Riemann integration: usefulness of partitions

Question

I saw this old mathoverflow question coming up again recently. Arguments in favor of Riemann integration stress its geometric transparency, its simplicity and no-nonsense character.

Q: Why then do we not dispense with all the business about partitions and sub- vs. supersums (sometimes called "lower" and "upper Darboux sums") and just define the integral $\int_a^b f$ of a bounded (piecewise) continuous function $f:[a,b] \to \mathbb{R}$ as the limit $$\lim_{n \to \infty}\frac{b-a}{n}\sum_{j=1}^n f\left(\frac{(n-j)a+jb}{n}\right)$$ or the limit $$\lim_{n \to \infty}\frac{1}{n}\sum_{j=\lceil na \rceil}^{\lfloor nb \rfloor} f\left(\frac{j}{n}\right)$$ or something similar?

Proving $\int_a^b f + \int_a^b g = \int_a^b (f+g)$ is much simpler with these definitions as it is for the Riemann integral. The proof of $\int_a^bf + \int_b^c f=\int_a^c f$ may have become more difficult with my first definition, but not with the second one. Proving the fundamental theorem of calculus and the change of variables theorem is also as easy with these definitions as it is for the Riemann integral. A basic version of Fubini's theorem is accessible with (a trivial generalization of) these definitions. These alternative definitions are equally transparent from a geometric point of view.

EDIT: of course this mode of integration is 'reliable' only for a limited class of functions, just as is the case for the Riemann integral. Staying true to the pragmatic spirit of the Riemann integral, I'm content if this modified version works fine for bounded piecewise continuous functions.

Answer 1

Since your fuctions are integrable, then the limit of any Riemann's Sum will approach the integral and your examples will do the job as well.

On the other hand if for specific Riemann's sum the limit approaches a number it does not mean that the number is the integral and the function is integrable.

You may be able to construct a function which is not integrable but $$\lim_{n \to \infty}\frac{1}{n}\sum_{j=\lceil na \rceil}^{\lfloor nb \rfloor} f\left(\frac{j}{n}\right)=0$$

Answer 2

So one of the preconditions of your question seems to be false. I don't have a copy of Riemann's original work in front of me, but I will say that as the Riemann integral is usually defined nowadays, one uses arbitrary partitions $a=x_0 < x_1 < .... < x_K=b$ and arbitrary choices of test points $t_i \in [x_{k-1},x_k]$, $k=1,...,K$, and one takes a funny limit with respect to a partial ordering on such choices. Not so transparent, or simple, or no-nonsense.

On top of that, if you wish to prove that a continuous function on a closed interval is integrable, then with the Riemann integral you'll still have to deal with that funny partial order limit, and you won't be able to avoid infs and sups either: those infs and sups are used to provide the upper and lower bounds to the Riemann sum that are needed for proving convergence of the limit of Riemann sums.

So, if what you want is a simple path from definition of integral to the integrability of continuous functions on closed intervals, the Darboux integral is designed to do that by baking the infs and sups directly into the definition, avoiding a lot of nonsense. I would say that it's a small change from Riemann to Darboux in a conceptual sense, and I don't think the payoff is gigantic, but I do think the payoff is positive.