Mixing
![Replace the string data from a file [closed]](https://lh3.googleusercontent.com/-k7lLl_7XAtM/AAAAAAAAAAI/AAAAAAAAAkI/UUNm0FKKX3w/s72-c/photo.jpg?sz=32)
$mathrm{ess, sup}(f)=text{sup}(f)$ for a continuous non negative function $f$
$begingroup$
My intuition tells me that the following proposition is true:
Let $f: Omega rightarrow mathbb{R}$ be a measurable function on a measurable set $Omega subseteq mathbb{R}^n$. If $f$ is continuous and non negative then $mathrm{ess, sup}(f)=text{sup}(f)$.
I'm using here the Borel sigma algebras on $Omega$ and $mathbb{R}$ respectively, and the Lebesgue measure $m$ on $Omega$.
My measure theory background is very poor, that being said I tried to give a proof for the case where $Omega$ is compact:
Let $A={a: m({ fgeq a })=0 }$. Therefore $mathrm{ess, sup}(f)= inf A = E$. Let also $M=max f = sup f$ (it exists because $f$ is continuous on a compact set). I will prove $E geq M$, and $E leq M$ (otherwise there would be a contradiction on the defition of $inf A$).
$E leq M$: if $a>M$ then ${ fgeq a }=emptyset$, so $m({ fgeq a })=0$, so $a in A$. Therefore $E=inf A leq M$
$E geq M$: let $a in A$, and suppose there's $x$ such that $f(x)>a$. By continuity of $f$ there also an open neighbourood $U(x)$ such that for every $yin U(x)$ $f(y)>a$. Therefore $m({ f>a})>0$ and this is impossible, since $m({ fgeq a })=0$ (because $a in A$), $Omega$ has finite measure since it is (closed and) bounded and ${ f<a } $ and ${ f > a }$ are disjoint.
Provided the above reasoning holds I would like to extend from here the conclusion to a generic measurable $Omega$: anybody could give me a hint in this direction? (or show me why the general proposition is wrong?)
Thanks!
Edit
Thanks to everybody for the help. Maybe it could be more interesting to require $Omega$ non empty and connected.
general-topology measure-theory lebesgue-measure
asked Jan 11 at 13:31
LeonardoLeonardo
3449
$endgroup$
add a comment |
$begingroup$
My intuition tells me that the following proposition is true:
Let $f: Omega rightarrow mathbb{R}$ be a measurable function on a measurable set $Omega subseteq mathbb{R}^n$. If $f$ is continuous and non negative then $mathrm{ess, sup}(f)=text{sup}(f)$.
I'm using here the Borel sigma algebras on $Omega$ and $mathbb{R}$ respectively, and the Lebesgue measure $m$ on $Omega$.
My measure theory background is very poor, that being said I tried to give a proof for the case where $Omega$ is compact:
Let $A={a: m({ fgeq a })=0 }$. Therefore $mathrm{ess, sup}(f)= inf A = E$. Let also $M=max f = sup f$ (it exists because $f$ is continuous on a compact set). I will prove $E geq M$, and $E leq M$ (otherwise there would be a contradiction on the defition of $inf A$).
$E leq M$: if $a>M$ then ${ fgeq a }=emptyset$, so $m({ fgeq a })=0$, so $a in A$. Therefore $E=inf A leq M$
$E geq M$: let $a in A$, and suppose there's $x$ such that $f(x)>a$. By continuity of $f$ there also an open neighbourood $U(x)$ such that for every $yin U(x)$ $f(y)>a$. Therefore $m({ f>a})>0$ and this is impossible, since $m({ fgeq a })=0$ (because $a in A$), $Omega$ has finite measure since it is (closed and) bounded and ${ f<a } $ and ${ f > a }$ are disjoint.
Provided the above reasoning holds I would like to extend from here the conclusion to a generic measurable $Omega$: anybody could give me a hint in this direction? (or show me why the general proposition is wrong?)
Thanks!
