Timeline for Why isn't integral defined as the area under the graph of function?
Current License: CC BY-SA 4.0
33 events
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| Apr 12, 2019 at 5:59 | answer | added | cgodfrey | timeline score: 5 | |
| Apr 11, 2019 at 19:17 | answer | added | Max M | timeline score: 7 | |
| Jan 30, 2019 at 5:21 | answer | added | Nate Eldredge | timeline score: 21 | |
| Jan 30, 2019 at 5:15 | comment | added | Nate Eldredge | Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space. From the usual definitions via simple functions, this is fairly straightforward, and one can prove the standard theorems (dominated convergence, etc) without too much trouble. But if you want to use a definition as "area under the graph", you have to take a detour and construct Lebesgue measure on $\mathbb{R}$, which is a fair amount of work. | |
| Jan 30, 2019 at 1:45 | history | edited | Piotr Hajlasz |
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| Jan 29, 2019 at 22:01 | comment | added | Todd Trimble | The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers. | |
| Jan 29, 2019 at 14:02 | comment | added | user57888 | @mathreadler You are right, I edited the question, to remove possible uncertainty as to the intended reading. | |
| Jan 29, 2019 at 13:58 | history | edited | user57888 | CC BY-SA 4.0 |
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| Jan 29, 2019 at 13:46 | comment | added | mathreadler | @Qfwfq maybe we should ask the OP this? | |
| Jan 29, 2019 at 13:44 | answer | added | Dave L Renfro | timeline score: 17 | |
| Jan 29, 2019 at 13:20 | comment | added | Qfwfq | @mathreadler: what do you mean by "graphing" a function? | |
| Jan 29, 2019 at 12:31 | comment | added | mathreadler | It is too sloppy definition. Very many integrable functions can't even be graphed to any precision. | |
| Jan 29, 2019 at 9:56 | history | edited | user57888 | CC BY-SA 4.0 |
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| Jan 29, 2019 at 9:51 | vote | accept | user57888 | ||
| Jan 28, 2019 at 20:29 | comment | added | Qfwfq | Oh yes, it's basically the same, upvoted (both answers) | |
| Jan 28, 2019 at 20:24 | comment | added | user57888 | @Qfwfq You're right. This is essentially the answer of Nik Weaver. I missed it, because I expected some counterexample. | |
| Jan 28, 2019 at 20:10 | comment | added | Qfwfq | Is your question why Lebesgue integral is defined as an increasing limit of areas of pluri-rectangles (with possibly contable pieces with measurable "base") instead of directly as the measure of the subgraph? It's basically the same thing, once you've proved the theorem in Lebesgue measure theory that $\mu(E)=\lim_n \mu(E_n)$ when good $E$ is approximated by good $E_n$ from the inside ($E$ increasing union of $E_n$). Take $E$ as the subgraph and $E_n$ the pluri-rectangles. | |
| S Jan 28, 2019 at 20:01 | history | suggested | CommunityBot | CC BY-SA 4.0 |
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| Jan 28, 2019 at 20:00 | answer | added | Piotr Hajlasz | timeline score: 93 | |
| Jan 28, 2019 at 19:45 | comment | added | user57888 | @ArturoMagidin Ok, then you want to apply it to some examples, where Riemann integration (or Riemann integration modulo a set of measure 0) doesn't apply. I would be happy to see some examples, where Lebesgue integral is easier to compute than the measure of the are under graph. This is a part of what I'm asking. | |
| Jan 28, 2019 at 19:42 | comment | added | Arturo Magidin | @user57888: That only lets you calculate integrals for those functions that you already knew how to compute integrals. Why bother developing measure theory at all, then? Lebesgue integration is stronger than Riemann integration (larger class of functions, more robust convergence theorems). But if the only ones you can calculate are the ones you could calculate before, then you haven't really done anything except spin your wheels. | |
| Jan 28, 2019 at 19:41 | review | Suggested edits | |||
| S Jan 28, 2019 at 20:01 | |||||
| Jan 28, 2019 at 19:40 | vote | accept | user57888 | ||
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| Jan 28, 2019 at 19:37 | comment | added | user57888 | @ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral. | |
| Jan 28, 2019 at 19:37 | answer | added | Nik Weaver | timeline score: 30 | |
| Jan 28, 2019 at 19:37 | history | edited | user57888 | CC BY-SA 4.0 |
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| Jan 28, 2019 at 19:31 | comment | added | Arturo Magidin | @user57888: you define measure theory because you are trying to define the Lebesgue integral. How would you actually compute the value of integrals using your idea? | |
| Jan 28, 2019 at 19:29 | comment | added | user57888 | @ArturoMagidin Well, once you defined Lebesgue measure, you can define area under graph as measure of the set of points under graph. EDIT: I've written a couple of seconds too late. | |
| Jan 28, 2019 at 19:29 | comment | added | Nik Weaver | I think the question is why develop integration theory in addition to measure theory. Once you have measure theory, just define the integral to be the measure of the region under the graph. Right? | |
| Jan 28, 2019 at 19:25 | review | Close votes | |||
| Jan 29, 2019 at 17:18 | |||||
| Jan 28, 2019 at 19:22 | comment | added | Arturo Magidin | You do realize that the point of defining the integral is to come up with a notion of "area under the graph". This notion does not exist a priori. In reality, we can't geometrically compute areas except of very specific figures, mainly rectilinear ones. As to why we have to define measure theory, perhaps look at the answer on this question: math.stackexchange.com/questions/7436/lebesgue-integral-basics | |
| Jan 28, 2019 at 19:14 | history | edited | user57888 | CC BY-SA 4.0 |
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| Jan 28, 2019 at 19:03 | history | asked | user57888 | CC BY-SA 4.0 |