Measure Theory Problem: $lim_{ntoinfty} n cdot mu(E_n) = 0$
up vote
1
down vote
favorite
$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.
If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then
$$lim_{ntoinfty} n cdot mu(E_n) = 0$$
I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!
measure-theory
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
add a comment |
up vote
1
down vote
favorite
$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.
If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then
$$lim_{ntoinfty} n cdot mu(E_n) = 0$$
I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!
measure-theory
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49
4
Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53
Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54
To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.
If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then
$$lim_{ntoinfty} n cdot mu(E_n) = 0$$
I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!
measure-theory
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
$(mathbf X,M,mu)$ measure space, $f : mathbf X to [0;infty]$ a measurable function such that Integral $int_X f , dmu < infty$.
If $E_n = {x in mathbf X : n + 1 ge f(x) > n}$ then
$$lim_{ntoinfty} n cdot mu(E_n) = 0$$
I can't figure it out: I thought about the fact that $E_n$ are a partition of $X$, so I could use countable additivity there, but I don't know if $mu$ is a finite measure so I can't show any convergence. Thank you for your help!
measure-theory
measure-theory
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
edited Nov 27 at 12:26
Martin Sleziak
44.6k7115269
44.6k7115269
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
asked Sep 12 '15 at 18:43
Tommaso Guerrini
1567
1567
How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49
4
Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53
Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54
To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24
add a comment |
How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49
4
Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53
Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54
To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24
How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49
How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49
4
4
Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53
Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53
Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54
Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54
To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24
To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24
add a comment |
active
oldest
votes
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',
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%2f1432518%2fmeasure-theory-problem-lim-n-to-infty-n-cdot-mue-n-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1432518%2fmeasure-theory-problem-lim-n-to-infty-n-cdot-mue-n-0%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
How can you rewrite $E_{n}$ using $f^{-1}$ ?
– krirkrirk
Sep 12 '15 at 18:49
4
Start with $infty>int_X f,dmu =sum_{n=0}^infty int_{E_{n}}f,dmuge sum_{n=0}^inftyint_{E_{n}}n,dmu=sum_{n=1}^infty ncdotmu(E_{n})$.
– John Dawkins
Sep 12 '15 at 18:53
Thank you guys! It always seems so trivial when you see the solution! Have a nice day :)
– Tommaso Guerrini
Sep 12 '15 at 18:54
To conceptually connect your comment (that you don't know if $mu$ is a finite measure) to the given solutions: We do know that $$nu(E):=int_E fdmu$$ is a finite measure on $M$. Then since $(E_n)_{n=1}^infty$ are pairwise disjoint, $nu(E_n)to 0$. But $nu(E_n)geqslant nmu(E_n)$.
– bangs
Nov 27 at 13:24