Mixing
Why isn't integral defined as the area under the graph of function?
$begingroup$
In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way?
The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).
What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?
Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by Fubini's theorem, so the two integrals shouldn't behave very differently.
EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points $(x,y) in E times mathbb{R}$ where $E$ is a space with measure and $y leq f(x)$. I assume that $f$ is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.
The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.
reference-request real-analysis measure-theory mathematics-education
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
asked Jan 28 at 19:03
user57888user57888
595510
$endgroup$
|
show 11 more comments
$begingroup$
In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way?
The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).
What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?
Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by Fubini's theorem, so the two integrals shouldn't behave very differently.
EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points $(x,y) in E times mathbb{R}$ where $E$ is a space with measure and $y leq f(x)$. I assume that $f$ is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.
The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.
reference-request real-analysis measure-theory mathematics-education
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
asked Jan 28 at 19:03
user57888user57888
595510
$endgroup$
10
$begingroup$
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
$endgroup$
– Arturo Magidin
Jan 28 at 19:22
20
$begingroup$
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?
$endgroup$
– Nik Weaver
Jan 28 at 19:29
2
$begingroup$
@ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral.
$endgroup$
– user57888
Jan 28 at 19:37
5
$begingroup$
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.
$endgroup$
– Qfwfq
Jan 28 at 20:10
10
$begingroup$
The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers.
$endgroup$
– Todd Trimble♦
Jan 29 at 22:01
|
show 11 more comments
$begingroup$
In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way?
The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).
What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?
Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by Fubini's theorem, so the two integrals shouldn't behave very differently.
EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points $(x,y) in E times mathbb{R}$ where $E$ is a space with measure and $y leq f(x)$. I assume that $f$ is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.
The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.
reference-request real-analysis measure-theory mathematics-education
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
asked Jan 28 at 19:03
user57888user57888
595510
$endgroup$
In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way?
The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).
What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?
Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by Fubini's theorem, so the two integrals shouldn't behave very differently.
EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points $(x,y) in E times mathbb{R}$ where $E$ is a space with measure and $y leq f(x)$. I assume that $f$ is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.
The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.
reference-request real-analysis measure-theory mathematics-education
reference-request real-analysis measure-theory mathematics-education
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
asked Jan 28 at 19:03
user57888user57888
595510
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
asked Jan 28 at 19:03
user57888user57888
595510
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
edited Jan 30 at 1:45
Piotr Hajlasz
10.1k43975
10.1k43975
asked Jan 28 at 19:03
user57888user57888
595510
asked Jan 28 at 19:03
user57888user57888
595510
asked Jan 28 at 19:03
user57888user57888
595510
595510
10
$begingroup$
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
$endgroup$
– Arturo Magidin
Jan 28 at 19:22
20
$begingroup$
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?
$endgroup$
– Nik Weaver
Jan 28 at 19:29
2
$begingroup$
@ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral.
$endgroup$
– user57888
Jan 28 at 19:37
5
$begingroup$
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.
$endgroup$
– Qfwfq
Jan 28 at 20:10
10
$begingroup$
The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers.
$endgroup$
– Todd Trimble♦
Jan 29 at 22:01
|
show 11 more comments
10
$begingroup$
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
$endgroup$
– Arturo Magidin
Jan 28 at 19:22
20
$begingroup$
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?
$endgroup$
– Nik Weaver
Jan 28 at 19:29
2
$begingroup$
@ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral.
$endgroup$
– user57888
Jan 28 at 19:37
5
$begingroup$
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.
$endgroup$
– Qfwfq
Jan 28 at 20:10
10
$begingroup$
The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers.
$endgroup$
– Todd Trimble♦
Jan 29 at 22:01
10
10
$begingroup$
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
$endgroup$
– Arturo Magidin
Jan 28 at 19:22
$begingroup$
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
$endgroup$
– Arturo Magidin
Jan 28 at 19:22
20
20
$begingroup$
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?
$endgroup$
– Nik Weaver
Jan 28 at 19:29
$begingroup$
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?