Edit
Thanks to everybody for the help. Maybe it could be more interesting to require $Omega$ non empty and connected.
general-topology measure-theory lebesgue-measure
asked Jan 11 at 13:31
LeonardoLeonardo
3449
$endgroup$
1
$begingroup$
$mathrm{ess, sup}$ has three consecutive $s$'s in defiance of the usual rules of spelling. Try using$mathrm{ess, sup}$.
$endgroup$
– Umberto P.
Jan 11 at 16:53
$begingroup$
Does $f: mathbb{R}^n supseteq Omega rightarrow mathbb{R}$ mean $f: Omega rightarrow mathbb{R}$ such that $Omega subseteq mathbb{R}^n$? If so the proposition seems false.
$endgroup$
– 6005
Jan 11 at 18:25
$begingroup$
@6005 Yes, can you give me a counterexample? And I would really appreciate some additional hypothesis to make the statement true
$endgroup$
– Leonardo
Jan 11 at 18:29
$begingroup$
@Leonardo I have given an answer with a counterexample, but my counterexample feels too simple so I may have missed something in the question
$endgroup$
– 6005
Jan 11 at 18:30
add a comment |
$begingroup$
My intuition tells me that the following proposition is true:
Let $f: Omega rightarrow mathbb{R}$ be a measurable function on a measurable set $Omega subseteq mathbb{R}^n$. If $f$ is continuous and non negative then $mathrm{ess, sup}(f)=text{sup}(f)$.
I'm using here the Borel sigma algebras on $Omega$ and $mathbb{R}$ respectively, and the Lebesgue measure $m$ on $Omega$.
My measure theory background is very poor, that being said I tried to give a proof for the case where $Omega$ is compact:
Let $A={a: m({ fgeq a })=0 }$. Therefore $mathrm{ess, sup}(f)= inf A = E$. Let also $M=max f = sup f$ (it exists because $f$ is continuous on a compact set). I will prove $E geq M$, and $E leq M$ (otherwise there would be a contradiction on the defition of $inf A$).
$E leq M$: if $a>M$ then ${ fgeq a }=emptyset$, so $m({ fgeq a })=0$, so $a in A$. Therefore $E=inf A leq M$
$E geq M$: let $a in A$, and suppose there's $x$ such that $f(x)>a$. By continuity of $f$ there also an open neighbourood $U(x)$ such that for every $yin U(x)$ $f(y)>a$. Therefore $m({ f>a})>0$ and this is impossible, since $m({ fgeq a })=0$ (because $a in A$), $Omega$ has finite measure since it is (closed and) bounded and ${ f<a } $ and ${ f > a }$ are disjoint.
Provided the above reasoning holds I would like to extend from here the conclusion to a generic measurable $Omega$: anybody could give me a hint in this direction? (or show me why the general proposition is wrong?)
Thanks!
Edit
Thanks to everybody for the help. Maybe it could be more interesting to require $Omega$ non empty and connected.
general-topology measure-theory lebesgue-measure
asked Jan 11 at 13:31
LeonardoLeonardo
3449
$endgroup$
My intuition tells me that the following proposition is true:
Let $f: Omega rightarrow mathbb{R}$ be a measurable function on a measurable set $Omega subseteq mathbb{R}^n$. If $f$ is continuous and non negative then $mathrm{ess, sup}(f)=text{sup}(f)$.
I'm using here the Borel sigma algebras on $Omega$ and $mathbb{R}$ respectively, and the Lebesgue measure $m$ on $Omega$.
My measure theory background is very poor, that being said I tried to give a proof for the case where $Omega$ is compact:
Let $A={a: m({ fgeq a })=0 }$. Therefore $mathrm{ess, sup}(f)= inf A = E$. Let also $M=max f = sup f$ (it exists because $f$ is continuous on a compact set). I will prove $E geq M$, and $E leq M$ (otherwise there would be a contradiction on the defition of $inf A$).
