the Riemann's criterion does not hold. So the function is not Riemann inegrable.
But if you use the Lebesgue theorem, bounded function on [a, b] is Riemann integrable if and only if the set of discontinuities of $f\left(x\right)$ has measure zero.
We know that (1) every finite set has measure zero, and (2) every countable subset of $\mathbb{R}$ has measure zero.
The rational numbers $\mathbb{Q}$ is a countable subset of $\mathbb{R}$, and the rational numbers $\mathbb{R}\setminus\mathbb{Q}$ is a countable subset of $\mathbb{R}$.
Hence, the set of discontinuities of $f\left(x\right)$ on [0, 1] has measure zero. Therefore $f\left(x\right)$ is Riemann integrable.
What are the mistakes here? Thank you very much.
Best Answer
well, your discontinuities are a lot more than $\mathbb{Q}\cap [0,1]$ they are in fact all of $[0,1]$, since it is the boundary that matters. or show me one point apart from $0$ where your function is continuous!