Countable open cover up to a null set
Given a metric space $(X,d)$,a probability measure $mu$ (on the Borel sigma algebra) and an open cover $C:={A_i}_{iin I}$ of $X$, is it always possible to find a countable subset of $C$ that covers $X$ up to a null set i.e. does there exists countable $C'subseteq C$ such that $mu(Xbackslash cup_{Ain C'} A)=0$?
Naively one would suspect that this is true because if we consider a "minimum" $C'subseteq C$ that covers $X$ up to a null set, then for any $Ain C'$ we have $mu(Abackslash cup_{Aneq A'in C'} A')>0$. Hence if such $C'$ is uncountable it's routine to show that $mu(X)=infty$, a contradiction. Obviously this argument doesn't necessarily work because such $C'$ may not exists.
For the sake of completeness I should mention that this is a natural question that one might want to answer to solve Exercise 2.2.3 of Einsiedler and Ward's Ergodic Theory text: Let $(X,d)$ be a metric space, $T:Xto X$ Borel measurable, and $mu$ be a $T-$invariant probability measure. Prove that for $mu-$almost every $xin X$ there is a sequence $n_ktoinfty$ with $T^{n_k}(x)to x$ as $kto infty$
measure-theory metric-spaces ergodic-theory borel-sets
edited Dec 9 at 17:04
Yanko
6,104724
asked Dec 9 at 14:48
MathStudent
133
add a comment |
Given a metric space $(X,d)$,a probability measure $mu$ (on the Borel sigma algebra) and an open cover $C:={A_i}_{iin I}$ of $X$, is it always possible to find a countable subset of $C$ that covers $X$ up to a null set i.e. does there exists countable $C'subseteq C$ such that $mu(Xbackslash cup_{Ain C'} A)=0$?
Naively one would suspect that this is true because if we consider a "minimum" $C'subseteq C$ that covers $X$ up to a null set, then for any $Ain C'$ we have $mu(Abackslash cup_{Aneq A'in C'} A')>0$. Hence if such $C'$ is uncountable it's routine to show that $mu(X)=infty$, a contradiction. Obviously this argument doesn't necessarily work because such $C'$ may not exists.
For the sake of completeness I should mention that this is a natural question that one might want to answer to solve Exercise 2.2.3 of Einsiedler and Ward's Ergodic Theory text: Let $(X,d)$ be a metric space, $T:Xto X$ Borel measurable, and $mu$ be a $T-$invariant probability measure. Prove that for $mu-$almost every $xin X$ there is a sequence $n_ktoinfty$ with $T^{n_k}(x)to x$ as $kto infty$
measure-theory metric-spaces ergodic-theory borel-sets
edited Dec 9 at 17:04
Yanko
6,104724
asked Dec 9 at 14:48
MathStudent
133
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
– Yanko
Dec 9 at 14:52
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
– Yanko
Dec 9 at 14:53
1
No I didn't assume that $X$ is separable nor that $mu$ gives an open set positive measure. The original question that I'm working on is easy if $X$ is separable/second-countable/compact and in fact for this question you directly get countable/finite subcover (rendering the up to null set property useless). Is there any good reason that I should assume that $mu$ gives open sets positive measure? I am aware that in that case we can't have uncountably many disjoint open sets but that doesn't seem too bad.
– MathStudent
Dec 9 at 15:01
It turns out that in the introduction of the book it was mentioned that metric spaces are separable unless stated otherwise, which means that this problem is easy. This is why one should read introductions carefully.
– MathStudent
Dec 26 at 23:08
add a comment |
Given a metric space $(X,d)$,a probability measure $mu$ (on the Borel sigma algebra) and an open cover $C:={A_i}_{iin I}$ of $X$, is it always possible to find a countable subset of $C$ that covers $X$ up to a null set i.e. does there exists countable $C'subseteq C$ such that $mu(Xbackslash cup_{Ain C'} A)=0$?
Naively one would suspect that this is true because if we consider a "minimum" $C'subseteq C$ that covers $X$ up to a null set, then for any $Ain C'$ we have $mu(Abackslash cup_{Aneq A'in C'} A')>0$. Hence if such $C'$ is uncountable it's routine to show that $mu(X)=infty$, a contradiction. Obviously this argument doesn't necessarily work because such $C'$ may not exists.
