Let $f\colon[a,b]\to\mathbb{R}$ be a bounded function, and let $\mathcal{P}_{n}$ be a sequence of partitions (not necessarily evenly spaced) with mesh $m(\mathcal{P}_{n}) \to 0$ as $n\to \infty$. Then $$|S_{L}(\mathcal{P}_{n})-S_{R}(\mathcal{P}_{n})|\to 0 \text{ for } n\to\infty$$
Note:
$$S_{L}(\mathcal{P}_{n}) = \sum^{n}_{k=1}f(c_{k-1})[x_{k}-x_{k-1}] \text{ and }S_{R}(\mathcal{P}_{n})=\sum^{n}_{k=1}f(c_{k})[x_{k}-x_{k-1}]$$ where $c_{k}$ is an internal point, ie. $c_{k}\in[x_{k-1},x_{k}]$
Ok so I thought of the function $f\colon[0,1]\to\mathbb{R}$: $$f(x) = \begin{cases} 1, & \text{if } x\in\mathbb{Q}, \\ 0, & \text{if } x \notin \mathbb{Q}.\end{cases}$$
But I'm not sure how to progress with showing the difference between the Riemann sums doesn't converge to zero, i.e. showing the function is non-integrable.
How would I do this using only the information above?
For every partition, produce a finer partition an take all the $c_i$ to be rational, then take the same refinement partition and take all $c_i$ irrational.
One sum will be $0$ the other $1$.