Countable open cover up to a null set
$begingroup$
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 '18 at 17:04
Yanko
6,7931529
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
$endgroup$
add a comment |
$begingroup$
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 '18 at 17:04
Yanko
6,7931529
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
$endgroup$
$begingroup$
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
$endgroup$
– Yanko
Dec 9 '18 at 14:52
$begingroup$
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
$endgroup$
– Yanko
Dec 9 '18 at 14:53
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 9 '18 at 15:01
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 23:08
add a comment |
$begingroup$
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 '18 at 17:04
Yanko
6,7931529
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
$endgroup$
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 '18 at 17:04
Yanko
6,7931529
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
edited Dec 9 '18 at 17:04
Yanko
6,7931529
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
edited Dec 9 '18 at 17:04
Yanko
6,7931529
edited Dec 9 '18 at 17:04
Yanko
6,7931529
edited Dec 9 '18 at 17:04
Yanko
6,7931529
6,7931529
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
asked Dec 9 '18 at 14:48
MathStudentMathStudent
133
133
$begingroup$
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
$endgroup$
– Yanko
Dec 9 '18 at 14:52
$begingroup$
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
$endgroup$
– Yanko
Dec 9 '18 at 14:53
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 9 '18 at 15:01
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 23:08
add a comment |
$begingroup$
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
$endgroup$
– Yanko
Dec 9 '18 at 14:52
$begingroup$
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
$endgroup$
– Yanko
Dec 9 '18 at 14:53
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 9 '18 at 15:01
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 23:08
$begingroup$
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
$endgroup$
– Yanko
Dec 9 '18 at 14:52
$begingroup$
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
$endgroup$
– Yanko
Dec 9 '18 at 14:52
$begingroup$
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
$endgroup$
– Yanko
Dec 9 '18 at 14:53
$begingroup$
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
$endgroup$
– Yanko
Dec 9 '18 at 14:53
1
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 9 '18 at 15:01
$begingroup$
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.
$endgroup$
– MathStudent
Dec 9 '18 at 15:01
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 23:08
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 23:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 '18 at 22:59
BlackbirdBlackbird
1,465711
$endgroup$
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 '18 at 22:59
BlackbirdBlackbird
1,465711
$endgroup$
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
add a comment |
$begingroup$
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 '18 at 22:59
BlackbirdBlackbird
1,465711
$endgroup$
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
add a comment |
$begingroup$
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 '18 at 22:59
BlackbirdBlackbird
1,465711
$endgroup$
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 '18 at 22:59
BlackbirdBlackbird
1,465711
answered Dec 12 '18 at 22:59
BlackbirdBlackbird
1,465711
answered Dec 12 '18 at 22:59
BlackbirdBlackbird
1,465711
answered Dec 12 '18 at 22:59
BlackbirdBlackbird
1,465711
1,465711
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
add a comment |
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
1
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 21:26
add a comment |
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$begingroup$
Do you assume that $X$ is separable? Do you assume that $mu$ gives open sets positive measure?
$endgroup$
– Yanko
Dec 9 '18 at 14:52
$begingroup$
I remember that I once read that every separable probability space is measure-equivalent to a compact metric space.
$endgroup$
– Yanko
Dec 9 '18 at 14:53
1
$begingroup$
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.
$endgroup$
– MathStudent
Dec 9 '18 at 15:01
$begingroup$
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.
$endgroup$
– MathStudent
Dec 26 '18 at 23:08