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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.

Hecatonchires
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    Geometric transparency, simplicity, and no-nonsensicality are in the eye of the beholder. For example, what's so special about partitioning $[a,b]$ into equal length subintervals? Why get hung up about guaranteeing equality of the length of the subintervals? – Lee Mosher Nov 10 '19 at 20:41
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    These limits also exist for some non-Riemann-integrable functions. And they have some unwanted properties. Consider $\chi_{\mathbb{Q}}$ and your first suggestion. Take an interval with rational endpoints, you get $b-a$. Take an interval with one rational and one irrational endpoint, you get $0$. Not nice. Okay, you said "piecewise continuous", which that isn't. But we also want the integral for less regular functions. – Daniel Fischer Nov 10 '19 at 20:41
  • @DanielFischer: Engineers don't know that function. They know only continuous functions, perhaps here and there interrupted by a jump discontinuity. – Hecatonchires Nov 10 '19 at 20:43
  • @LeeMosher, because we can prove certain basic theorems more quickly like that. – Hecatonchires Nov 10 '19 at 20:49
  • I wouldn't say it's quicker, but then it's not my question. But if you already have the answer to your question, what are you asking? I worry that this becomes just an argumentative discussion without anything an answerer can provide. – Lee Mosher Nov 10 '19 at 20:52
  • Objections of course. Aspects that I have overlooked. Why did Riemann opt for these partitions? Indeed merely an aesthetical choice, or does he have better reasons? – Hecatonchires Nov 10 '19 at 20:53
  • A couple of your points seem odd to me. On the practical application side of things, we rarely if ever have continuous functions; rather, we have data. We sample said data and then use any number of numerical integration techniques to apply the theory to our application, and often our data will be unevenly sampled. Second, your comment on an answer below that Lebegue's integral leads to "bewildering" sets is misplaced if we're only applying each integral to piecewise continuous functions -- with such an assumption, we don't need the full power of measure theory. – Brian Moehring Nov 11 '19 at 00:47

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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.

Lee Mosher
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  • The intuitive simplicity of Riemann vs. Lebesgue lies in its verticality (fit adjacent vertical rectangles under the graph of the function and sum their areas) vs. Lebesgue's horizontality and the strange and bewildering geometry of some the sets we have to work with in the Borel sigma algebra of the latter theory. – Hecatonchires Nov 10 '19 at 22:44
  • Let me re-iterate what I want: an integration theory that still stays true to this "vertical intuition" mentioned in the previous comment, but takes less pages for proofs and elaboration than Riemann's theory does. The premise of my question is that you can do this by skipping the use of partitions altogether (the question where the "test-point" within one part should be is also no longer pertinent if you just evaluate the function equidistantially as I propose to do) – Hecatonchires Nov 10 '19 at 22:48
  • See also the first answer here: https://math.stackexchange.com/questions/1642034/eli5-riemann-integrable-vs-lebesgue-integrable – Hecatonchires Nov 10 '19 at 22:55
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    Well, I guess I'll just re-iterate my answer to your re-iteration, after which I'll leave the battlefield: the Darboux integral, with partitions and sups and infs, takes less pages for proofs and elaboration then the actual Riemann integral with arbitrary partitions and test points, and it also takes less pages for proofs and elaboration then your more restricted Riemann integral, with even partitions and endpoints for test points. – Lee Mosher Nov 11 '19 at 03:47
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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 $$\lim_{n \to \infty}\frac{1}{n}\sum_{j=\lceil na \rceil}^{\lfloor nb \rfloor} f\left(\frac{j}{n}\right)=0$$

Mohammad Riazi-Kermani
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