A double integral being finite implies the corresponding Borel-measure is $= 0$ on singletons
$begingroup$
Let $mu$ be a positive Borel measure on $mathbb R^d$ with $0 < mu(mathbb R^d) < infty$. Let $I_s(mu) < infty$, where $I_s(mu)$ is defined by
$$I_s(mu) := int_{mathbb R^d} int_{mathbb R^d} frac{d mu(y)}{|x - y|^s} d mu(x)$$
I now want to show that $mu({x}) = 0$ for all $x in mathbb R^d$.
I started by assuming for the sake of contradication that there is an $x_0 in mathbb R^d$ with $mu({x_0}) =: c > 0$. Then we have $infty geq mu(A) geq c$ for all Borel sets $A subseteq mathbb R^d$ with $x in A$. I presume I can use this to evaluate first the inner and then the outer integral and arrive at the conclusion that $I_s(mu) = infty$, but I'm a little stuck on the details. I don't really know any explicite values of $mu$, so how do I calculate/solve the integrals here?
integration analysis measure-theory borel-measures
asked Dec 5 '18 at 22:51
moranmoran
1,392717
$endgroup$
add a comment |
$begingroup$
Let $mu$ be a positive Borel measure on $mathbb R^d$ with $0 < mu(mathbb R^d) < infty$. Let $I_s(mu) < infty$, where $I_s(mu)$ is defined by
$$I_s(mu) := int_{mathbb R^d} int_{mathbb R^d} frac{d mu(y)}{|x - y|^s} d mu(x)$$
I now want to show that $mu({x}) = 0$ for all $x in mathbb R^d$.
I started by assuming for the sake of contradication that there is an $x_0 in mathbb R^d$ with $mu({x_0}) =: c > 0$. Then we have $infty geq mu(A) geq c$ for all Borel sets $A subseteq mathbb R^d$ with $x in A$. I presume I can use this to evaluate first the inner and then the outer integral and arrive at the conclusion that $I_s(mu) = infty$, but I'm a little stuck on the details. I don't really know any explicite values of $mu$, so how do I calculate/solve the integrals here?
integration analysis measure-theory borel-measures
asked Dec 5 '18 at 22:51
moranmoran
1,392717
$endgroup$
1
$begingroup$
If $mu$ assigns positive measure to some ${ x_0 }$ then $I_s(mu) geq mu({ x_0})^2 f(x_0,x_0)$ where $f(x,y)=|x-y|^{-s}$ off the diagonal and $+infty$ on the diagonal.
$endgroup$
– Ian
Dec 5 '18 at 22:57
$begingroup$
@Ian Thank you, but could you maybe explain how we derive that inequality? If I understand correctly then the $mu({x_0})^2$ on the RHS comes from because we can write $mu(mathbb R^d) geq mu({x_0})$, but why can we set $f(x, x) = + infty$ for this function $f$ when the integral expression only contains the $|x - y|^{-s}$-part which is undefined for $x = y$?
$endgroup$
– moran
Dec 5 '18 at 23:32
1
$begingroup$
The RHS is the integral of $f$ over ${ x_0 } times { x_0 }$, and $f$ is a positive function. And without extending $f$ to the diagonal in some manner, this question doesn't make any sense.
$endgroup$
– Ian
Dec 5 '18 at 23:36
add a comment |
$begingroup$
Let $mu$ be a positive Borel measure on $mathbb R^d$ with $0 < mu(mathbb R^d) < infty$. Let $I_s(mu) < infty$, where $I_s(mu)$ is defined by
$$I_s(mu) := int_{mathbb R^d} int_{mathbb R^d} frac{d mu(y)}{|x - y|^s} d mu(x)$$
I now want to show that $mu({x}) = 0$ for all $x in mathbb R^d$.
I started by assuming for the sake of contradication that there is an $x_0 in mathbb R^d$ with $mu({x_0}) =: c > 0$. Then we have $infty geq mu(A) geq c$ for all Borel sets $A subseteq mathbb R^d$ with $x in A$. I presume I can use this to evaluate first the inner and then the outer integral and arrive at the conclusion that $I_s(mu) = infty$, but I'm a little stuck on the details. I don't really know any explicite values of $mu$, so how do I calculate/solve the integrals here?
integration analysis measure-theory borel-measures
asked Dec 5 '18 at 22:51
moranmoran
1,392717
$endgroup$
Let $mu$ be a positive Borel measure on $mathbb R^d$ with $0 < mu(mathbb R^d) < infty$. Let $I_s(mu) < infty$, where $I_s(mu)$ is defined by
$$I_s(mu) := int_{mathbb R^d} int_{mathbb R^d} frac{d mu(y)}{|x - y|^s} d mu(x)$$
I now want to show that $mu({x}) = 0$ for all $x in mathbb R^d$.
