Example where The Lebesgue Integral is Better
$begingroup$
What is an example that involves a fuction on an interval of the real numbers where the Lebesgue integral is better than the Riemann integral.
By better, it probably means that the Lebesgue intregral is defined while the Riemann integral is not.
By an example, I mean an example that has importance in mathematics for reasons other than just showing that the Lebesgue integral is more general than the Riemann integral. If the importance of the example is not clear, then please explain why it is important.
A possible answer could be an example where the use of the dominated convergence theorem plays a role.
integration measure-theory lebesgue-integral riemann-integration
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
$endgroup$
add a comment |
$begingroup$
What is an example that involves a fuction on an interval of the real numbers where the Lebesgue integral is better than the Riemann integral.
By better, it probably means that the Lebesgue intregral is defined while the Riemann integral is not.
By an example, I mean an example that has importance in mathematics for reasons other than just showing that the Lebesgue integral is more general than the Riemann integral. If the importance of the example is not clear, then please explain why it is important.
A possible answer could be an example where the use of the dominated convergence theorem plays a role.
integration measure-theory lebesgue-integral riemann-integration
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
$endgroup$
3
$begingroup$
I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have.
$endgroup$
– Snake707
Jan 8 at 16:02
$begingroup$
You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details.
$endgroup$
– p4sch
Jan 13 at 21:30
add a comment |
$begingroup$
What is an example that involves a fuction on an interval of the real numbers where the Lebesgue integral is better than the Riemann integral.
By better, it probably means that the Lebesgue intregral is defined while the Riemann integral is not.
By an example, I mean an example that has importance in mathematics for reasons other than just showing that the Lebesgue integral is more general than the Riemann integral. If the importance of the example is not clear, then please explain why it is important.
A possible answer could be an example where the use of the dominated convergence theorem plays a role.
integration measure-theory lebesgue-integral riemann-integration
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
$endgroup$
What is an example that involves a fuction on an interval of the real numbers where the Lebesgue integral is better than the Riemann integral.
By better, it probably means that the Lebesgue intregral is defined while the Riemann integral is not.
By an example, I mean an example that has importance in mathematics for reasons other than just showing that the Lebesgue integral is more general than the Riemann integral. If the importance of the example is not clear, then please explain why it is important.
A possible answer could be an example where the use of the dominated convergence theorem plays a role.
integration measure-theory lebesgue-integral riemann-integration
integration measure-theory lebesgue-integral riemann-integration
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
edited Jan 14 at 10:46
José Carlos Santos
156k22126227
156k22126227
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
asked Jan 8 at 15:55
LinearGuyLinearGuy
13711
13711
3
$begingroup$
I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have.
$endgroup$
– Snake707
Jan 8 at 16:02
$begingroup$
You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details.
$endgroup$
– p4sch
Jan 13 at 21:30
add a comment |
3
$begingroup$
I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have.
$endgroup$
– Snake707
Jan 8 at 16:02
$begingroup$
You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details.
$endgroup$
– p4sch
Jan 13 at 21:30
3
3
$begingroup$
I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have.
$endgroup$
– Snake707
Jan 8 at 16:02
$begingroup$
I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have.
$endgroup$
– Snake707
Jan 8 at 16:02
$begingroup$
You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details.
$endgroup$
– p4sch
Jan 13 at 21:30
$begingroup$
You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details.
$endgroup$
– p4sch
Jan 13 at 21:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Take the Dirichlet function (restricted to, say, $[0,1]$), for instance:
- it is Lebesgue-integrable;
- it is not Riemann integrable;
- it was defined by Dirichlet before Lebesgue was even born.
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
In fact the intuition behind Lebesgue integrability is that how likely the surface under a non-negative function is defined and how much is its surface (note that Lebesgue integrability is defined for non-negative functions). For example define the surface of a $1times 1$ solid square (I mean the perimeter and inside the square). If any time we remove vertical lines from the solid square (that is, removing a line from a surface) the intuition says as if "nothing" has been removed from the solid square and the area must be preserved. The intuition is show below:
As you can see, the white lines are step by step removed from the dark shaded solid square, leaving a "reduced" square which is "almost" as of the primary square (by "almost" we mean a set whose Lebesgue measure is equal to that of the primary square i.e. each time a zero-set has been subtracted from the square). Now let a function be defined as $1$ on the domain of the "reduced" square. For example let$$f_n(x)=begin{cases}1&,quad x={1over k} , text{for some }1le kle n\0 &quad text{elsewhere}end{cases}$$whose underneath surface is as follows:
Surely, $f_{infty}(x)$ exists and is Reimann un-integrable in $[0,1]$ but this is unfair under the generalization to Lebesgue integrability since the surface is "almost" the same except some zero-set infinitely many distinct lines counted and subtracted from the surface of the function. Probably, this intuition was the main lead Lebesgue to establish probably his most prominent theory on measure-sets.
