Mixing
Proving $epsilon$-$delta$ convergence of Lebesgue integrals just from the definition
$begingroup$
I am pretty lost with how to do this.
The problem is to show that, in a measure space $(X,M,mu)$ with a measurable function $f: X to [0,infty)$ such that $int_X f dmu < infty$. Then, for all $epsilon > 0$, there is a $delta > 0$ such that for all $E in M$ if $mu(E) < delta$ then
$$int_E f dmu < epsilon$$
Apparently, I should do this from just the definition, which is
$$int_Efdmu = sup_{0 leq s leq f} int_E s dmu$$
and
$$int_Esdmu = sum_{j=1}^n alpha_j mu(A_j cap E)$$
But I have no idea how to continue
real-analysis general-topology measure-theory lebesgue-integral
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
$endgroup$
add a comment |
$begingroup$
I am pretty lost with how to do this.
The problem is to show that, in a measure space $(X,M,mu)$ with a measurable function $f: X to [0,infty)$ such that $int_X f dmu < infty$. Then, for all $epsilon > 0$, there is a $delta > 0$ such that for all $E in M$ if $mu(E) < delta$ then
$$int_E f dmu < epsilon$$
Apparently, I should do this from just the definition, which is
$$int_Efdmu = sup_{0 leq s leq f} int_E s dmu$$
and
$$int_Esdmu = sum_{j=1}^n alpha_j mu(A_j cap E)$$
But I have no idea how to continue
real-analysis general-topology measure-theory lebesgue-integral
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
$endgroup$
$begingroup$
I forgot to add it! Do you mean $mu(E) < delta$?
$endgroup$
– The Bosco
Feb 1 at 10:10
$begingroup$
Do you have the monotone convergence Theorem at your disposal? I am asking because using your definition it can be shown in approximately one lecture. Here is a wiki link to that theorem: en.wikipedia.org/wiki/…
$endgroup$
– humanStampedist
Feb 1 at 10:36
$begingroup$
The instructions say no limit theorems :/ so I think that the monotone convergence theorem was not allowed
$endgroup$
– The Bosco
Feb 1 at 10:38
add a comment |
$begingroup$
I am pretty lost with how to do this.
The problem is to show that, in a measure space $(X,M,mu)$ with a measurable function $f: X to [0,infty)$ such that $int_X f dmu < infty$. Then, for all $epsilon > 0$, there is a $delta > 0$ such that for all $E in M$ if $mu(E) < delta$ then
$$int_E f dmu < epsilon$$
Apparently, I should do this from just the definition, which is
$$int_Efdmu = sup_{0 leq s leq f} int_E s dmu$$
and
$$int_Esdmu = sum_{j=1}^n alpha_j mu(A_j cap E)$$
But I have no idea how to continue
real-analysis general-topology measure-theory lebesgue-integral
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
$endgroup$
I am pretty lost with how to do this.
The problem is to show that, in a measure space $(X,M,mu)$ with a measurable function $f: X to [0,infty)$ such that $int_X f dmu < infty$. Then, for all $epsilon > 0$, there is a $delta > 0$ such that for all $E in M$ if $mu(E) < delta$ then
$$int_E f dmu < epsilon$$
Apparently, I should do this from just the definition, which is
$$int_Efdmu = sup_{0 leq s leq f} int_E s dmu$$
and
$$int_Esdmu = sum_{j=1}^n alpha_j mu(A_j cap E)$$
But I have no idea how to continue
real-analysis general-topology measure-theory lebesgue-integral
real-analysis general-topology measure-theory lebesgue-integral
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
edited Feb 1 at 14:20
Asaf Karagila♦
308k33441775
308k33441775
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
asked Feb 1 at 10:03
The BoscoThe Bosco
608212
608212
$begingroup$
I forgot to add it! Do you mean $mu(E) < delta$?
