Is the concept of integral developed so much in higher analysis so that integral of every everywhere discontinuous real function of a real variable defined on the segment exists and is unique?
If not so much, how much it has been generalized?
Asked
Active
Viewed 125 times
-1
-
Could you further specify your question? Integrable in which way or sense? Riemann? Lebesgue? There are of course integrable discontinuous functions and non integrable discontinous functions, for both Lebesgue and Riemann integrals – Luke Mar 12 '20 at 12:21
-
Yes, that´s true. @Luke – Mar 12 '20 at 12:24
-
1See: Ralph Henstock, A short history of integration theory, Southeast Asian Bulletin of Mathematics 12 #2 (1988), 75-95 (has 262 references) AND Peter Bullen, Non-absolute integrals in the twentieth century, AMS Special Session on Nonabsolute Integration, 23-24 September 2000, 27 pages (has 195 references) AND the references given here. – Dave L. Renfro Mar 12 '20 at 12:27
-
1https://www.researchgate.net/publication/325357838_List_of_names_of_integrals_in_alphabetic_order – Jean Marie Mar 12 '20 at 12:29
-
1See also Integrals in analysis and category theory AND Expressing the Lebesgue integral using categories + the difficulty of describing estimates in category theory AND Why do we denote (co)ends with integral notation (beyond Fubini's Theorem)? – Dave L. Renfro Mar 12 '20 at 12:35
-
@DaveL.Renfro Thany you for all the effort, those two comments could be very useful and good answer. – Mar 12 '20 at 12:38
2 Answers
2
The Riemann integral has been generalised substantially, in different ways, and for different purposes. One extension is to the Lebesgue integral resulting in a theory with much better convergence properties. In a nutshell, the Riemann integral defines a metric structure on the space of continuous functions. Its completion is the theory of Lebesgue integration. In this sense, the extension from Riemann to Lebesgue is an instance of a general 'extension by metric completion' technique. It serves to 'fix' convergence deficiencies at the cost of introducing many more integrable functions beyond the continuous ones, some of which may be very wild. This is where considerations of the axiom of choice come in. Having said that, Lebesgue integration is intimately linked with measure theory which has its own merits beyond merely serving as a platform upon which to extend the Riemann integral. In any case, as powerful as it is, it does not achieve your goal of every function being integrable (again, with choice considerations).
Other approaches to integration that go beyond Riemann integration are reviewed nicely in "Theories of Integration" by Kurtz and Swartz. It is also worthwhile to mention an old approach to integration via the Daniel integral (wiki probably has a page on that). I would also mention the books "The theory of Lebesgue measure and integration" by Hartman and Mikusinski, "The Bochner Integral" by Mikusinski, and "An introduction to analysis" by Mikusinski and Mikusinski for their interesting portrayals of integration.
Ittay Weiss
- 79,840
- 7
- 141
- 236
1
No: there are (assuming the Axiom of Choice) non-measurable functions for which there is no reasonable definition of an integral.
Look up Lebesgue integration, and maybe Henstock-Kurzweil integral.
Robert Israel
- 448,999