$E leq M$: if $a>M$ then ${ fgeq a }=emptyset$, so $m({ fgeq a })=0$, so $a in A$. Therefore $E=inf A leq M$
$E geq M$: let $a in A$, and suppose there's $x$ such that $f(x)>a$. By continuity of $f$ there also an open neighbourood $U(x)$ such that for every $yin U(x)$ $f(y)>a$. Therefore $m({ f>a})>0$ and this is impossible, since $m({ fgeq a })=0$ (because $a in A$), $Omega$ has finite measure since it is (closed and) bounded and ${ f<a } $ and ${ f > a }$ are disjoint.
Provided the above reasoning holds I would like to extend from here the conclusion to a generic measurable $Omega$: anybody could give me a hint in this direction? (or show me why the general proposition is wrong?)
Thanks!
Edit
Thanks to everybody for the help. Maybe it could be more interesting to require $Omega$ non empty and connected.
general-topology measure-theory lebesgue-measure
general-topology measure-theory lebesgue-measure
asked Jan 11 at 13:31
LeonardoLeonardo
3449
asked Jan 11 at 13:31
LeonardoLeonardo
3449
edited Jan 11 at 19:16
Leonardo
asked Jan 11 at 13:31
LeonardoLeonardo
3449
asked Jan 11 at 13:31
LeonardoLeonardo
3449
asked Jan 11 at 13:31
LeonardoLeonardo
3449
3449
1
$begingroup$
$mathrm{ess, sup}$ has three consecutive $s$'s in defiance of the usual rules of spelling. Try using$mathrm{ess, sup}$.
$endgroup$
– Umberto P.
Jan 11 at 16:53
$begingroup$
Does $f: mathbb{R}^n supseteq Omega rightarrow mathbb{R}$ mean $f: Omega rightarrow mathbb{R}$ such that $Omega subseteq mathbb{R}^n$? If so the proposition seems false.
$endgroup$
– 6005
Jan 11 at 18:25
$begingroup$
@6005 Yes, can you give me a counterexample? And I would really appreciate some additional hypothesis to make the statement true
$endgroup$
– Leonardo
Jan 11 at 18:29
$begingroup$
@Leonardo I have given an answer with a counterexample, but my counterexample feels too simple so I may have missed something in the question
$endgroup$
– 6005
Jan 11 at 18:30
add a comment |
1
$begingroup$
$mathrm{ess, sup}$ has three consecutive $s$'s in defiance of the usual rules of spelling. Try using$mathrm{ess, sup}$.
$endgroup$
– Umberto P.
Jan 11 at 16:53
$begingroup$
Does $f: mathbb{R}^n supseteq Omega rightarrow mathbb{R}$ mean $f: Omega rightarrow mathbb{R}$ such that $Omega subseteq mathbb{R}^n$? If so the proposition seems false.
$endgroup$
– 6005
Jan 11 at 18:25
$begingroup$
@6005 Yes, can you give me a counterexample? And I would really appreciate some additional hypothesis to make the statement true
$endgroup$
– Leonardo
Jan 11 at 18:29
$begingroup$
@Leonardo I have given an answer with a counterexample, but my counterexample feels too simple so I may have missed something in the question
$endgroup$
– 6005
Jan 11 at 18:30
1
1
$begingroup$
$mathrm{ess, sup}$ has three consecutive $s$'s in defiance of the usual rules of spelling. Try using
$mathrm{ess, sup}$.$endgroup$
– Umberto P.
Jan 11 at 16:53
$begingroup$
$mathrm{ess, sup}$ has three consecutive $s$'s in defiance of the usual rules of spelling. Try using
$mathrm{ess, sup}$.$endgroup$
– Umberto P.
Jan 11 at 16:53
$begingroup$
Does $f: mathbb{R}^n supseteq Omega rightarrow mathbb{R}$ mean $f: Omega rightarrow mathbb{R}$ such that $Omega subseteq mathbb{R}^n$? If so the proposition seems false.