$endgroup$
– The Bosco
Feb 1 at 10:10
$begingroup$
Do you have the monotone convergence Theorem at your disposal? I am asking because using your definition it can be shown in approximately one lecture. Here is a wiki link to that theorem: en.wikipedia.org/wiki/…
$endgroup$
– humanStampedist
Feb 1 at 10:36
$begingroup$
The instructions say no limit theorems :/ so I think that the monotone convergence theorem was not allowed
$endgroup$
– The Bosco
Feb 1 at 10:38
add a comment |
$begingroup$
I forgot to add it! Do you mean $mu(E) < delta$?
$endgroup$
– The Bosco
Feb 1 at 10:10
$begingroup$
Do you have the monotone convergence Theorem at your disposal? I am asking because using your definition it can be shown in approximately one lecture. Here is a wiki link to that theorem: en.wikipedia.org/wiki/…
$endgroup$
– humanStampedist
Feb 1 at 10:36
$begingroup$
The instructions say no limit theorems :/ so I think that the monotone convergence theorem was not allowed
$endgroup$
– The Bosco
Feb 1 at 10:38
$begingroup$
I forgot to add it! Do you mean $mu(E) < delta$?
$endgroup$
– The Bosco
Feb 1 at 10:10
$begingroup$
I forgot to add it! Do you mean $mu(E) < delta$?
$endgroup$
– The Bosco
Feb 1 at 10:10
$begingroup$
Do you have the monotone convergence Theorem at your disposal? I am asking because using your definition it can be shown in approximately one lecture. Here is a wiki link to that theorem: en.wikipedia.org/wiki/…
$endgroup$
– humanStampedist
Feb 1 at 10:36
$begingroup$
Do you have the monotone convergence Theorem at your disposal? I am asking because using your definition it can be shown in approximately one lecture. Here is a wiki link to that theorem: en.wikipedia.org/wiki/…
$endgroup$
– humanStampedist
Feb 1 at 10:36
$begingroup$
The instructions say no limit theorems :/ so I think that the monotone convergence theorem was not allowed
$endgroup$
– The Bosco
Feb 1 at 10:38
$begingroup$
The instructions say no limit theorems :/ so I think that the monotone convergence theorem was not allowed
$endgroup$
– The Bosco
Feb 1 at 10:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $int_X f dmu =0$, the result is trivial. So suppose that $int_X f dmu >0$ and take $epsilon >0$. We can find a simple function $0 le s le f$ such that $$s = sum_{i=1}^n alpha_i chi_{A_i}$$ where the $A_i$ are disjoint measurable subsets and
$$0 < int_X f dmu - int_X s dmu = int_X f dmu - sum_{i=1}^n alpha_i mu(A_i) le epsilon/2$$
For $E in M$, you have
$$int_E s dmu = sum_{i=1}^n alpha_i mu(A_i cap E) le mu(E) left(sum_{i=1}^n alpha_i right).$$
Now take $delta=(epsilon/2)/left(sum_{i=1}^n alpha_i right)$ and suppose that $mu(E) le delta$.
You have
$$int_E f dmu = int_E (f-s) dmu + int_E s dmu le epsilon/2 + epsilon/2=epsilon$$
as desired.
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
There exists a simple function $s=sumlimits_{i=1}^n a_i chi_{E_i}$ satisfying $0leq s leq f$ and
$$int_X f dmu - int_X s dmu < frac{epsilon}{2}.$$
For this simple function $s$, it is easy to check that $int_E s dmu leq max{a_i ~|~ i=1,ldots,n }mu(E)$ for all $Ein M$.
Hence, there exists $delta >0$ satisfying
$$int_E s dmu < frac{epsilon}{2}$$
whenever $mu(E) < delta$.
Therefore, for all $Ein M$ satisfying $mu(E)<delta$, the inequality
$$ int_E f dmu = int_E (f-s)dmu + int_E s dmu
< int_X (f-s)dmu + frac{epsilon}{2}
< frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$
holds.