$endgroup$
– Nik Weaver
Jan 28 at 19:29
2
2
$begingroup$
@ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral.
$endgroup$
– user57888
Jan 28 at 19:37
$begingroup$
@ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral.
$endgroup$
– user57888
Jan 28 at 19:37
5
5
$begingroup$
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.
$endgroup$
– Qfwfq
Jan 28 at 20:10
$begingroup$
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.
$endgroup$
– Qfwfq
Jan 28 at 20:10
10
10
$begingroup$
The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers.
$endgroup$
– Todd Trimble♦
Jan 29 at 22:01
$begingroup$
The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers.
$endgroup$
– Todd Trimble♦
Jan 29 at 22:01
|
show 11 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Actually, in the following book the Lebesgue integral is defined the way you suggested:
Pugh, C. C. Real mathematical analysis.
Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.
First we define the planar Lebesgue measure $m_2$. Then we define the Lebesgue integral as follows:
Definition. The undergraph of $f:mathbb{R}to[0,infty)$ is $$ mathcal{U}f={(x,y)inmathbb{R}times [0,infty):0leq y<f(x)}. $$ The
function $f$ is Lebesgue measurable if $mathcal{U}f$ is Lebesgue measurable
with respect to the planar Lebesgue measure and then we define $$
int_{mathbb{R}} f=m_2(mathcal{U}f). $$
I find this approach quite nice if you want to have a quick introduction to the Lebesgue integration. For example:
You get the monotone convergence theorem for free: it is a
straightforward consequence of the fact that the measure of the union
of an increasing sequence of sets is the limit of measures.
As pointed out by Nik Weaver, the equality $int(f+g)=int f+int g$ is not obvious, but it can be proved quickly with the following trick:
$$
T_f:(x,y)mapsto (x,f(x)+y)
$$
maps the set $mathcal{U}g$ to a set disjoint from $mathcal{U}f$,
$$
mathcal{U}(f+g)=mathcal{U}f sqcup T_f(mathcal{Ug})
$$
and then
$$ int_{mathbb{R}} f+g= int_{mathbb{R}} f +int_{mathbb{R}} g $$
follows immediately once you prove that the sets $mathcal{U}(g)$ and
$T_f(mathcal{U}g)$ have the same measure. Pugh proves it on one page.
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
$endgroup$
5
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
1
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
2
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
1
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
|
show 6 more comments
$begingroup$
If $f: mathbb{R} to [0,infty)$ is Borel (or Lebesgue) measurable, then for each rational $a > 0$ define $X_a = f^{-1}([a,infty)) times [0,a)$. Then each $X_a$ is measurable and their union is exactly the region under the graph. So the region under the graph is measurable.
I think the reason why we develop the Lebesgue integral in the usual way is because it provides a powerful technique (characteristic functions --> simple functions --> arbitrary measurable functions) for deriving the basic theory of the integral. Even simple things like $int f + int g = int (f + g)$ aren't obvious if you take "measure of the region under the graph" as the definition.
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
$endgroup$
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
3
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
1
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
1
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
|
show 2 more comments
$begingroup$
The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text Measure and Integral. An Introduction to Real Analysis, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).
(second sentence of Chapter 5, on p. 64) The approach we have chosen [for the integral of a nonnegative function $f:E rightarrow [0, +infty],$ where $E subseteq {mathbb R}^n$ is measurable] is based on the notion that the integral of a nonnegative $f$ should represent the volume of the region under the graph of $f.$
I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
$endgroup$
add a comment |
$begingroup$
Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space $(X,mu)$. 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.
Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of $mathbb{R}$. But for finding the product measure of the region under the graph of a function $f : X to mathbb{R}$ via a Fubini-like approach, the only subsets of $mathbb{R}$ you really need to measure are intervals, of the form $[0,f(x)]$. We don't need any fancy measure theory to tell us the length of such sets.
Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on $mathbb{R}$ anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
$endgroup$
1
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f321916%2fwhy-isnt-integral-defined-as-the-area-under-the-graph-of-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Actually, in the following book the Lebesgue integral is defined the way you suggested:
Pugh, C. C. Real mathematical analysis.
Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.
First we define the planar Lebesgue measure $m_2$. Then we define the Lebesgue integral as follows:
Definition. The undergraph of $f:mathbb{R}to[0,infty)$ is $$ mathcal{U}f={(x,y)inmathbb{R}times [0,infty):0leq y<f(x)}. $$ The
function $f$ is Lebesgue measurable if $mathcal{U}f$ is Lebesgue measurable
with respect to the planar Lebesgue measure and then we define $$
int_{mathbb{R}} f=m_2(mathcal{U}f). $$
I find this approach quite nice if you want to have a quick introduction to the Lebesgue integration. For example:
You get the monotone convergence theorem for free: it is a
straightforward consequence of the fact that the measure of the union
of an increasing sequence of sets is the limit of measures.
As pointed out by Nik Weaver, the equality $int(f+g)=int f+int g$ is not obvious, but it can be proved quickly with the following trick:
$$
T_f:(x,y)mapsto (x,f(x)+y)
$$
maps the set $mathcal{U}g$ to a set disjoint from $mathcal{U}f$,
$$
mathcal{U}(f+g)=mathcal{U}f sqcup T_f(mathcal{Ug})
$$
and then
$$ int_{mathbb{R}} f+g= int_{mathbb{R}} f +int_{mathbb{R}} g $$
follows immediately once you prove that the sets $mathcal{U}(g)$ and
$T_f(mathcal{U}g)$ have the same measure. Pugh proves it on one page.
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
$endgroup$
5
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
1
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
2
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
1
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
|
show 6 more comments
$begingroup$
Actually, in the following book the Lebesgue integral is defined the way you suggested:
Pugh, C. C. Real mathematical analysis.
Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.
First we define the planar Lebesgue measure $m_2$. Then we define the Lebesgue integral as follows:
Definition. The undergraph of $f:mathbb{R}to[0,infty)$ is $$ mathcal{U}f={(x,y)inmathbb{R}times [0,infty):0leq y<f(x)}. $$ The
function $f$ is Lebesgue measurable if $mathcal{U}f$ is Lebesgue measurable
with respect to the planar Lebesgue measure and then we define $$
int_{mathbb{R}} f=m_2(mathcal{U}f). $$
I find this approach quite nice if you want to have a quick introduction to the Lebesgue integration. For example:
You get the monotone convergence theorem for free: it is a
straightforward consequence of the fact that the measure of the union
of an increasing sequence of sets is the limit of measures.
As pointed out by Nik Weaver, the equality $int(f+g)=int f+int g$ is not obvious, but it can be proved quickly with the following trick:
$$
T_f:(x,y)mapsto (x,f(x)+y)
$$
maps the set $mathcal{U}g$ to a set disjoint from $mathcal{U}f$,
$$
mathcal{U}(f+g)=mathcal{U}f sqcup T_f(mathcal{Ug})
$$
and then
$$ int_{mathbb{R}} f+g= int_{mathbb{R}} f +int_{mathbb{R}} g $$
follows immediately once you prove that the sets $mathcal{U}(g)$ and
$T_f(mathcal{U}g)$ have the same measure. Pugh proves it on one page.
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
$endgroup$
5
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
1
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
2
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
1
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
|
show 6 more comments
$begingroup$
Actually, in the following book the Lebesgue integral is defined the way you suggested:
Pugh, C. C. Real mathematical analysis.
Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.
First we define the planar Lebesgue measure $m_2$. Then we define the Lebesgue integral as follows:
Definition. The undergraph of $f:mathbb{R}to[0,infty)$ is $$ mathcal{U}f={(x,y)inmathbb{R}times [0,infty):0leq y<f(x)}. $$ The
function $f$ is Lebesgue measurable if $mathcal{U}f$ is Lebesgue measurable
with respect to the planar Lebesgue measure and then we define $$
int_{mathbb{R}} f=m_2(mathcal{U}f). $$
I find this approach quite nice if you want to have a quick introduction to the Lebesgue integration. For example:
You get the monotone convergence theorem for free: it is a
straightforward consequence of the fact that the measure of the union
of an increasing sequence of sets is the limit of measures.