An interesting interpretation is that Reimann integral counts the surface beneath a function vertically while Lebesgue integral counts it horizontally.
Hope I could have clarified enough so far why such a generalization exists. You are highly encouraged to check out Lebesgue integration on Wikipedia.
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
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: "69"
};
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%2fmath.stackexchange.com%2fquestions%2f3066343%2fexample-where-the-lebesgue-integral-is-better%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take the Dirichlet function (restricted to, say, $[0,1]$), for instance:
- it is Lebesgue-integrable;
- it is not Riemann integrable;
- it was defined by Dirichlet before Lebesgue was even born.
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
Take the Dirichlet function (restricted to, say, $[0,1]$), for instance:
- it is Lebesgue-integrable;
- it is not Riemann integrable;
- it was defined by Dirichlet before Lebesgue was even born.
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
add a comment |
$begingroup$
Take the Dirichlet function (restricted to, say, $[0,1]$), for instance:
- it is Lebesgue-integrable;
- it is not Riemann integrable;
- it was defined by Dirichlet before Lebesgue was even born.
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
$endgroup$
Take the Dirichlet function (restricted to, say, $[0,1]$), for instance:
- it is Lebesgue-integrable;
- it is not Riemann integrable;
- it was defined by Dirichlet before Lebesgue was even born.
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
answered Jan 8 at 16:01
José Carlos SantosJosé Carlos Santos
156k22126227
156k22126227
add a comment |
add a comment |
$begingroup$
In fact the intuition behind Lebesgue integrability is that how likely the surface under a non-negative function is defined and how much is its surface (note that Lebesgue integrability is defined for non-negative functions). For example define the surface of a $1times 1$ solid square (I mean the perimeter and inside the square). If any time we remove vertical lines from the solid square (that is, removing a line from a surface) the intuition says as if "nothing" has been removed from the solid square and the area must be preserved. The intuition is show below:
As you can see, the white lines are step by step removed from the dark shaded solid square, leaving a "reduced" square which is "almost" as of the primary square (by "almost" we mean a set whose Lebesgue measure is equal to that of the primary square i.e. each time a zero-set has been subtracted from the square). Now let a function be defined as $1$ on the domain of the "reduced" square. For example let$$f_n(x)=begin{cases}1&,quad x={1over k} , text{for some }1le kle n\0 &quad text{elsewhere}end{cases}$$whose underneath surface is as follows:
Surely, $f_{infty}(x)$ exists and is Reimann un-integrable in $[0,1]$ but this is unfair under the generalization to Lebesgue integrability since the surface is "almost" the same except some zero-set infinitely many distinct lines counted and subtracted from the surface of the function. Probably, this intuition was the main lead Lebesgue to establish probably his most prominent theory on measure-sets.
An interesting interpretation is that Reimann integral counts the surface beneath a function vertically while Lebesgue integral counts it horizontally.
Hope I could have clarified enough so far why such a generalization exists. You are highly encouraged to check out Lebesgue integration on Wikipedia.
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
In fact the intuition behind Lebesgue integrability is that how likely the surface under a non-negative function is defined and how much is its surface (note that Lebesgue integrability is defined for non-negative functions). For example define the surface of a $1times 1$ solid square (I mean the perimeter and inside the square). If any time we remove vertical lines from the solid square (that is, removing a line from a surface) the intuition says as if "nothing" has been removed from the solid square and the area must be preserved. The intuition is show below:
As you can see, the white lines are step by step removed from the dark shaded solid square, leaving a "reduced" square which is "almost" as of the primary square (by "almost" we mean a set whose Lebesgue measure is equal to that of the primary square i.e. each time a zero-set has been subtracted from the square). Now let a function be defined as $1$ on the domain of the "reduced" square. For example let$$f_n(x)=begin{cases}1&,quad x={1over k} , text{for some }1le kle n\0 &quad text{elsewhere}end{cases}$$whose underneath surface is as follows:
Surely, $f_{infty}(x)$ exists and is Reimann un-integrable in $[0,1]$ but this is unfair under the generalization to Lebesgue integrability since the surface is "almost" the same except some zero-set infinitely many distinct lines counted and subtracted from the surface of the function. Probably, this intuition was the main lead Lebesgue to establish probably his most prominent theory on measure-sets.
