I need to show that the following function is not Riemann Integrable on $[0,1]$,
$f(x) = \begin{cases} x & x\in \mathbb{Q} \\ \ 0 & x\in\mathbb{R}\backslash \mathbb{Q} \end{cases}$
My solution is:
Assume $f$ is Riemmann integrrable. Thus by Darboux's Criterion, given an $\epsilon \gt 0$ there exists a partition $P = \{x_{0},...,x_{n}\}$ of $[0,1]$ such that
$U(f,P) - L(f,P) \lt \epsilon \\ \iff \sum_{i=0}^{n}M_{i}(x_{i}-x_{i-1}) - 0 \lt \epsilon \\$
Now,
$M_{n}(x_{n} - x_{n-1}) \lt \sum_{i=0}^{n}M_{i}(x_{i}-x_{i-1})$ , since the interval has only positive values.
Thus,
$M_{n}(x_{n} - x_{n-1}) \lt \epsilon$
but, $M_{n} = x_{n} = 1$,
So,
$1 \lt \epsilon + x_{n-1}$
Am I on the right track, because I'm not sure where to got from here?
For a general partition $(x_0,x_1, \ldots,x_n)$ we have $M_j = \sup_{x \in [x_{j-1},x_j]}f(x) = x_j,$ even if $x_j$ is not rational. This follows because the rationals are dense and for any $\delta > 0$ there exists a rational $r$ such that $x_j - \delta < r = f(r) \leqslant x_j$. Unless $M_j = x_j$ we get a contradiction.
Consequently, the upper sum is
$$U(P,f) = \sum_{j=1}^n x_j(x_j - x_{j-1}) \\= \frac{1}{2}\sum_{j=1}^n x_j(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n x_j(x_j - x_{j-1}) \\ > \frac{1}{2}\sum_{j=1}^n x_j(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n x_{j-1}(x_j - x_{j-1}) \\ = \frac{1}{2} \sum_{j=1}^n (x_j^2 - x_{j-1}^2) = \frac{1}{2}(x_n^2 - x_0^2 ) = \frac{1}{2}.$$
Since the irrationals are dense, we have $L(P,f) = 0$. Thus, $U(P,f) - L(P,f) > \frac{1}{2}$ for every partition $P$ proving that $f$ is not Riemann integrable.