Show $f$ : integrable $Leftrightarrow$ $ sum_{n=1}^{infty} muleft( { x in E : |f(x)| ge n } right) < infty...
$begingroup$
Let $f : E to [0, infty]$ measurable, where $E$ is a finite measure space.
Show : $f$ is integrable if and only if
$$
sum_{n=1}^{infty} muleft( { x in E : |f(x)| ge n } right) < infty
$$
Try
($Rightarrow$)
Let $E_n = { x in E : |f(x)| ge n }$.
Since $E_1 supset E_2 supset cdots $, we have
$$
E = (Esetminus E_1) sqcup (E_1 setminus E_2) sqcup cdots
$$
Therefore, $int_E f dmu = int_{Esetminus E_1} f dmu + int_{E_1setminus E_2} f dmu + cdots$
But I cannot proceed from here.
($Leftarrow$)
We have $sum_{n=1}^{infty} muleft( E_n right) = sum_{n=1}^{infty} int chi_{E_n} dmu$,
By Levi's convergence theorem, $sum_{n=1}^{infty} chi_{E_n}$ converges a.e. in $E$.
But I'm stuck at how I can relate this to the integrability of $f$.
real-analysis measure-theory
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
$endgroup$
add a comment |
$begingroup$
Let $f : E to [0, infty]$ measurable, where $E$ is a finite measure space.
Show : $f$ is integrable if and only if
$$
sum_{n=1}^{infty} muleft( { x in E : |f(x)| ge n } right) < infty
$$
Try
($Rightarrow$)
Let $E_n = { x in E : |f(x)| ge n }$.
Since $E_1 supset E_2 supset cdots $, we have
$$
E = (Esetminus E_1) sqcup (E_1 setminus E_2) sqcup cdots
$$
Therefore, $int_E f dmu = int_{Esetminus E_1} f dmu + int_{E_1setminus E_2} f dmu + cdots$
But I cannot proceed from here.
($Leftarrow$)
We have $sum_{n=1}^{infty} muleft( E_n right) = sum_{n=1}^{infty} int chi_{E_n} dmu$,
By Levi's convergence theorem, $sum_{n=1}^{infty} chi_{E_n}$ converges a.e. in $E$.
But I'm stuck at how I can relate this to the integrability of $f$.
real-analysis measure-theory
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
$endgroup$
add a comment |
$begingroup$
Let $f : E to [0, infty]$ measurable, where $E$ is a finite measure space.
Show : $f$ is integrable if and only if
$$
sum_{n=1}^{infty} muleft( { x in E : |f(x)| ge n } right) < infty
$$
Try
($Rightarrow$)
Let $E_n = { x in E : |f(x)| ge n }$.
Since $E_1 supset E_2 supset cdots $, we have
$$
E = (Esetminus E_1) sqcup (E_1 setminus E_2) sqcup cdots
$$
Therefore, $int_E f dmu = int_{Esetminus E_1} f dmu + int_{E_1setminus E_2} f dmu + cdots$
But I cannot proceed from here.
($Leftarrow$)
We have $sum_{n=1}^{infty} muleft( E_n right) = sum_{n=1}^{infty} int chi_{E_n} dmu$,
By Levi's convergence theorem, $sum_{n=1}^{infty} chi_{E_n}$ converges a.e. in $E$.
But I'm stuck at how I can relate this to the integrability of $f$.
real-analysis measure-theory
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
$endgroup$
Let $f : E to [0, infty]$ measurable, where $E$ is a finite measure space.
Show : $f$ is integrable if and only if
$$
sum_{n=1}^{infty} muleft( { x in E : |f(x)| ge n } right) < infty
$$
Try
($Rightarrow$)
Let $E_n = { x in E : |f(x)| ge n }$.
Since $E_1 supset E_2 supset cdots $, we have
$$
E = (Esetminus E_1) sqcup (E_1 setminus E_2) sqcup cdots
$$
Therefore, $int_E f dmu = int_{Esetminus E_1} f dmu + int_{E_1setminus E_2} f dmu + cdots$
But I cannot proceed from here.
($Leftarrow$)
We have $sum_{n=1}^{infty} muleft( E_n right) = sum_{n=1}^{infty} int chi_{E_n} dmu$,
By Levi's convergence theorem, $sum_{n=1}^{infty} chi_{E_n}$ converges a.e. in $E$.
But I'm stuck at how I can relate this to the integrability of $f$.
real-analysis measure-theory
real-analysis measure-theory
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
asked Dec 21 '18 at 2:33
MoreblueMoreblue
9151317
9151317
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Note that $|f| ge sum_{n ge 1} chi_{E_n}$ pointwise.
So if $f$ is integrable, then $$sum_{n ge 1} mu(E_n) = sum_{n ge 1} int chi_{E_n} , dmu = int sum_{n ge 1} chi_{E_n} , dmu le int |f| , dmu < infty$$ where the interchange of integral and sum is due to the monotone convergence theorem.
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
$endgroup$
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%2f3048126%2fshow-f-integrable-leftrightarrow-sum-n-1-infty-mu-left-x-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Note that $|f| ge sum_{n ge 1} chi_{E_n}$ pointwise.
So if $f$ is integrable, then $$sum_{n ge 1} mu(E_n) = sum_{n ge 1} int chi_{E_n} , dmu = int sum_{n ge 1} chi_{E_n} , dmu le int |f| , dmu < infty$$ where the interchange of integral and sum is due to the monotone convergence theorem.
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
$endgroup$
add a comment |
$begingroup$
Hint: Note that $|f| ge sum_{n ge 1} chi_{E_n}$ pointwise.
So if $f$ is integrable, then $$sum_{n ge 1} mu(E_n) = sum_{n ge 1} int chi_{E_n} , dmu = int sum_{n ge 1} chi_{E_n} , dmu le int |f| , dmu < infty$$ where the interchange of integral and sum is due to the monotone convergence theorem.
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
$endgroup$
add a comment |
$begingroup$
Hint: Note that $|f| ge sum_{n ge 1} chi_{E_n}$ pointwise.
So if $f$ is integrable, then $$sum_{n ge 1} mu(E_n) = sum_{n ge 1} int chi_{E_n} , dmu = int sum_{n ge 1} chi_{E_n} , dmu le int |f| , dmu < infty$$ where the interchange of integral and sum is due to the monotone convergence theorem.
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
$endgroup$
Hint: Note that $|f| ge sum_{n ge 1} chi_{E_n}$ pointwise.
So if $f$ is integrable, then $$sum_{n ge 1} mu(E_n) = sum_{n ge 1} int chi_{E_n} , dmu = int sum_{n ge 1} chi_{E_n} , dmu le int |f| , dmu < infty$$ where the interchange of integral and sum is due to the monotone convergence theorem.
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
answered Dec 21 '18 at 2:48
angryavianangryavian
42.6k23481
42.6k23481
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%2f3048126%2fshow-f-integrable-leftrightarrow-sum-n-1-infty-mu-left-x-in%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