For the sake of completeness I should mention that this is a natural question that one might want to answer to solve Exercise 2.2.3 of Einsiedler and Ward's Ergodic Theory text: Let $(X,d)$ be a metric space, $T:Xto X$ Borel measurable, and $mu$ be a $T-$invariant probability measure. Prove that for $mu-$almost every $xin X$ there is a sequence $n_ktoinfty$ with $T^{n_k}(x)to x$ as $kto infty$
measure-theory metric-spaces ergodic-theory borel-sets
edited Dec 9 at 17:04
Yanko
6,104724
asked Dec 9 at 14:48
MathStudent
133
Given a metric space $(X,d)$,a probability measure $mu$ (on the Borel sigma algebra) and an open cover $C:={A_i}_{iin I}$ of $X$, is it always possible to find a countable subset of $C$ that covers $X$ up to a null set i.e. does there exists countable $C'subseteq C$ such that $mu(Xbackslash cup_{Ain C'} A)=0$?
Naively one would suspect that this is true because if we consider a "minimum" $C'subseteq C$ that covers $X$ up to a null set, then for any $Ain C'$ we have $mu(Abackslash cup_{Aneq A'in C'} A')>0$. Hence if such $C'$ is uncountable it's routine to show that $mu(X)=infty$, a contradiction. Obviously this argument doesn't necessarily work because such $C'$ may not exists.
For the sake of completeness I should mention that this is a natural question that one might want to answer to solve Exercise 2.2.3 of Einsiedler and Ward's Ergodic Theory text: Let $(X,d)$ be a metric space, $T:Xto X$ Borel measurable, and $mu$ be a $T-$invariant probability measure. Prove that for $mu-$almost every $xin X$ there is a sequence $n_ktoinfty$ with $T^{n_k}(x)to x$ as $kto infty$
measure-theory metric-spaces ergodic-theory borel-sets
measure-theory metric-spaces ergodic-theory borel-sets
edited Dec 9 at 17:04
Yanko
6,104724
asked Dec 9 at 14:48
MathStudent
133
edited Dec 9 at 17:04
Yanko
6,104724
asked Dec 9 at 14:48
MathStudent
133
edited Dec 9 at 17:04
Yanko
6,104724
edited Dec 9 at 17:04
Yanko
6,104724
edited Dec 9 at 17:04
Yanko
6,104724
6,104724
asked Dec 9 at 14:48
MathStudent
133
asked Dec 9 at 14:48
MathStudent
133
asked Dec 9 at 14:48
MathStudent
133
133
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
– Yanko
Dec 9 at 14:52
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
– Yanko
Dec 9 at 14:53
1
No I didn't assume that $X$ is separable nor that $mu$ gives an open set positive measure. The original question that I'm working on is easy if $X$ is separable/second-countable/compact and in fact for this question you directly get countable/finite subcover (rendering the up to null set property useless). Is there any good reason that I should assume that $mu$ gives open sets positive measure? I am aware that in that case we can't have uncountably many disjoint open sets but that doesn't seem too bad.
– MathStudent
Dec 9 at 15:01
It turns out that in the introduction of the book it was mentioned that metric spaces are separable unless stated otherwise, which means that this problem is easy. This is why one should read introductions carefully.
– MathStudent
Dec 26 at 23:08
add a comment |
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
– Yanko
Dec 9 at 14:52
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
– Yanko
Dec 9 at 14:53
1
No I didn't assume that $X$ is separable nor that $mu$ gives an open set positive measure. The original question that I'm working on is easy if $X$ is separable/second-countable/compact and in fact for this question you directly get countable/finite subcover (rendering the up to null set property useless). Is there any good reason that I should assume that $mu$ gives open sets positive measure? I am aware that in that case we can't have uncountably many disjoint open sets but that doesn't seem too bad.
– MathStudent
Dec 9 at 15:01
It turns out that in the introduction of the book it was mentioned that metric spaces are separable unless stated otherwise, which means that this problem is easy. This is why one should read introductions carefully.
– MathStudent
Dec 26 at 23:08
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
– Yanko
Dec 9 at 14:52
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
– Yanko
Dec 9 at 14:52
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
– Yanko
Dec 9 at 14:53
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
– Yanko
Dec 9 at 14:53
1
1
No I didn't assume that $X$ is separable nor that $mu$ gives an open set positive measure. The original question that I'm working on is easy if $X$ is separable/second-countable/compact and in fact for this question you directly get countable/finite subcover (rendering the up to null set property useless). Is there any good reason that I should assume that $mu$ gives open sets positive measure? I am aware that in that case we can't have uncountably many disjoint open sets but that doesn't seem too bad.
