Prove that $$\begin{align}f\colon [0,1]&\to \mathbb{R}\\x&\mapsto f(x)=\begin{cases}0, & x\notin \mathbb{Q}\\ \frac{1}{q}, & x=\frac{p}{q} \text{ in lowest terms.}\\ \end{cases}\end{align}$$ is Riemann integrable on $[0,1]$.
My proof (is it a valid one?):
All I know about this function is that $\displaystyle\forall a\in(0,1),\; \lim_{x\to a}f(x)=0.$
That is, we have that for all $a\in(0,1)$, for every $\varepsilon>0$, there is a $\delta>0$ such that, given $x\in[a-\delta,a+\delta]\setminus \{a\}$ we have $0\leq f(x)<\varepsilon$.
Let $r$ be irrational (so $\frac{r}{n}$ is irrational for all natural $n$) and consider the partition $$ P_n=\left\{0,\frac{r}{n},\frac{r}{n}+\frac{1}{n},\dots , \frac{r}{n}+\frac{i}{n},\dots ,\frac{r}{n}+\frac{n-1}{n},1\right\}.$$ Note that the number on the center of every interval $\left[(P_n)_i,(P_n)_{i+1}\right]$ ought be irrational.
Let $n$ be such that $(P_n)_{i+1}-(P_n)_i<2\delta$ (so that $[(P_n)_i,(P_n)_{i+1}]$ is of the form $[a-\delta,a+\delta]$ for an irrational $a$). Now, since $f(a)=0$ (for $a$ is irrational), we have that $\forall x\in [(P_n)_i,(P_n)_{i+1}],\; 0\leq f(x)<\varepsilon$ and thus $\sup\{f(x)\mid x\in[(P_n)_i,(P_n)_{i+1}] \}\leq\varepsilon$. Since this can be true for all epsilon, we have that the upper sum can be made as close to $0$ as desired. The lower sum is always $0$. We conclude that $f$ is integrable on $[0,1]$ $\square$.
Is the proof valid?
I claim $\forall i,\; \forall x\in [(P_n)_i,(P_n)_{i+1}],\; f(x)<\varepsilon$. This can´t be. What happened to the rational values? lol
