Do regular partitions suffice for Riemann integrability of a real-valued function on a closed interval? Does the following condition for a bounded function $f: [a, b] \to \mathbb R$ suffice that it be Riemann integrable on $[a,b]$, with Riemann integral $I$?
For every $\epsilon > 0$, there exists a positive integer $N$ such that, for every positive integer $n \geq N$:  for each $i = 1, 2, \dots, n$, for each number $c_{i} \in [x_{i-1}, x_{i}]$ — the $i$th subinterval in the $n$th regular partition of $[a, b]$ whose subintervals have length $\Delta x = (b-a)/n$ — we have
$\left| \sum_{i=1}^{n} f\bigl(c_{i})\,\Delta x - I \right| < \epsilon$.
The condition may be stated equivalently as:
For every $\epsilon > 0$, there exists a $\delta > 0$ such that, for every positive integer $n$ with $\Delta x = (b - a)/n < \delta$:  for each $i = 1, 2, \dots, n$, for each number $c_{i} \in [x_{i-1}, x_{i}]$ — the $i$th subinterval in the $n$th regular partition of $[a, b]$ — we have
$\left| \sum_{i=1}^{n} f\bigl(c_{i})\,\Delta x - I \right| < \epsilon$.
In short, do regular partitions suffice for Riemann integrability? 
Notes:


*

*The usual condition for Riemann integrability involves arbitrary partitions, not just those that are regular, and then of course it involves the mesh of partitions being $< \delta$. Of course for a regular partition with $n$ subintervals, that mesh is $\Delta x = (b-a)/n$.
The only thing I intend to change in the usual condition is to restrict partitions just to those that are regular.

*Note that the condition as stated allows the sample points to be chosen arbitrarily in the subintervals. It's well known that it is not enough to allow just endpoints.
 A: Regular partitions are enough !
The problem is not banal: see


*

*Charles G. Denlinger Elements of Real Analysis (2011), p. 378 ;

*Jingcheng Ton Partitions of the interval in the definition of Riemann's integral Journal of Math. Educ. in Sc. and Tech. 32 (2001), pp. 788-793 (theorem 3).
It is sufficient to use any fixed sequence of partitions with the norm converging to zero (the fact is not much known).
A: Take the following Dirichlet's function:
$$D(x):=\begin{cases}1\;,\;\;x\in\Bbb Q\cap [0,1]\\{}\\0\;,\;\;x\in\Bbb Q^c\cap [0,1]\end{cases}$$
Then, for any $\;\epsilon>0\;$ and any $\;\delta>0\;$ there exists an irrational number in any non-trivial subinterval of $\;[0,1]\;$ , and thus the Riemann sum you wrote is zero (and thus $\;I=0\;$ in this case), which would mean the above function is Riemann integrable in the unit interval...but it isn't, of course.
The problem here is, imo, not the regular partitions you mention but in fact that you require only "the existence" of some points $\;c_i\;$ in those regular subintervals, whereas the usual definition requires that the sum exists in the limit for any choice of these points $\;c_i\;$ in the subintervals of the partition, as you mention at the very end of your question.