$endgroup$
– 6005
Jan 11 at 18:25
$begingroup$
Does $f: mathbb{R}^n supseteq Omega rightarrow mathbb{R}$ mean $f: Omega rightarrow mathbb{R}$ such that $Omega subseteq mathbb{R}^n$? If so the proposition seems false.
$endgroup$
– 6005
Jan 11 at 18:25
$begingroup$
@6005 Yes, can you give me a counterexample? And I would really appreciate some additional hypothesis to make the statement true
$endgroup$
– Leonardo
Jan 11 at 18:29
$begingroup$
@6005 Yes, can you give me a counterexample? And I would really appreciate some additional hypothesis to make the statement true
$endgroup$
– Leonardo
Jan 11 at 18:29
$begingroup$
@Leonardo I have given an answer with a counterexample, but my counterexample feels too simple so I may have missed something in the question
$endgroup$
– 6005
Jan 11 at 18:30
$begingroup$
@Leonardo I have given an answer with a counterexample, but my counterexample feels too simple so I may have missed something in the question
$endgroup$
– 6005
Jan 11 at 18:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For a counterexample to this proposition, take
$Omega = [0,1] cup {2}$ in $mathbb{R}$ with $f: Omega to mathbb{R}$ given by $f(x) = x$. This function is continuous.
Then $sup f = 2$, whereas $mathrm{ess,sup} f = 1$, since ${f ge 1} = {1,2}$ and $lambda({1,2}) = 0$ where $lambda$ is Lebesgue measure.
Additional condition to make it true:
Let's additionally assume that
$Omega ne varnothing$, and for all $x in Omega$, for any open ball $B$ around $x$, $$lambda(B cap Omega) > 0. tag{1}$$
Then, in this case your proposition is true.
For the proof, we already know $mathrm{ess,sup} f le sup f$, so we need to show $mathrm{ess,sup} f ge sup f$.
Equivalently, fix any $a < sup f$; we want to show that $lambda({f ge a}) > 0$. Since $a < sup f$, there exists $x_0$ such that $f(x_0) > a$.
Since $f$ is continuous, there exists an open ball $B$ containing $x_0$ such that for $x in B cap Omega$, $f(x) > a$. By our assumption (1) above, $lambda(B cap Omega) > 0$. So $lambda({f ge a}) > 0$.
Therefore, $mathrm{ess,sup} f = sup f$.
answered Jan 11 at 18:29
60056005
36.2k751125
$endgroup$
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
add a comment |
$begingroup$
If $m(Omega)=0$ then $mathrm{ess, sup}_{Omega} f=-infty$ so the claim is false.
The result does hold if we assume that $m(Qcap Omega)>0$ for almost every $xin Omega$, and each open cube $Q$ containing $x$. (Without this condition, one of the other answers shows the claim is still false.)
In general, $sup fge mathrm{ess, sup} f$. So, suppose $b:=sup f>mathrm{ess, sup}f:=a$ and without loss of generality, assume that $f$ is bounded above (if not, consider $Omegacap Q_n$ for compact cubes $Q_n$ of edge length $n$).
Then, there is an $x_0in Omega$ and an $a'$ such that $a<a'<f(x_0)<b$ and a $delta >0$ and a cube $Qni x_0$ such that $f(Q)subseteq (a',b).$ But then, $m({ fgeq a' })ge m(Qcap Omega)>0,$ which is a contradiction.
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
$endgroup$
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
|
show 3 more comments
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%2f3069828%2fmathrmess-supf-textsupf-for-a-continuous-non-negative-function-f%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
$begingroup$
For a counterexample to this proposition, take
$Omega = [0,1] cup {2}$ in $mathbb{R}$ with $f: Omega to mathbb{R}$ given by $f(x) = x$. This function is continuous.
Then $sup f = 2$, whereas $mathrm{ess,sup} f = 1$, since ${f ge 1} = {1,2}$ and $lambda({1,2}) = 0$ where $lambda$ is Lebesgue measure.