– MathStudent
Dec 9 at 15:01
No I didn't assume that $X$ is separable nor that $mu$ gives an open set positive measure. The original question that I'm working on is easy if $X$ is separable/second-countable/compact and in fact for this question you directly get countable/finite subcover (rendering the up to null set property useless). Is there any good reason that I should assume that $mu$ gives open sets positive measure? I am aware that in that case we can't have uncountably many disjoint open sets but that doesn't seem too bad.
– MathStudent
Dec 9 at 15:01
It turns out that in the introduction of the book it was mentioned that metric spaces are separable unless stated otherwise, which means that this problem is easy. This is why one should read introductions carefully.
– MathStudent
Dec 26 at 23:08
It turns out that in the introduction of the book it was mentioned that metric spaces are separable unless stated otherwise, which means that this problem is easy. This is why one should read introductions carefully.
– MathStudent
Dec 26 at 23:08
add a comment |
1 Answer
1
active
oldest
votes
I don't have a complete answer, but I will argue that a positive answer to your question would refute a statement "widely believed by set theorists", namely the existence of measurable cardinals.
Let $X$ be an uncountable set with metric
begin{align}
d(x,y) &=
begin{cases}
1 & text{if $xneq y$,} \
0 & text{if $x=y$,}
end{cases}
end{align}
so that $X$ has the discrete topology.
Suppose we can find a probability measure $mu$ such that $mu({x})=0$ for each $xin X$. Then, this would provide a negative answer to your question, namely, the open cover ${B_{1/2}(x): xin X}$ would have no countable subfamily that covers $X$ up to a null set.
According to this answer and the Wikipedia page it cites, the existence of such a measure space cannot be proven from ZFC (assuming the consistency of ZFC). If you can prove in ZFC that countable subcovers modulo null sets always exist, you would get a proof of the inconsistency of the existence of measurable cardinals with ZFC. This is a possibility as far as I understand, but it seems that most set theorists do not consider it a likely possibility (see e.g. here or Volume 5II, Chapter 54 or Measure Theory by Fremlin).
answered Dec 12 at 22:59
Blackbird
1,410711
1
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
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%2f3032446%2fcountable-open-cover-up-to-a-null-set%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
I don't have a complete answer, but I will argue that a positive answer to your question would refute a statement "widely believed by set theorists", namely the existence of measurable cardinals.
Let $X$ be an uncountable set with metric
begin{align}
d(x,y) &=
begin{cases}
1 & text{if $xneq y$,} \
0 & text{if $x=y$,}
end{cases}
end{align}
so that $X$ has the discrete topology.
Suppose we can find a probability measure $mu$ such that $mu({x})=0$ for each $xin X$. Then, this would provide a negative answer to your question, namely, the open cover ${B_{1/2}(x): xin X}$ would have no countable subfamily that covers $X$ up to a null set.
According to this answer and the Wikipedia page it cites, the existence of such a measure space cannot be proven from ZFC (assuming the consistency of ZFC). If you can prove in ZFC that countable subcovers modulo null sets always exist, you would get a proof of the inconsistency of the existence of measurable cardinals with ZFC. This is a possibility as far as I understand, but it seems that most set theorists do not consider it a likely possibility (see e.g. here or Volume 5II, Chapter 54 or Measure Theory by Fremlin).
answered Dec 12 at 22:59
Blackbird
1,410711
1
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
add a comment |
I don't have a complete answer, but I will argue that a positive answer to your question would refute a statement "widely believed by set theorists", namely the existence of measurable cardinals.
Let $X$ be an uncountable set with metric
begin{align}
d(x,y) &=
begin{cases}
1 & text{if $xneq y$,} \
0 & text{if $x=y$,}
end{cases}
end{align}
so that $X$ has the discrete topology.
Suppose we can find a probability measure $mu$ such that $mu({x})=0$ for each $xin X$. Then, this would provide a negative answer to your question, namely, the open cover ${B_{1/2}(x): xin X}$ would have no countable subfamily that covers $X$ up to a null set.
