Mixing
What is the measure of $(infty,infty)$
$begingroup$
I'm dealing with a counter-example. Specifically, let $E = cap_{k=1}^{infty}E_k$ and $E_{k+1} subset E_k$. We have
$$m(E) = lim_{n to infty} m(E_n)$$
only if $m(E) < infty$. My book says $E_n = (n, infty)$ is a counter-example when $m(E)$ is not less than $infty$. $E = cap_{n=1}^{infty}E_n$ is empty so it has measure zero. Then what is $lim_{n to infty} m(E_n)$? The way that I think about it is that you need to cover $lim_{n to infty} E_n$ with $(infty, infty)$, and the measure of $(infty, infty)$ is $infty$? Is that correct?
measure-theory lebesgue-measure
asked Jan 29 at 3:39
user1691278user1691278
49939
$endgroup$
add a comment |
$begingroup$
I'm dealing with a counter-example. Specifically, let $E = cap_{k=1}^{infty}E_k$ and $E_{k+1} subset E_k$. We have
$$m(E) = lim_{n to infty} m(E_n)$$
only if $m(E) < infty$. My book says $E_n = (n, infty)$ is a counter-example when $m(E)$ is not less than $infty$. $E = cap_{n=1}^{infty}E_n$ is empty so it has measure zero. Then what is $lim_{n to infty} m(E_n)$? The way that I think about it is that you need to cover $lim_{n to infty} E_n$ with $(infty, infty)$, and the measure of $(infty, infty)$ is $infty$? Is that correct?
measure-theory lebesgue-measure
asked Jan 29 at 3:39
user1691278user1691278
49939
$endgroup$
add a comment |
$begingroup$
I'm dealing with a counter-example. Specifically, let $E = cap_{k=1}^{infty}E_k$ and $E_{k+1} subset E_k$. We have
$$m(E) = lim_{n to infty} m(E_n)$$
only if $m(E) < infty$. My book says $E_n = (n, infty)$ is a counter-example when $m(E)$ is not less than $infty$. $E = cap_{n=1}^{infty}E_n$ is empty so it has measure zero. Then what is $lim_{n to infty} m(E_n)$? The way that I think about it is that you need to cover $lim_{n to infty} E_n$ with $(infty, infty)$, and the measure of $(infty, infty)$ is $infty$? Is that correct?
measure-theory lebesgue-measure
asked Jan 29 at 3:39
user1691278user1691278
49939
$endgroup$
I'm dealing with a counter-example. Specifically, let $E = cap_{k=1}^{infty}E_k$ and $E_{k+1} subset E_k$. We have
$$m(E) = lim_{n to infty} m(E_n)$$
only if $m(E) < infty$. My book says $E_n = (n, infty)$ is a counter-example when $m(E)$ is not less than $infty$. $E = cap_{n=1}^{infty}E_n$ is empty so it has measure zero. Then what is $lim_{n to infty} m(E_n)$? The way that I think about it is that you need to cover $lim_{n to infty} E_n$ with $(infty, infty)$, and the measure of $(infty, infty)$ is $infty$? Is that correct?
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Jan 29 at 3:39
user1691278user1691278
49939
asked Jan 29 at 3:39
user1691278user1691278
49939
asked Jan 29 at 3:39
user1691278user1691278
49939
asked Jan 29 at 3:39
user1691278user1691278
49939
asked Jan 29 at 3:39
user1691278user1691278
49939
49939
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $M>0$ be arbitrary. Then $m(E_n) > m((n, n+M)) = M$. Since this is true for any $n$,
$$lim_{ntoinfty} m(E_n) =infty$$
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
$endgroup$
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
1
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
add a comment |
$begingroup$
$m(E_n)=infty$ for each $n$ so $lim m(E_n)=infty$.
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
In this context, $infty$ is a well-defined mathematical object that comprises a single point at the "end" of the real number line - that is, for every other real number $x$, $x < infty$. The same total ordering properties apply as before, and thus there are no "numbers" strictly between $infty$ and itself. This means $(infty, infty)$ is an empty interval, thus its measure is $0$.
On the other hand, the measure of every interval $(x, infty)$ for finite $x$ is $infty$. So the limit of measures as $x$ goes to $infty$ is $infty$, however the measure of the "final" interval $E = (infty, infty)$ is, as above, $0$, thus a counterexample for the case where the "approximating" intervals do have infinite measure.
