Timeline for About the Riemann integrability of composite functions
Current License: CC BY-SA 4.0
9 events
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| Apr 2, 2023 at 15:22 | comment | added | wangtwo | For this type of question, we can not forget Gelbaum and Olmsted. See Chapter 8, Example 34 in Gelbaum, Bernard R. and Olmsted, John M. H.'s Counterexamples in analysis. (Corrected reprint of the second (1965) edition. Dover Publications, Inc., Mineola, NY, 2003. xxiv+195 pp.) I can not find the 1965 edition, but I can see the example in the 1980 Chinese translation and 2003 editions. | |
| Sep 27, 2022 at 22:12 | history | edited | CommunityBot |
replaced http://math.uga.edu/~pete with http://alpha.math.uga.edu/~pete
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| Jun 28, 2022 at 3:46 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added a Wayback Machine link for the dead link.
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| Jun 22, 2022 at 8:13 | history | edited | CommunityBot |
replaced http://math.uga.edu/~pete with http://alpha.math.uga.edu/~pete
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| Apr 4, 2019 at 11:29 | comment | added | Pietro Majer | As to the last statement, more generally, if $g$ is a homeomorphism with absolutely continuous inverse $g^{-1}$, then $f\circ g$ is Riemann integrable whenever $f$ is. Indeed, $f\circ g$ is bounded, and its discontinuity set is $g^{-1}$(discontinuity set of $f$). | |
| Jan 1, 2016 at 21:01 | comment | added | user46189 | I know this is late to the game, but I found myself wondering about a case similar to (3) from the OP: What conditions on $g$ would we need to ensure that $f\circ g$ is Riemann integrable? Rudin shows that if $g$ is differentiable, strictly increasing, and $g'$ is integrable, then $f\circ g$ is Riemann integrable. The final result you post nudged me to the idea that you really want $g$ to carry a partition to a `quasi'-partition. Thus $g$ strictly-increasing/strictly-decreasing, differentiable with bounded derivative is sufficient. Also, $g$ strictly-monotonic and Lipschitz works. | |
| Apr 1, 2010 at 18:13 | comment | added | Qiaochu Yuan | I googled something like "composition of Riemann integrable functions is integrable" because I hoped that an article discussing this issue would discuss the issue I was interested in. So I guess I got lucky. | |
| Apr 1, 2010 at 9:57 | vote | accept | X.M. Du | ||
| Apr 1, 2010 at 9:57 | |||||
| Apr 1, 2010 at 8:46 | history | answered | Pete L. Clark | CC BY-SA 2.5 |