Additional condition to make it true:
Let's additionally assume that
$Omega ne varnothing$, and for all $x in Omega$, for any open ball $B$ around $x$, $$lambda(B cap Omega) > 0. tag{1}$$
Then, in this case your proposition is true.
For the proof, we already know $mathrm{ess,sup} f le sup f$, so we need to show $mathrm{ess,sup} f ge sup f$.
Equivalently, fix any $a < sup f$; we want to show that $lambda({f ge a}) > 0$. Since $a < sup f$, there exists $x_0$ such that $f(x_0) > a$.
Since $f$ is continuous, there exists an open ball $B$ containing $x_0$ such that for $x in B cap Omega$, $f(x) > a$. By our assumption (1) above, $lambda(B cap Omega) > 0$. So $lambda({f ge a}) > 0$.
Therefore, $mathrm{ess,sup} f = sup f$.
answered Jan 11 at 18:29
60056005
36.2k751125
$endgroup$
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
add a comment |
$begingroup$
For a counterexample to this proposition, take
$Omega = [0,1] cup {2}$ in $mathbb{R}$ with $f: Omega to mathbb{R}$ given by $f(x) = x$. This function is continuous.
Then $sup f = 2$, whereas $mathrm{ess,sup} f = 1$, since ${f ge 1} = {1,2}$ and $lambda({1,2}) = 0$ where $lambda$ is Lebesgue measure.
Additional condition to make it true:
Let's additionally assume that
$Omega ne varnothing$, and for all $x in Omega$, for any open ball $B$ around $x$, $$lambda(B cap Omega) > 0. tag{1}$$
Then, in this case your proposition is true.
For the proof, we already know $mathrm{ess,sup} f le sup f$, so we need to show $mathrm{ess,sup} f ge sup f$.
Equivalently, fix any $a < sup f$; we want to show that $lambda({f ge a}) > 0$. Since $a < sup f$, there exists $x_0$ such that $f(x_0) > a$.
Since $f$ is continuous, there exists an open ball $B$ containing $x_0$ such that for $x in B cap Omega$, $f(x) > a$. By our assumption (1) above, $lambda(B cap Omega) > 0$. So $lambda({f ge a}) > 0$.
Therefore, $mathrm{ess,sup} f = sup f$.
answered Jan 11 at 18:29
60056005
36.2k751125
$endgroup$
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
add a comment |
$begingroup$
For a counterexample to this proposition, take
$Omega = [0,1] cup {2}$ in $mathbb{R}$ with $f: Omega to mathbb{R}$ given by $f(x) = x$. This function is continuous.
Then $sup f = 2$, whereas $mathrm{ess,sup} f = 1$, since ${f ge 1} = {1,2}$ and $lambda({1,2}) = 0$ where $lambda$ is Lebesgue measure.
Additional condition to make it true:
Let's additionally assume that
$Omega ne varnothing$, and for all $x in Omega$, for any open ball $B$ around $x$, $$lambda(B cap Omega) > 0. tag{1}$$
Then, in this case your proposition is true.
For the proof, we already know $mathrm{ess,sup} f le sup f$, so we need to show $mathrm{ess,sup} f ge sup f$.
Equivalently, fix any $a < sup f$; we want to show that $lambda({f ge a}) > 0$. Since $a < sup f$, there exists $x_0$ such that $f(x_0) > a$.
Since $f$ is continuous, there exists an open ball $B$ containing $x_0$ such that for $x in B cap Omega$, $f(x) > a$. By our assumption (1) above, $lambda(B cap Omega) > 0$. So $lambda({f ge a}) > 0$.
Therefore, $mathrm{ess,sup} f = sup f$.
answered Jan 11 at 18:29
60056005
36.2k751125
$endgroup$
For a counterexample to this proposition, take
$Omega = [0,1] cup {2}$ in $mathbb{R}$ with $f: Omega to mathbb{R}$ given by $f(x) = x$. This function is continuous.
Then $sup f = 2$, whereas $mathrm{ess,sup} f = 1$, since ${f ge 1} = {1,2}$ and $lambda({1,2}) = 0$ where $lambda$ is Lebesgue measure.