I started by assuming for the sake of contradication that there is an $x_0 in mathbb R^d$ with $mu({x_0}) =: c > 0$. Then we have $infty geq mu(A) geq c$ for all Borel sets $A subseteq mathbb R^d$ with $x in A$. I presume I can use this to evaluate first the inner and then the outer integral and arrive at the conclusion that $I_s(mu) = infty$, but I'm a little stuck on the details. I don't really know any explicite values of $mu$, so how do I calculate/solve the integrals here?
integration analysis measure-theory borel-measures
integration analysis measure-theory borel-measures
asked Dec 5 '18 at 22:51
moranmoran
1,392717
asked Dec 5 '18 at 22:51
moranmoran
1,392717
asked Dec 5 '18 at 22:51
moranmoran
1,392717
asked Dec 5 '18 at 22:51
moranmoran
1,392717
asked Dec 5 '18 at 22:51
moranmoran
1,392717
1,392717
1
$begingroup$
If $mu$ assigns positive measure to some ${ x_0 }$ then $I_s(mu) geq mu({ x_0})^2 f(x_0,x_0)$ where $f(x,y)=|x-y|^{-s}$ off the diagonal and $+infty$ on the diagonal.
$endgroup$
– Ian
Dec 5 '18 at 22:57
$begingroup$
@Ian Thank you, but could you maybe explain how we derive that inequality? If I understand correctly then the $mu({x_0})^2$ on the RHS comes from because we can write $mu(mathbb R^d) geq mu({x_0})$, but why can we set $f(x, x) = + infty$ for this function $f$ when the integral expression only contains the $|x - y|^{-s}$-part which is undefined for $x = y$?
$endgroup$
– moran
Dec 5 '18 at 23:32
1
$begingroup$
The RHS is the integral of $f$ over ${ x_0 } times { x_0 }$, and $f$ is a positive function. And without extending $f$ to the diagonal in some manner, this question doesn't make any sense.
$endgroup$
– Ian
Dec 5 '18 at 23:36
add a comment |
1
$begingroup$
If $mu$ assigns positive measure to some ${ x_0 }$ then $I_s(mu) geq mu({ x_0})^2 f(x_0,x_0)$ where $f(x,y)=|x-y|^{-s}$ off the diagonal and $+infty$ on the diagonal.
$endgroup$
– Ian
Dec 5 '18 at 22:57
$begingroup$
@Ian Thank you, but could you maybe explain how we derive that inequality? If I understand correctly then the $mu({x_0})^2$ on the RHS comes from because we can write $mu(mathbb R^d) geq mu({x_0})$, but why can we set $f(x, x) = + infty$ for this function $f$ when the integral expression only contains the $|x - y|^{-s}$-part which is undefined for $x = y$?
$endgroup$
– moran
Dec 5 '18 at 23:32
1
$begingroup$
The RHS is the integral of $f$ over ${ x_0 } times { x_0 }$, and $f$ is a positive function. And without extending $f$ to the diagonal in some manner, this question doesn't make any sense.
$endgroup$
– Ian
Dec 5 '18 at 23:36
1
1
$begingroup$
If $mu$ assigns positive measure to some ${ x_0 }$ then $I_s(mu) geq mu({ x_0})^2 f(x_0,x_0)$ where $f(x,y)=|x-y|^{-s}$ off the diagonal and $+infty$ on the diagonal.
$endgroup$
– Ian
Dec 5 '18 at 22:57
$begingroup$
If $mu$ assigns positive measure to some ${ x_0 }$ then $I_s(mu) geq mu({ x_0})^2 f(x_0,x_0)$ where $f(x,y)=|x-y|^{-s}$ off the diagonal and $+infty$ on the diagonal.
$endgroup$
– Ian
Dec 5 '18 at 22:57
$begingroup$
@Ian Thank you, but could you maybe explain how we derive that inequality? If I understand correctly then the $mu({x_0})^2$ on the RHS comes from because we can write $mu(mathbb R^d) geq mu({x_0})$, but why can we set $f(x, x) = + infty$ for this function $f$ when the integral expression only contains the $|x - y|^{-s}$-part which is undefined for $x = y$?