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
$endgroup$
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%2f3096049%2fproving-epsilon-delta-convergence-of-lebesgue-integrals-just-from-the-defi%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
$begingroup$
If $int_X f dmu =0$, the result is trivial. So suppose that $int_X f dmu >0$ and take $epsilon >0$. We can find a simple function $0 le s le f$ such that $$s = sum_{i=1}^n alpha_i chi_{A_i}$$ where the $A_i$ are disjoint measurable subsets and
$$0 < int_X f dmu - int_X s dmu = int_X f dmu - sum_{i=1}^n alpha_i mu(A_i) le epsilon/2$$
For $E in M$, you have
$$int_E s dmu = sum_{i=1}^n alpha_i mu(A_i cap E) le mu(E) left(sum_{i=1}^n alpha_i right).$$
Now take $delta=(epsilon/2)/left(sum_{i=1}^n alpha_i right)$ and suppose that $mu(E) le delta$.
You have
$$int_E f dmu = int_E (f-s) dmu + int_E s dmu le epsilon/2 + epsilon/2=epsilon$$
as desired.
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
If $int_X f dmu =0$, the result is trivial. So suppose that $int_X f dmu >0$ and take $epsilon >0$. We can find a simple function $0 le s le f$ such that $$s = sum_{i=1}^n alpha_i chi_{A_i}$$ where the $A_i$ are disjoint measurable subsets and
$$0 < int_X f dmu - int_X s dmu = int_X f dmu - sum_{i=1}^n alpha_i mu(A_i) le epsilon/2$$
For $E in M$, you have
$$int_E s dmu = sum_{i=1}^n alpha_i mu(A_i cap E) le mu(E) left(sum_{i=1}^n alpha_i right).$$
Now take $delta=(epsilon/2)/left(sum_{i=1}^n alpha_i right)$ and suppose that $mu(E) le delta$.
You have
$$int_E f dmu = int_E (f-s) dmu + int_E s dmu le epsilon/2 + epsilon/2=epsilon$$
as desired.
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
If $int_X f dmu =0$, the result is trivial. So suppose that $int_X f dmu >0$ and take $epsilon >0$. We can find a simple function $0 le s le f$ such that $$s = sum_{i=1}^n alpha_i chi_{A_i}$$ where the $A_i$ are disjoint measurable subsets and
$$0 < int_X f dmu - int_X s dmu = int_X f dmu - sum_{i=1}^n alpha_i mu(A_i) le epsilon/2$$
For $E in M$, you have
$$int_E s dmu = sum_{i=1}^n alpha_i mu(A_i cap E) le mu(E) left(sum_{i=1}^n alpha_i right).$$
Now take $delta=(epsilon/2)/left(sum_{i=1}^n alpha_i right)$ and suppose that $mu(E) le delta$.
You have
$$int_E f dmu = int_E (f-s) dmu + int_E s dmu le epsilon/2 + epsilon/2=epsilon$$
as desired.
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
If $int_X f dmu =0$, the result is trivial. So suppose that $int_X f dmu >0$ and take $epsilon >0$. We can find a simple function $0 le s le f$ such that $$s = sum_{i=1}^n alpha_i chi_{A_i}$$ where the $A_i$ are disjoint measurable subsets and
$$0 < int_X f dmu - int_X s dmu = int_X f dmu - sum_{i=1}^n alpha_i mu(A_i) le epsilon/2$$
For $E in M$, you have
$$int_E s dmu = sum_{i=1}^n alpha_i mu(A_i cap E) le mu(E) left(sum_{i=1}^n alpha_i right).$$
Now take $delta=(epsilon/2)/left(sum_{i=1}^n alpha_i right)$ and suppose that $mu(E) le delta$.
You have
$$int_E f dmu = int_E (f-s) dmu + int_E s dmu le epsilon/2 + epsilon/2=epsilon$$
as desired.