An interesting interpretation is that Reimann integral counts the surface beneath a function vertically while Lebesgue integral counts it horizontally.
Hope I could have clarified enough so far why such a generalization exists. You are highly encouraged to check out Lebesgue integration on Wikipedia.
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
In fact the intuition behind Lebesgue integrability is that how likely the surface under a non-negative function is defined and how much is its surface (note that Lebesgue integrability is defined for non-negative functions). For example define the surface of a $1times 1$ solid square (I mean the perimeter and inside the square). If any time we remove vertical lines from the solid square (that is, removing a line from a surface) the intuition says as if "nothing" has been removed from the solid square and the area must be preserved. The intuition is show below:
As you can see, the white lines are step by step removed from the dark shaded solid square, leaving a "reduced" square which is "almost" as of the primary square (by "almost" we mean a set whose Lebesgue measure is equal to that of the primary square i.e. each time a zero-set has been subtracted from the square). Now let a function be defined as $1$ on the domain of the "reduced" square. For example let$$f_n(x)=begin{cases}1&,quad x={1over k} , text{for some }1le kle n\0 &quad text{elsewhere}end{cases}$$whose underneath surface is as follows:
Surely, $f_{infty}(x)$ exists and is Reimann un-integrable in $[0,1]$ but this is unfair under the generalization to Lebesgue integrability since the surface is "almost" the same except some zero-set infinitely many distinct lines counted and subtracted from the surface of the function. Probably, this intuition was the main lead Lebesgue to establish probably his most prominent theory on measure-sets.
An interesting interpretation is that Reimann integral counts the surface beneath a function vertically while Lebesgue integral counts it horizontally.
Hope I could have clarified enough so far why such a generalization exists. You are highly encouraged to check out Lebesgue integration on Wikipedia.
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
In fact the intuition behind Lebesgue integrability is that how likely the surface under a non-negative function is defined and how much is its surface (note that Lebesgue integrability is defined for non-negative functions). For example define the surface of a $1times 1$ solid square (I mean the perimeter and inside the square). If any time we remove vertical lines from the solid square (that is, removing a line from a surface) the intuition says as if "nothing" has been removed from the solid square and the area must be preserved. The intuition is show below:
As you can see, the white lines are step by step removed from the dark shaded solid square, leaving a "reduced" square which is "almost" as of the primary square (by "almost" we mean a set whose Lebesgue measure is equal to that of the primary square i.e. each time a zero-set has been subtracted from the square). Now let a function be defined as $1$ on the domain of the "reduced" square. For example let$$f_n(x)=begin{cases}1&,quad x={1over k} , text{for some }1le kle n\0 &quad text{elsewhere}end{cases}$$whose underneath surface is as follows:
Surely, $f_{infty}(x)$ exists and is Reimann un-integrable in $[0,1]$ but this is unfair under the generalization to Lebesgue integrability since the surface is "almost" the same except some zero-set infinitely many distinct lines counted and subtracted from the surface of the function. Probably, this intuition was the main lead Lebesgue to establish probably his most prominent theory on measure-sets.
An interesting interpretation is that Reimann integral counts the surface beneath a function vertically while Lebesgue integral counts it horizontally.
Hope I could have clarified enough so far why such a generalization exists. You are highly encouraged to check out Lebesgue integration on Wikipedia.
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
edited Jan 14 at 20:49
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 14 at 20:35
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f3066343%2fexample-where-the-lebesgue-integral-is-better%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
3
$begingroup$
I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have.
$endgroup$
– Snake707
Jan 8 at 16:02
$begingroup$
You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details.
$endgroup$
– p4sch
Jan 13 at 21:30