Why isn't integral defined as the area under the graph of function?












4












$begingroup$


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.



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.






real-analysis measure-theory mathematics-education







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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
    48 mins ago






  • 4




    $begingroup$
    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
    41 mins ago










  • $begingroup$
    @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.
    $endgroup$
    – user57888
    40 mins ago












  • $begingroup$
    @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?
    $endgroup$
    – Arturo Magidin
    39 mins ago










  • $begingroup$
    @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
    32 mins ago


















4












$begingroup$


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.



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.






real-analysis measure-theory mathematics-education







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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
    48 mins ago






  • 4




    $begingroup$
    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
    41 mins ago










  • $begingroup$
    @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.
    $endgroup$
    – user57888
    40 mins ago












  • $begingroup$
    @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?
    $endgroup$
    – Arturo Magidin
    39 mins ago










  • $begingroup$
    @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
    32 mins ago
















4












4








4





$begingroup$


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.



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.






real-analysis measure-theory mathematics-education







share|cite|improve this question











$endgroup$




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.



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.






real-analysis measure-theory mathematics-education




real-analysis measure-theory mathematics-education






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 9 mins ago









Community♦

123




123










asked 1 hour ago









user57888user57888

30827




30827








  • 1




    $begingroup$
    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
    48 mins ago






  • 4




    $begingroup$
    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
    41 mins ago










  • $begingroup$
    @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.
    $endgroup$
    – user57888
    40 mins ago












  • $begingroup$
    @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?
    $endgroup$
    – Arturo Magidin
    39 mins ago










  • $begingroup$
    @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
    32 mins ago
















  • 1




    $begingroup$
    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
    48 mins ago






  • 4




    $begingroup$
    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
    41 mins ago










  • $begingroup$
    @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.
    $endgroup$
    – user57888
    40 mins ago












  • $begingroup$
    @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?
    $endgroup$
    – Arturo Magidin
    39 mins ago










  • $begingroup$
    @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
    32 mins ago










1




1




$begingroup$
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
48 mins ago




$begingroup$
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
48 mins ago




4




4




$begingroup$
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
41 mins ago




$begingroup$
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
41 mins ago












$begingroup$
@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.
$endgroup$
– user57888
40 mins ago






$begingroup$
@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.
$endgroup$
– user57888
40 mins ago














$begingroup$
@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?
$endgroup$
– Arturo Magidin
39 mins ago




$begingroup$
@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?
$endgroup$
– Arturo Magidin
39 mins ago












$begingroup$
@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
32 mins ago






$begingroup$
@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
32 mins ago












2 Answers
2






active

oldest

votes


















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There are uncountably many of $X_a$, so I'm not sure i follow your argument.
    $endgroup$
    – user57888
    29 mins ago










  • $begingroup$
    I said rational $a > 0$.
    $endgroup$
    – Nik Weaver
    28 mins ago










  • $begingroup$
    Right, right, right. Ok, sorry for my mistake.
    $endgroup$
    – user57888
    27 mins ago










  • $begingroup$
    Don't worry about it!
    $endgroup$
    – Nik Weaver
    26 mins ago



















4












$begingroup$

Actually, in the book:



Pugh, C. C. Real mathematical analysis. Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.



The Lebesgue integral is defined the way you suggested. 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 intoroduction 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.






share|cite











$endgroup$













  • $begingroup$
    Wow! I'm going to have to look that up.
    $endgroup$
    – Nik Weaver
    6 mins ago










  • $begingroup$
    How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
    $endgroup$
    – user57888
    5 mins ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There are uncountably many of $X_a$, so I'm not sure i follow your argument.
    $endgroup$
    – user57888
    29 mins ago










  • $begingroup$
    I said rational $a > 0$.
    $endgroup$
    – Nik Weaver
    28 mins ago










  • $begingroup$
    Right, right, right. Ok, sorry for my mistake.
    $endgroup$
    – user57888
    27 mins ago










  • $begingroup$
    Don't worry about it!
    $endgroup$
    – Nik Weaver
    26 mins ago
















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    There are uncountably many of $X_a$, so I'm not sure i follow your argument.
    $endgroup$
    – user57888
    29 mins ago










  • $begingroup$
    I said rational $a > 0$.
    $endgroup$
    – Nik Weaver
    28 mins ago