Additional condition to make it true:
Let's additionally assume that
$Omega ne varnothing$, and for all $x in Omega$, for any open ball $B$ around $x$, $$lambda(B cap Omega) > 0. tag{1}$$
Then, in this case your proposition is true.
For the proof, we already know $mathrm{ess,sup} f le sup f$, so we need to show $mathrm{ess,sup} f ge sup f$.
Equivalently, fix any $a < sup f$; we want to show that $lambda({f ge a}) > 0$. Since $a < sup f$, there exists $x_0$ such that $f(x_0) > a$.
Since $f$ is continuous, there exists an open ball $B$ containing $x_0$ such that for $x in B cap Omega$, $f(x) > a$. By our assumption (1) above, $lambda(B cap Omega) > 0$. So $lambda({f ge a}) > 0$.
Therefore, $mathrm{ess,sup} f = sup f$.
answered Jan 11 at 18:29
60056005
36.2k751125
edited Jan 11 at 18:46
answered Jan 11 at 18:29
60056005
36.2k751125
answered Jan 11 at 18:29
60056005
36.2k751125
answered Jan 11 at 18:29
60056005
36.2k751125
36.2k751125
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
add a comment |
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Thanks, this is very helpful and extends my work to non empty open sets. I added some hypothesis at the end of the post, to see if this can be further extended.
$endgroup$
– Leonardo
Jan 11 at 19:24
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
$begingroup$
Somehow I was implicitly taking $Omega$ to be open when I thought up my answer, so I wasn't thinking carefully enough. The proposition that the complement of a zero set is dense holds precisely if the condition you invoke is true, so my answer is useless and I've deleted it accordingly. +1 Nice answer!
$endgroup$
– user159517
Jan 11 at 21:41
add a comment |
$begingroup$
If $m(Omega)=0$ then $mathrm{ess, sup}_{Omega} f=-infty$ so the claim is false.
The result does hold if we assume that $m(Qcap Omega)>0$ for almost every $xin Omega$, and each open cube $Q$ containing $x$. (Without this condition, one of the other answers shows the claim is still false.)
In general, $sup fge mathrm{ess, sup} f$. So, suppose $b:=sup f>mathrm{ess, sup}f:=a$ and without loss of generality, assume that $f$ is bounded above (if not, consider $Omegacap Q_n$ for compact cubes $Q_n$ of edge length $n$).
Then, there is an $x_0in Omega$ and an $a'$ such that $a<a'<f(x_0)<b$ and a $delta >0$ and a cube $Qni x_0$ such that $f(Q)subseteq (a',b).$ But then, $m({ fgeq a' })ge m(Qcap Omega)>0,$ which is a contradiction.
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
$endgroup$
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
|
show 3 more comments
$begingroup$
If $m(Omega)=0$ then $mathrm{ess, sup}_{Omega} f=-infty$ so the claim is false.
The result does hold if we assume that $m(Qcap Omega)>0$ for almost every $xin Omega$, and each open cube $Q$ containing $x$. (Without this condition, one of the other answers shows the claim is still false.)
In general, $sup fge mathrm{ess, sup} f$. So, suppose $b:=sup f>mathrm{ess, sup}f:=a$ and without loss of generality, assume that $f$ is bounded above (if not, consider $Omegacap Q_n$ for compact cubes $Q_n$ of edge length $n$).
Then, there is an $x_0in Omega$ and an $a'$ such that $a<a'<f(x_0)<b$ and a $delta >0$ and a cube $Qni x_0$ such that $f(Q)subseteq (a',b).$ But then, $m({ fgeq a' })ge m(Qcap Omega)>0,$ which is a contradiction.
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
$endgroup$
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
|
show 3 more comments
$begingroup$
If $m(Omega)=0$ then $mathrm{ess, sup}_{Omega} f=-infty$ so the claim is false.