As pointed out by Nik Weaver, the equality $int(f+g)=int f+int g$ is not obvious, but it can be proved quickly with the following trick:
$$
T_f:(x,y)mapsto (x,f(x)+y)
$$
maps the set $mathcal{U}g$ to a set disjoint from $mathcal{U}f$,
$$
mathcal{U}(f+g)=mathcal{U}f sqcup T_f(mathcal{Ug})
$$
and then
$$ int_{mathbb{R}} f+g= int_{mathbb{R}} f +int_{mathbb{R}} g $$
follows immediately once you prove that the sets $mathcal{U}(g)$ and
$T_f(mathcal{U}g)$ have the same measure. Pugh proves it on one page.
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
$endgroup$
Actually, in the following book the Lebesgue integral is defined the way you suggested:
Pugh, C. C. Real mathematical analysis.
Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.
First we define the planar Lebesgue measure $m_2$. Then we define the Lebesgue integral as follows:
Definition. The undergraph of $f:mathbb{R}to[0,infty)$ is $$ mathcal{U}f={(x,y)inmathbb{R}times [0,infty):0leq y<f(x)}. $$ The
function $f$ is Lebesgue measurable if $mathcal{U}f$ is Lebesgue measurable
with respect to the planar Lebesgue measure and then we define $$
int_{mathbb{R}} f=m_2(mathcal{U}f). $$
I find this approach quite nice if you want to have a quick introduction to the Lebesgue integration. For example:
You get the monotone convergence theorem for free: it is a
straightforward consequence of the fact that the measure of the union
of an increasing sequence of sets is the limit of measures.
As pointed out by Nik Weaver, the equality $int(f+g)=int f+int g$ is not obvious, but it can be proved quickly with the following trick:
$$
T_f:(x,y)mapsto (x,f(x)+y)
$$
maps the set $mathcal{U}g$ to a set disjoint from $mathcal{U}f$,
$$
mathcal{U}(f+g)=mathcal{U}f sqcup T_f(mathcal{Ug})
$$
and then
$$ int_{mathbb{R}} f+g= int_{mathbb{R}} f +int_{mathbb{R}} g $$
follows immediately once you prove that the sets $mathcal{U}(g)$ and
$T_f(mathcal{U}g)$ have the same measure. Pugh proves it on one page.
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
edited Mar 8 at 2:04
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
answered Jan 28 at 20:00
Piotr HajlaszPiotr Hajlasz
10.1k43975
10.1k43975
5
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
1
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
2
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
1
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
|
show 6 more comments
5
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
1
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
2
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
1
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
5
5
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
Jan 28 at 20:04
1
1
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
$begingroup$
@user57888: a fair chunk of the relevant material is available online here. It skips the page where he defines the integral but you can get a good idea of how it plays out on pages 407+.
$endgroup$
– Nik Weaver
Jan 28 at 20:14
2
2
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
$begingroup$
Well, on that last point the issue is that you have to prove that $mathcal{U}g$ and $T_f(mathcal{U}g)$ have the same measure ...
$endgroup$
– Nik Weaver
Jan 28 at 20:50
1
1
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
$begingroup$
@Basj Yes, it takes sometime to develop measure. From the distance, it might seem that developing integral from that point is immediate. From my very recent teaching experience, when I taught this stuff for the first time: it's really not. You have to add a whole new layer of the theory, prove some facts about measurable functions, and only then you are allowed to integrate. My question was really whether this process cannot be streamlined significantly..
$endgroup$
– user57888
Jan 29 at 11:41
|
show 6 more comments
$begingroup$
If $f: mathbb{R} to [0,infty)$ is Borel (or Lebesgue) measurable, then for each rational $a > 0$ define $X_a = f^{-1}([a,infty)) times [0,a)$. Then each $X_a$ is measurable and their union is exactly the region under the graph. So the region under the graph is measurable.
