Timeline for Can the Riemann integral be defined through a closure/completion process?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 18, 2022 at 18:42 | answer | added | Willie Wong | timeline score: 3 | |
| May 15, 2021 at 18:27 | answer | added | Willie Wong | timeline score: 1 | |
| Apr 12, 2019 at 17:06 | comment | added | Pietro Majer | @ Francois Ziegler for instance, Riemann integrable functions are a Banach space with the uniform norm. Indeed, $f$ is R-integrable iff $f$ is bounded and continuous a.e., and clearly these properties are preserved under uniform convergence. | |
| Dec 14, 2017 at 17:09 | comment | added | Francois Ziegler | This seems to require that the space of Riemann integrable functions admit a norm or uniform structure in which it is complete. Does it? | |
| Dec 14, 2017 at 16:00 | answer | added | Kusma | timeline score: 1 | |
| Dec 14, 2017 at 9:08 | comment | added | Gro-Tsen | @Chaitanya But here I'm just considering functions on $[0,1]$, so we have $\|•\|_1\leq\|•\|_\infty$. Of course, the completions will be different, as you point out, but that's just the point of the question: describing R-integrable functions as the completion of step functions for something (not necessarily a norm, though: could be a Fréchet space for example). | |
| Dec 14, 2017 at 5:01 | comment | added | Chaitanya | I think it would be difficult to find a 'mix' between the uniform (L-infinity) and the L-1 norm. The reason is that two inequivalent norms on a Banach space are incomparable, as can be seen by using the open mapping theorem for the identity map (which is clearly surjective). | |
| Dec 12, 2017 at 21:57 | comment | added | user54321 | Whoops, in the first comment I meant the metric to be $\inf \{ \mu(K) + \lVert f - g \rVert_{\ell^\infty(K^c)}\}$, with the complement of $K$. | |
| Dec 12, 2017 at 19:49 | comment | added | user54321 | /continued In this "metric", $f_n \to f$ if, except for very small sets which may vary with $n$, $f_n \to f$ in the $\ell^\infty$ sense. If $f$ is Riemann integrable it is clear that it is a limit of step functions in this metric, but hopefully the converse also holds. | |
| Dec 12, 2017 at 19:49 | comment | added | user54321 | I wonder if the following helps. Riemann-integrable functions must be bounded by definition, so let's consider only those bounded by $1$ and supported in the unit interval. The $\ell^1$ norm is the weakest norm under which the integral is continuous, by definition, so any candidate norm that solves this problem must be stronger than the $\ell^1$ norm. By the example of $\sin 1/x$, it must also be weaker than the $\ell^\infty$ norm. Consider the quasimetric between functions $d(f,g) = \inf \{ \mu(K) + \lVert f - g \rVert_{\ell^\infty(K)} \mid K \text{ compact}\}$ /continued below | |
| Dec 12, 2017 at 16:16 | comment | added | Gro-Tsen | @NateEldredge I think the simplest example is $x\mapsto \sin\frac{1}{x}$ (extended arbitrarily at $0$): it is not regulated because it has no right limit at $0$, but it is Riemann-integrable because it is bounded and discontinuous only at $0$. | |
| Dec 12, 2017 at 16:05 | comment | added | Nate Eldredge | Just to help others like me who aren't so familiar with these concepts, here you can find some examples of Riemann integrable functions which are not regulated. | |
| Dec 12, 2017 at 14:41 | comment | added | Chaitanya | Regard the space of step functions as embedded in L^infinity, If we devise a family of bounded functionals on L^infinity such that the intersection of their kernels consists precisely of the Riemann integrable functions, then we can regard that space as the closure of the step functions in the topology. generated by the corresponding seminorms. Mimicing the weak topology, I am trying to cook up two such functionals using one-sided limsup minus liminf (and the sup norm). The idea is that Riemann integrable means continuous 'almost everywhere'. | |
| Dec 12, 2017 at 12:22 | comment | added | Gro-Tsen | @Corbennick This is precisely what I translated as "regulated function". The term (and perhaps the concept itself) is, I think, due to Dieudonné. | |
| Dec 12, 2017 at 12:08 | comment | added | user1688 | The french school does integration theory with the ''fonctions réglée'', which are uniform limits of step functions. | |
| Dec 12, 2017 at 11:40 | history | asked | Gro-Tsen | CC BY-SA 3.0 |