  • $begingroup$
    Right, right, right. Ok, sorry for my mistake.
    $endgroup$
    – user57888
    27 mins ago










  • $begingroup$
    Don't worry about it!
    $endgroup$
    – Nik Weaver
    26 mins ago














6












6








6





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 33 mins ago









Nik WeaverNik Weaver

20.6k147127




20.6k147127












  • $begingroup$
    There are uncountably many of $X_a$, so I'm not sure i follow your argument.
    $endgroup$
    – user57888
    29 mins ago










  • $begingroup$
    I said rational $a > 0$.
    $endgroup$
    – Nik Weaver
    28 mins ago










  • $begingroup$
    Right, right, right. Ok, sorry for my mistake.
    $endgroup$
    – user57888
    27 mins ago










  • $begingroup$
    Don't worry about it!
    $endgroup$
    – Nik Weaver
    26 mins ago


















  • $begingroup$
    There are uncountably many of $X_a$, so I'm not sure i follow your argument.
    $endgroup$
    – user57888
    29 mins ago










  • $begingroup$
    I said rational $a > 0$.
    $endgroup$
    – Nik Weaver
    28 mins ago










  • $begingroup$
    Right, right, right. Ok, sorry for my mistake.
    $endgroup$
    – user57888
    27 mins ago










  • $begingroup$
    Don't worry about it!
    $endgroup$
    – Nik Weaver
    26 mins ago
















$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
29 mins ago




$begingroup$
There are uncountably many of $X_a$, so I'm not sure i follow your argument.
$endgroup$
– user57888
29 mins ago












$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
28 mins ago




$begingroup$
I said rational $a > 0$.
$endgroup$
– Nik Weaver
28 mins ago












$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
27 mins ago




$begingroup$
Right, right, right. Ok, sorry for my mistake.
$endgroup$
– user57888
27 mins ago












$begingroup$
Don't worry about it!
$endgroup$
– Nik Weaver
26 mins ago




$begingroup$
Don't worry about it!
$endgroup$
– Nik Weaver
26 mins ago











4












$begingroup$

Actually, in the book:



Pugh, C. C. Real mathematical analysis. Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.



The Lebesgue integral is defined the way you suggested. 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 intoroduction 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.






share|cite











$endgroup$













  • $begingroup$
    Wow! I'm going to have to look that up.
    $endgroup$
    – Nik Weaver
    6 mins ago










  • $begingroup$
    How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
    $endgroup$
    – user57888
    5 mins ago
















4












$begingroup$

Actually, in the book:



Pugh, C. C. Real mathematical analysis. Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.



The Lebesgue integral is defined the way you suggested. 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 intoroduction 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.






share|cite











$endgroup$













  • $begingroup$
    Wow! I'm going to have to look that up.
    $endgroup$
    – Nik Weaver
    6 mins ago










  • $begingroup$
    How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
    $endgroup$
    – user57888
    5 mins ago














4












4








4





$begingroup$

Actually, in the book:



Pugh, C. C. Real mathematical analysis. Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.



The Lebesgue integral is defined the way you suggested. 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 intoroduction 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.






share|cite











$endgroup$



Actually, in the book:



Pugh, C. C. Real mathematical analysis. Second edition. Undergraduate Texts in Mathematics. Springer, Cham, 2015.



The Lebesgue integral is defined the way you suggested. 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 intoroduction 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.







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edited 3 mins ago

























answered 9 mins ago









Piotr HajlaszPiotr Hajlasz

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  • $begingroup$
    Wow! I'm going to have to look that up.
    $endgroup$
    – Nik Weaver
    6 mins ago










  • $begingroup$
    How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
    $endgroup$
    – user57888
    5 mins ago


















  • $begingroup$
    Wow! I'm going to have to look that up.
    $endgroup$
    – Nik Weaver
    6 mins ago










  • $begingroup$
    How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
    $endgroup$
    – user57888
    5 mins ago
















$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
6 mins ago




$begingroup$
Wow! I'm going to have to look that up.
$endgroup$
– Nik Weaver
6 mins ago












$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
5 mins ago




$begingroup$
How do you find it in practice (if you know the book)? Does the author struggle with some added difficulties?
$endgroup$
– user57888
5 mins ago


















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