I think the reason why we develop the Lebesgue integral in the usual way is because it provides a powerful technique (characteristic functions --> simple functions --> arbitrary measurable functions) for deriving the basic theory of the integral. Even simple things like $int f + int g = int (f + g)$ aren't obvious if you take "measure of the region under the graph" as the definition.
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
$endgroup$
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
3
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
1
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
1
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
|
show 2 more comments
$begingroup$
If $f: mathbb{R} to [0,infty)$ is Borel (or Lebesgue) measurable, then for each rational $a > 0$ define $X_a = f^{-1}([a,infty)) times [0,a)$. Then each $X_a$ is measurable and their union is exactly the region under the graph. So the region under the graph is measurable.
I think the reason why we develop the Lebesgue integral in the usual way is because it provides a powerful technique (characteristic functions --> simple functions --> arbitrary measurable functions) for deriving the basic theory of the integral. Even simple things like $int f + int g = int (f + g)$ aren't obvious if you take "measure of the region under the graph" as the definition.
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
$endgroup$
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
3
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
1
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
1
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
|
show 2 more comments
$begingroup$
If $f: mathbb{R} to [0,infty)$ is Borel (or Lebesgue) measurable, then for each rational $a > 0$ define $X_a = f^{-1}([a,infty)) times [0,a)$. Then each $X_a$ is measurable and their union is exactly the region under the graph. So the region under the graph is measurable.
I think the reason why we develop the Lebesgue integral in the usual way is because it provides a powerful technique (characteristic functions --> simple functions --> arbitrary measurable functions) for deriving the basic theory of the integral. Even simple things like $int f + int g = int (f + g)$ aren't obvious if you take "measure of the region under the graph" as the definition.
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
$endgroup$
If $f: mathbb{R} to [0,infty)$ is Borel (or Lebesgue) measurable, then for each rational $a > 0$ define $X_a = f^{-1}([a,infty)) times [0,a)$. Then each $X_a$ is measurable and their union is exactly the region under the graph. So the region under the graph is measurable.
I think the reason why we develop the Lebesgue integral in the usual way is because it provides a powerful technique (characteristic functions --> simple functions --> arbitrary measurable functions) for deriving the basic theory of the integral. Even simple things like $int f + int g = int (f + g)$ aren't obvious if you take "measure of the region under the graph" as the definition.
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
answered Jan 28 at 19:37
Nik WeaverNik Weaver
22k150131
22k150131
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
3
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
1
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
1
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
|
show 2 more comments
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
3
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
1
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
1
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
Jan 28 at 19:41
3
3
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
Jan 28 at 19:41
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
Jan 28 at 19:42
1
1
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
$begingroup$
On the other hand, additivity of integral isn't obvious under the standard definition as well, since you first have to prove that a sum of two measurable functions is measurable and I think that the additivity for areas under the graph could involve a similar argument.
$endgroup$
– user57888
Jan 28 at 21:08
1
1
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
$begingroup$
Well, I guess you have to prove that in either case (not that it's difficult). Additivity follows from additivity for simple functions --- an easy calculation --- plus the MCT.
$endgroup$
– Nik Weaver
Jan 28 at 21:50
|
show 2 more comments
$begingroup$
The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text Measure and Integral. An Introduction to Real Analysis, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).
(second sentence of Chapter 5, on p. 64) The approach we have chosen [for the integral of a nonnegative function $f:E rightarrow [0, +infty],$ where $E subseteq {mathbb R}^n$ is measurable] is based on the notion that the integral of a nonnegative $f$ should represent the volume of the region under the graph of $f.$
I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
$endgroup$
add a comment |
$begingroup$
The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text Measure and Integral. An Introduction to Real Analysis, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).
(second sentence of Chapter 5, on p. 64) The approach we have chosen [for the integral of a nonnegative function $f:E rightarrow [0, +infty],$ where $E subseteq {mathbb R}^n$ is measurable] is based on the notion that the integral of a nonnegative $f$ should represent the volume of the region under the graph of $f.$
I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
$endgroup$
add a comment |
$begingroup$
The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text Measure and Integral. An Introduction to Real Analysis, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).