The result does hold if we assume that $m(Qcap Omega)>0$ for almost every $xin Omega$, and each open cube $Q$ containing $x$. (Without this condition, one of the other answers shows the claim is still false.)
In general, $sup fge mathrm{ess, sup} f$. So, suppose $b:=sup f>mathrm{ess, sup}f:=a$ and without loss of generality, assume that $f$ is bounded above (if not, consider $Omegacap Q_n$ for compact cubes $Q_n$ of edge length $n$).
Then, there is an $x_0in Omega$ and an $a'$ such that $a<a'<f(x_0)<b$ and a $delta >0$ and a cube $Qni x_0$ such that $f(Q)subseteq (a',b).$ But then, $m({ fgeq a' })ge m(Qcap Omega)>0,$ which is a contradiction.
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
$endgroup$
If $m(Omega)=0$ then $mathrm{ess, sup}_{Omega} f=-infty$ so the claim is false.
The result does hold if we assume that $m(Qcap Omega)>0$ for almost every $xin Omega$, and each open cube $Q$ containing $x$. (Without this condition, one of the other answers shows the claim is still false.)
In general, $sup fge mathrm{ess, sup} f$. So, suppose $b:=sup f>mathrm{ess, sup}f:=a$ and without loss of generality, assume that $f$ is bounded above (if not, consider $Omegacap Q_n$ for compact cubes $Q_n$ of edge length $n$).
Then, there is an $x_0in Omega$ and an $a'$ such that $a<a'<f(x_0)<b$ and a $delta >0$ and a cube $Qni x_0$ such that $f(Q)subseteq (a',b).$ But then, $m({ fgeq a' })ge m(Qcap Omega)>0,$ which is a contradiction.
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
edited Jan 11 at 21:31
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
answered Jan 11 at 17:39
MatematletaMatematleta
10.8k2918
10.8k2918
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
|
show 3 more comments
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
Is $m(Q cap Omega) > 0$ for each $x in Omega$ is an additional assumption you are making? I have given a counterexample to the proposition in my answer.
$endgroup$
– 6005
Jan 11 at 18:47
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
The additional assumption is that $m(Omega)>0$ which implies that for all open cubes for which $Omegacap Qneq emptyset$, one has $m(Omegacap Q)>0$. As you point out, the result is false if $m(Omega)=0.$
$endgroup$
– Matematleta
Jan 11 at 18:55
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
No, my counterexample has $m(Omega) = 1$. Your condition is stronger than that because you are saying that for all $x in Omega$ and for all open cubes $Q$ around $x$, $m(Q cap Q) > 0$.
$endgroup$
– 6005
Jan 11 at 18:58
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Yes, that is always true, as I have said in my answer.
$endgroup$
– 6005
Jan 11 at 19:04
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
$begingroup$
Do you agree that $[0,1] cup {2}$ is an $Omega$ that does not satisfy your open cube condition, and that in my answer I give a counterexample using this $Omega$?
$endgroup$
– 6005
Jan 11 at 19:05
|
show 3 more comments
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%2f3069828%2fmathrmess-supf-textsupf-for-a-continuous-non-negative-function-f%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
$begingroup$
$mathrm{ess, sup}$ has three consecutive $s$'s in defiance of the usual rules of spelling. Try using
$mathrm{ess, sup}$.$endgroup$
– Umberto P.
Jan 11 at 16:53
$begingroup$
Does $f: mathbb{R}^n supseteq Omega rightarrow mathbb{R}$ mean $f: Omega rightarrow mathbb{R}$ such that $Omega subseteq mathbb{R}^n$? If so the proposition seems false.
$endgroup$
– 6005
Jan 11 at 18:25
$begingroup$
@6005 Yes, can you give me a counterexample? And I would really appreciate some additional hypothesis to make the statement true
$endgroup$
– Leonardo
Jan 11 at 18:29
$begingroup$
@Leonardo I have given an answer with a counterexample, but my counterexample feels too simple so I may have missed something in the question
$endgroup$
– 6005
Jan 11 at 18:30