Let $f$ be defined as:
$$ f(x) = \begin{cases} x & \text{ if } x\in\mathbb{Q}; \\\\ 0 & \text{ if }x\notin\mathbb{Q}. \end{cases} $$ Is $f$ Riemann integrable on $[0,1]$? Prove it.
We know that the upper sum $U = x$ and lower sum $L=0$.
Since $\lim U\neq \lim L$, $f(x)$ is not Riemann integrable.
Is this sufficient for the proof?
Start from the basic. Divide the interval [0,1] into $n$ sub-intervals, each of length $\frac{1}{n}$, i.e. $[0,\frac{1}{n}], [\frac{1}{n}, \frac{2}{n}], \cdots, [\frac{n-1}{n}, 1]$, Note that, at each sub-interval of the form $[\frac{i}{n}, \frac{i+1}{n}]$, maximum value of the function = $\frac{i+1}{n}$ and minimum value is 0, as each sub-interval has at-least one rational and one irrational number. Then show that Lower integral converges to 0, but upper integral does not which will show that the function is not Riemann Integrable.