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Feb 1 at 11:27
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Feb 1 at 11:12
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
$begingroup$
There exists a simple function $s=sumlimits_{i=1}^n a_i chi_{E_i}$ satisfying $0leq s leq f$ and
$$int_X f dmu - int_X s dmu < frac{epsilon}{2}.$$
For this simple function $s$, it is easy to check that $int_E s dmu leq max{a_i ~|~ i=1,ldots,n }mu(E)$ for all $Ein M$.
Hence, there exists $delta >0$ satisfying
$$int_E s dmu < frac{epsilon}{2}$$
whenever $mu(E) < delta$.
Therefore, for all $Ein M$ satisfying $mu(E)<delta$, the inequality
$$ int_E f dmu = int_E (f-s)dmu + int_E s dmu
< int_X (f-s)dmu + frac{epsilon}{2}
< frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$
holds.
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
$endgroup$
add a comment |
$begingroup$
There exists a simple function $s=sumlimits_{i=1}^n a_i chi_{E_i}$ satisfying $0leq s leq f$ and
$$int_X f dmu - int_X s dmu < frac{epsilon}{2}.$$
For this simple function $s$, it is easy to check that $int_E s dmu leq max{a_i ~|~ i=1,ldots,n }mu(E)$ for all $Ein M$.
Hence, there exists $delta >0$ satisfying
$$int_E s dmu < frac{epsilon}{2}$$
whenever $mu(E) < delta$.
Therefore, for all $Ein M$ satisfying $mu(E)<delta$, the inequality
$$ int_E f dmu = int_E (f-s)dmu + int_E s dmu
< int_X (f-s)dmu + frac{epsilon}{2}
< frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$
holds.
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
$endgroup$
add a comment |
$begingroup$
There exists a simple function $s=sumlimits_{i=1}^n a_i chi_{E_i}$ satisfying $0leq s leq f$ and
$$int_X f dmu - int_X s dmu < frac{epsilon}{2}.$$
For this simple function $s$, it is easy to check that $int_E s dmu leq max{a_i ~|~ i=1,ldots,n }mu(E)$ for all $Ein M$.
Hence, there exists $delta >0$ satisfying
$$int_E s dmu < frac{epsilon}{2}$$
whenever $mu(E) < delta$.
Therefore, for all $Ein M$ satisfying $mu(E)<delta$, the inequality
$$ int_E f dmu = int_E (f-s)dmu + int_E s dmu
< int_X (f-s)dmu + frac{epsilon}{2}
< frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$
holds.
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
$endgroup$
There exists a simple function $s=sumlimits_{i=1}^n a_i chi_{E_i}$ satisfying $0leq s leq f$ and
$$int_X f dmu - int_X s dmu < frac{epsilon}{2}.$$
For this simple function $s$, it is easy to check that $int_E s dmu leq max{a_i ~|~ i=1,ldots,n }mu(E)$ for all $Ein M$.
Hence, there exists $delta >0$ satisfying
$$int_E s dmu < frac{epsilon}{2}$$
whenever $mu(E) < delta$.
Therefore, for all $Ein M$ satisfying $mu(E)<delta$, the inequality
$$ int_E f dmu = int_E (f-s)dmu + int_E s dmu
< int_X (f-s)dmu + frac{epsilon}{2}
< frac{epsilon}{2} + frac{epsilon}{2} = epsilon $$
holds.
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
answered Feb 1 at 11:05
Doyun NamDoyun Nam
68119
68119
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%2f3096049%2fproving-epsilon-delta-convergence-of-lebesgue-integrals-just-from-the-defi%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
$begingroup$
I forgot to add it! Do you mean $mu(E) < delta$?
$endgroup$
– The Bosco
Feb 1 at 10:10
$begingroup$
Do you have the monotone convergence Theorem at your disposal? I am asking because using your definition it can be shown in approximately one lecture. Here is a wiki link to that theorem: en.wikipedia.org/wiki/…
$endgroup$
– humanStampedist
Feb 1 at 10:36
$begingroup$
The instructions say no limit theorems :/ so I think that the monotone convergence theorem was not allowed
$endgroup$
– The Bosco
Feb 1 at 10:38