Definition talk:Definite Integral/Riemann

The new edits that have just appeared that make this definition even more complicated than it already was: can we confirm that this is indeed an improvement? It appears that the language of sample points is already handled in Definition:Riemann Sum. the cited link https://www.encyclopediaofmath.org/index.php/Riemann_integral does not have that "every samplepoint sequence" bit, and (I am prepared to bet) neither was there anything about it in the Bartle and Sherbert.

Can this new stuff be implemented as a different definition, and an equivalence proof written? --prime mover (talk) 13:31, 9 December 2016 (EST)


 * My $\$0.02$? Rollback to the previous edit, but replace "Let $S \left({ f; \Delta }\right)$ denote the Riemann sum of $f$" with Let $S \left({ f; \Delta }\right)$ denote a Riemann sum of $f$. Then make it a theorem/definition/explanation/whatever that all Riemann Sums with such-and-such something converge to the same integral, and you can put that stuff about samplepoint sequences there. --GFauxPas (talk) 13:41, 9 December 2016 (EST)


 * Unfortunately there are always idiosyncrasies in the messy details establishing Riemann integration (subdivisions, "sample points"). Once you've seen them once, the rest is trivially equivalent. Here on, such a heuristic is undesired and we should create a presentation which does this nicely and rigorously. We can do better than selecting an arbitrary approach and declaring it sacrosanct. &mdash; Lord_Farin (talk) 16:30, 9 December 2016 (EST)


 * "a presentation which does this nicely and rigorously." I may be naive but isn't that what we had? --prime mover (talk) 16:47, 9 December 2016 (EST)

It went a long way, yes. Especially if the suggestion of GFauxPas is taken along. &mdash; Lord_Farin (talk) 04:54, 10 December 2016 (EST)