$endgroup$
– moran
Dec 5 '18 at 23:32
$begingroup$
@Ian Thank you, but could you maybe explain how we derive that inequality? If I understand correctly then the $mu({x_0})^2$ on the RHS comes from because we can write $mu(mathbb R^d) geq mu({x_0})$, but why can we set $f(x, x) = + infty$ for this function $f$ when the integral expression only contains the $|x - y|^{-s}$-part which is undefined for $x = y$?
$endgroup$
– moran
Dec 5 '18 at 23:32
1
1
$begingroup$
The RHS is the integral of $f$ over ${ x_0 } times { x_0 }$, and $f$ is a positive function. And without extending $f$ to the diagonal in some manner, this question doesn't make any sense.
$endgroup$
– Ian
Dec 5 '18 at 23:36
$begingroup$
The RHS is the integral of $f$ over ${ x_0 } times { x_0 }$, and $f$ is a positive function. And without extending $f$ to the diagonal in some manner, this question doesn't make any sense.
$endgroup$
– Ian
Dec 5 '18 at 23:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for almost every $y$ by Fubini's Theorem. [$f$ is the integrand in the double integral]. Suppose $mu (E)=0$ and $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for $y notin E$. Celarly $mu {y}=0$ if $y in E$. If $y in E^{c}$ I leave it to you to show (by considering the behavior of $f$ near $y$) that we must have $mu {y}=0$. [Basically I have reduced the double integral to a single integral. I hope this makes it easy for you].
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
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%2f3027771%2fa-double-integral-being-finite-implies-the-corresponding-borel-measure-is-0%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: $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for almost every $y$ by Fubini's Theorem. [$f$ is the integrand in the double integral]. Suppose $mu (E)=0$ and $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for $y notin E$. Celarly $mu {y}=0$ if $y in E$. If $y in E^{c}$ I leave it to you to show (by considering the behavior of $f$ near $y$) that we must have $mu {y}=0$. [Basically I have reduced the double integral to a single integral. I hope this makes it easy for you].
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
add a comment |
$begingroup$
Hint: $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for almost every $y$ by Fubini's Theorem. [$f$ is the integrand in the double integral]. Suppose $mu (E)=0$ and $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for $y notin E$. Celarly $mu {y}=0$ if $y in E$. If $y in E^{c}$ I leave it to you to show (by considering the behavior of $f$ near $y$) that we must have $mu {y}=0$. [Basically I have reduced the double integral to a single integral. I hope this makes it easy for you].
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
add a comment |
$begingroup$
Hint: $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for almost every $y$ by Fubini's Theorem. [$f$ is the integrand in the double integral]. Suppose $mu (E)=0$ and $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for $y notin E$. Celarly $mu {y}=0$ if $y in E$. If $y in E^{c}$ I leave it to you to show (by considering the behavior of $f$ near $y$) that we must have $mu {y}=0$. [Basically I have reduced the double integral to a single integral. I hope this makes it easy for you].
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
$endgroup$
Hint: $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for almost every $y$ by Fubini's Theorem. [$f$ is the integrand in the double integral]. Suppose $mu (E)=0$ and $int_{mathbb R^{d}} f(x,y),dmu (x)<infty$ for $y notin E$. Celarly $mu {y}=0$ if $y in E$. If $y in E^{c}$ I leave it to you to show (by considering the behavior of $f$ near $y$) that we must have $mu {y}=0$. [Basically I have reduced the double integral to a single integral. I hope this makes it easy for you].
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
answered Dec 5 '18 at 23:45
Kavi Rama MurthyKavi Rama Murthy
55.9k42158
55.9k42158
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%2f3027771%2fa-double-integral-being-finite-implies-the-corresponding-borel-measure-is-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
1
$begingroup$
If $mu$ assigns positive measure to some ${ x_0 }$ then $I_s(mu) geq mu({ x_0})^2 f(x_0,x_0)$ where $f(x,y)=|x-y|^{-s}$ off the diagonal and $+infty$ on the diagonal.
$endgroup$
– Ian
Dec 5 '18 at 22:57
$begingroup$
@Ian Thank you, but could you maybe explain how we derive that inequality? If I understand correctly then the $mu({x_0})^2$ on the RHS comes from because we can write $mu(mathbb R^d) geq mu({x_0})$, but why can we set $f(x, x) = + infty$ for this function $f$ when the integral expression only contains the $|x - y|^{-s}$-part which is undefined for $x = y$?
$endgroup$
– moran
Dec 5 '18 at 23:32
1
$begingroup$
The RHS is the integral of $f$ over ${ x_0 } times { x_0 }$, and $f$ is a positive function. And without extending $f$ to the diagonal in some manner, this question doesn't make any sense.
$endgroup$
– Ian
Dec 5 '18 at 23:36