(second sentence of Chapter 5, on p. 64) The approach we have chosen [for the integral of a nonnegative function $f:E rightarrow [0, +infty],$ where $E subseteq {mathbb R}^n$ is measurable] is based on the notion that the integral of a nonnegative $f$ should represent the volume of the region under the graph of $f.$
I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
$endgroup$
The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text Measure and Integral. An Introduction to Real Analysis, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).
(second sentence of Chapter 5, on p. 64) The approach we have chosen [for the integral of a nonnegative function $f:E rightarrow [0, +infty],$ where $E subseteq {mathbb R}^n$ is measurable] is based on the notion that the integral of a nonnegative $f$ should represent the volume of the region under the graph of $f.$
I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
answered Jan 29 at 13:44
Dave L RenfroDave L Renfro
1,7421511
1,7421511
add a comment |
add a comment |
$begingroup$
Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space $(X,mu)$. 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.
Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of $mathbb{R}$. But for finding the product measure of the region under the graph of a function $f : X to mathbb{R}$ via a Fubini-like approach, the only subsets of $mathbb{R}$ you really need to measure are intervals, of the form $[0,f(x)]$. We don't need any fancy measure theory to tell us the length of such sets.
Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on $mathbb{R}$ anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
$endgroup$
1
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
add a comment |
$begingroup$
Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space $(X,mu)$. 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.
Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of $mathbb{R}$. But for finding the product measure of the region under the graph of a function $f : X to mathbb{R}$ via a Fubini-like approach, the only subsets of $mathbb{R}$ you really need to measure are intervals, of the form $[0,f(x)]$. We don't need any fancy measure theory to tell us the length of such sets.
Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on $mathbb{R}$ anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
$endgroup$
1
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
add a comment |
$begingroup$
Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space $(X,mu)$. 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.
Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of $mathbb{R}$. But for finding the product measure of the region under the graph of a function $f : X to mathbb{R}$ via a Fubini-like approach, the only subsets of $mathbb{R}$ you really need to measure are intervals, of the form $[0,f(x)]$. We don't need any fancy measure theory to tell us the length of such sets.
Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on $mathbb{R}$ anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
$endgroup$
Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space $(X,mu)$. 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.
Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of $mathbb{R}$. But for finding the product measure of the region under the graph of a function $f : X to mathbb{R}$ via a Fubini-like approach, the only subsets of $mathbb{R}$ you really need to measure are intervals, of the form $[0,f(x)]$. We don't need any fancy measure theory to tell us the length of such sets.
Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on $mathbb{R}$ anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
answered Jan 30 at 5:21
Nate EldredgeNate Eldredge
20.2k371117
20.2k371117
1
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
add a comment |
1
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
1
1
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
$begingroup$
That's a fair point. Both as a student and as a teacher, I first had full presentation of Lebesgue measure, and only then of Lebesgue integral, so I missed that point. Do you know whether the route you sketched is actually followed somewhere or is it just a hypothetical scenario? I believe that probabilists might want it this way.
$endgroup$
– user57888
Jan 30 at 8:50
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f321916%2fwhy-isnt-integral-defined-as-the-area-under-the-graph-of-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
10
$begingroup$
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
$endgroup$
– Arturo Magidin
Jan 28 at 19:22
20
$begingroup$
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?
$endgroup$
– Nik Weaver
Jan 28 at 19:29
2
$begingroup$
@ArturoMagidin You show that your new integral is the same as Riemann integral whenever applicable, the same way you do it for Lebesgue integral.
$endgroup$
– user57888
Jan 28 at 19:37
5
$begingroup$
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.
$endgroup$
– Qfwfq
Jan 28 at 20:10
10
$begingroup$
The votes to close should be perhaps reconsidered. This is a serious question that has now received several serious answers.
$endgroup$
– Todd Trimble♦
Jan 29 at 22:01