According to this answer and the Wikipedia page it cites, the existence of such a measure space cannot be proven from ZFC (assuming the consistency of ZFC). If you can prove in ZFC that countable subcovers modulo null sets always exist, you would get a proof of the inconsistency of the existence of measurable cardinals with ZFC. This is a possibility as far as I understand, but it seems that most set theorists do not consider it a likely possibility (see e.g. here or Volume 5II, Chapter 54 or Measure Theory by Fremlin).
answered Dec 12 at 22:59
Blackbird
1,410711
1
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
add a comment |
I don't have a complete answer, but I will argue that a positive answer to your question would refute a statement "widely believed by set theorists", namely the existence of measurable cardinals.
Let $X$ be an uncountable set with metric
begin{align}
d(x,y) &=
begin{cases}
1 & text{if $xneq y$,} \
0 & text{if $x=y$,}
end{cases}
end{align}
so that $X$ has the discrete topology.
Suppose we can find a probability measure $mu$ such that $mu({x})=0$ for each $xin X$. Then, this would provide a negative answer to your question, namely, the open cover ${B_{1/2}(x): xin X}$ would have no countable subfamily that covers $X$ up to a null set.
According to this answer and the Wikipedia page it cites, the existence of such a measure space cannot be proven from ZFC (assuming the consistency of ZFC). If you can prove in ZFC that countable subcovers modulo null sets always exist, you would get a proof of the inconsistency of the existence of measurable cardinals with ZFC. This is a possibility as far as I understand, but it seems that most set theorists do not consider it a likely possibility (see e.g. here or Volume 5II, Chapter 54 or Measure Theory by Fremlin).
answered Dec 12 at 22:59
Blackbird
1,410711
I don't have a complete answer, but I will argue that a positive answer to your question would refute a statement "widely believed by set theorists", namely the existence of measurable cardinals.
Let $X$ be an uncountable set with metric
begin{align}
d(x,y) &=
begin{cases}
1 & text{if $xneq y$,} \
0 & text{if $x=y$,}
end{cases}
end{align}
so that $X$ has the discrete topology.
Suppose we can find a probability measure $mu$ such that $mu({x})=0$ for each $xin X$. Then, this would provide a negative answer to your question, namely, the open cover ${B_{1/2}(x): xin X}$ would have no countable subfamily that covers $X$ up to a null set.
According to this answer and the Wikipedia page it cites, the existence of such a measure space cannot be proven from ZFC (assuming the consistency of ZFC). If you can prove in ZFC that countable subcovers modulo null sets always exist, you would get a proof of the inconsistency of the existence of measurable cardinals with ZFC. This is a possibility as far as I understand, but it seems that most set theorists do not consider it a likely possibility (see e.g. here or Volume 5II, Chapter 54 or Measure Theory by Fremlin).
answered Dec 12 at 22:59
Blackbird
1,410711
answered Dec 12 at 22:59
Blackbird
1,410711
answered Dec 12 at 22:59
Blackbird
1,410711
answered Dec 12 at 22:59
Blackbird
1,410711
1,410711
1
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
add a comment |
1
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
1
1
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
I really appreciate the comment because this leads me to the right direction. I wanted to point out that this is much stronger than what we need. In particular apparently the given measure in this case is an outer measure, which is very strong. I believe it suffices to consider $mathbb{R}$ with the same metric as in your answer. The measurable set will be the countable and cocountable and the measure is 0 if it's countable and 1 if it's cocountable. By some easy work we can see that this is a counter example to both my question and the original question. Thank you.
– MathStudent
Dec 26 at 21:26
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.
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%2f3032446%2fcountable-open-cover-up-to-a-null-set%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
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
– Yanko
Dec 9 at 14:52
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
– Yanko
Dec 9 at 14:53
1
No I didn't assume that $X$ is separable nor that $mu$ gives an open set positive measure. The original question that I'm working on is easy if $X$ is separable/second-countable/compact and in fact for this question you directly get countable/finite subcover (rendering the up to null set property useless). Is there any good reason that I should assume that $mu$ gives open sets positive measure? I am aware that in that case we can't have uncountably many disjoint open sets but that doesn't seem too bad.
– MathStudent
Dec 9 at 15:01
It turns out that in the introduction of the book it was mentioned that metric spaces are separable unless stated otherwise, which means that this problem is easy. This is why one should read introductions carefully.
– MathStudent
Dec 26 at 23:08