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
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%2f3091708%2fwhat-is-the-measure-of-infty-infty%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $M>0$ be arbitrary. Then $m(E_n) > m((n, n+M)) = M$. Since this is true for any $n$,
$$lim_{ntoinfty} m(E_n) =infty$$
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
$endgroup$
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
1
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
add a comment |
$begingroup$
Let $M>0$ be arbitrary. Then $m(E_n) > m((n, n+M)) = M$. Since this is true for any $n$,
$$lim_{ntoinfty} m(E_n) =infty$$
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
$endgroup$
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
1
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
add a comment |
$begingroup$
Let $M>0$ be arbitrary. Then $m(E_n) > m((n, n+M)) = M$. Since this is true for any $n$,
$$lim_{ntoinfty} m(E_n) =infty$$
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
$endgroup$
Let $M>0$ be arbitrary. Then $m(E_n) > m((n, n+M)) = M$. Since this is true for any $n$,
$$lim_{ntoinfty} m(E_n) =infty$$
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
answered Jan 29 at 3:48
David PetersonDavid Peterson
8,87821935
8,87821935
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
1
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
add a comment |
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
1
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
$begingroup$
You get $infty$ as the limit because $M$ is arbitrary and $m(E_n)$ is greater than any number?
$endgroup$
– user1691278
Jan 29 at 4:10
1
1
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
$begingroup$
@user1691278 Unlike Calculus you shouldn't hesitate to use $infty$ in Lebesgue integration.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 5:52
add a comment |
$begingroup$
$m(E_n)=infty$ for each $n$ so $lim m(E_n)=infty$.
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
$m(E_n)=infty$ for each $n$ so $lim m(E_n)=infty$.
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
add a comment |
$begingroup$
$m(E_n)=infty$ for each $n$ so $lim m(E_n)=infty$.
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$m(E_n)=infty$ for each $n$ so $lim m(E_n)=infty$.
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:51
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
add a comment |
add a comment |
$begingroup$
In this context, $infty$ is a well-defined mathematical object that comprises a single point at the "end" of the real number line - that is, for every other real number $x$, $x < infty$. The same total ordering properties apply as before, and thus there are no "numbers" strictly between $infty$ and itself. This means $(infty, infty)$ is an empty interval, thus its measure is $0$.
On the other hand, the measure of every interval $(x, infty)$ for finite $x$ is $infty$. So the limit of measures as $x$ goes to $infty$ is $infty$, however the measure of the "final" interval $E = (infty, infty)$ is, as above, $0$, thus a counterexample for the case where the "approximating" intervals do have infinite measure.
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
add a comment |
$begingroup$
In this context, $infty$ is a well-defined mathematical object that comprises a single point at the "end" of the real number line - that is, for every other real number $x$, $x < infty$. The same total ordering properties apply as before, and thus there are no "numbers" strictly between $infty$ and itself. This means $(infty, infty)$ is an empty interval, thus its measure is $0$.
On the other hand, the measure of every interval $(x, infty)$ for finite $x$ is $infty$. So the limit of measures as $x$ goes to $infty$ is $infty$, however the measure of the "final" interval $E = (infty, infty)$ is, as above, $0$, thus a counterexample for the case where the "approximating" intervals do have infinite measure.
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
add a comment |
$begingroup$
In this context, $infty$ is a well-defined mathematical object that comprises a single point at the "end" of the real number line - that is, for every other real number $x$, $x < infty$. The same total ordering properties apply as before, and thus there are no "numbers" strictly between $infty$ and itself. This means $(infty, infty)$ is an empty interval, thus its measure is $0$.
On the other hand, the measure of every interval $(x, infty)$ for finite $x$ is $infty$. So the limit of measures as $x$ goes to $infty$ is $infty$, however the measure of the "final" interval $E = (infty, infty)$ is, as above, $0$, thus a counterexample for the case where the "approximating" intervals do have infinite measure.
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
$endgroup$
In this context, $infty$ is a well-defined mathematical object that comprises a single point at the "end" of the real number line - that is, for every other real number $x$, $x < infty$. The same total ordering properties apply as before, and thus there are no "numbers" strictly between $infty$ and itself. This means $(infty, infty)$ is an empty interval, thus its measure is $0$.
On the other hand, the measure of every interval $(x, infty)$ for finite $x$ is $infty$. So the limit of measures as $x$ goes to $infty$ is $infty$, however the measure of the "final" interval $E = (infty, infty)$ is, as above, $0$, thus a counterexample for the case where the "approximating" intervals do have infinite measure.
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
answered Jan 29 at 7:17
The_SympathizerThe_Sympathizer
7,8002246
7,8002246
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%2f3091708%2fwhat-is-the-measure-of-infty-infty%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
