Proving a function $f$ is not Riemann integrable

Question

I want to show that the function $f : [0, 1] \rightarrow \mathbb{R}$ defined by

$$f(x) \equiv \begin{cases} x & \text { if } x \in [0, 1] \text { is rational } \\ -x & \text{ if } x \in [0, 1] \text{ is not rational } \end{cases} $$

is not Riemann integrable. Here's my attempt:

Attempt:

Let $\{P_{n}\}$ be a sequence of partitions of $[0, 1]$ each of size $1/n$. By the density of $\mathbb{Q}$ and $\mathbb{R} - \mathbb{Q}$, it follows that

$$ \lim_{n\to\infty} U(f, P_{n}) = \lim_{n\to\infty} \left(\frac{1}{n}\right) \sum_{i = 1}^{n} x_{i} \\ = \lim_{n\to\infty}\left(\frac{1}{n}\right)\sum_{i=1}^{n} \frac{i}{n} \\ = \lim_{n\to\infty} \left(\frac{1}{n^2}\right)\left(n(n+1)/2\right) \\ = 1/2, $$

and

$$\lim_{n\to\infty} L(f, P_{n}) = \lim_{n\to\infty} \frac{1}{n} \sum_{i = 1}^{n} (-x_{i}) = - \lim_{n\to\infty} U(f, P_{n}) = -1/2.$$

So, the two limits are not equal, meaning that it is not integrable. Is my proof sufficient, or do I also need to show that there are no Archimedean sequences of $[0, 1]$ for $f$? If so, how would I go about doing that?

Answer 1

Your proof is fine. There are two equivalent conditions for Riemann integrability.

(1) For every $\epsilon > 0$ there exists a partition $P_\epsilon$ such for every refinement $P$ we have $U(P,f) - L(P,f) < \epsilon$.

(2) For every $\epsilon > 0$ there is a $\delta > 0$ such for partition $P$ with norm $\|P\| < \delta$ have $U(P,f) - L(P,f) < \epsilon$.

By producing a sequence of partitions $P_n$ such that $\|P_n\| = 1/n \to 0$, but $U(P_n,f) - L(P_n,f) \not\to 0$, condition (2) is violated and $f$ is not Riemann integrable.