Can we approximate a sigma-algebra by an increasing sequence of finite sigma-algebras?
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$newcommand{N}{mathbb N}$
$newcommand{mc}{mathcal}$
$newcommand{set}[1]{{#1}}$
Definition.
We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $set{E_n}_{nin N}$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.
Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $set{E_n}_{nin N}$, then for all $Fin mc F$, there is $Gin mc G$ such that $mu(FDelta G)=0$?
measure-theory
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
add a comment |
$begingroup$
$newcommand{N}{mathbb N}$
$newcommand{mc}{mathcal}$
$newcommand{set}[1]{{#1}}$
Definition.
We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $set{E_n}_{nin N}$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.
Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $set{E_n}_{nin N}$, then for all $Fin mc F$, there is $Gin mc G$ such that $mu(FDelta G)=0$?
measure-theory
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
add a comment |
$begingroup$
$newcommand{N}{mathbb N}$
$newcommand{mc}{mathcal}$
$newcommand{set}[1]{{#1}}$
Definition.
We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $set{E_n}_{nin N}$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.
Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $set{E_n}_{nin N}$, then for all $Fin mc F$, there is $Gin mc G$ such that $mu(FDelta G)=0$?
measure-theory
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
$newcommand{N}{mathbb N}$
$newcommand{mc}{mathcal}$
$newcommand{set}[1]{{#1}}$
Definition.
We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $set{E_n}_{nin N}$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.
Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $set{E_n}_{nin N}$, then for all $Fin mc F$, there is $Gin mc G$ such that $mu(FDelta G)=0$?
measure-theory
measure-theory
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
edited Aug 18 '18 at 17:09
caffeinemachine
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
asked Jul 23 '18 at 5:38
caffeinemachinecaffeinemachine
6,67121455
6,67121455
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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Let $mu (E_{n_{k}} Delta E)<frac 1 {2^{k}}$ for each $k$. Then $sum_k mu (E_{n_{k}} Delta E)<infty $ so $mu ( lim sup_k (E_{n_{k}} Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_{n_{k}}$.
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
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add a comment |
$begingroup$
Another way to see the result is the following:
Let $Ein mathcal F$ be fixed.
By hypothesis, we can find a sequence $n_1<n_2< n_3<cdots$ such that $chi_{E_{n_k}}to chi_E$ in $L^1$ as $kto infty$.
As is well-known, we may pass to a subsequence and assume that the convergence occurs pointwise a.e.
But then we have $lim_{kto infty} chi_{E_{n_k}} = limsup_{kto infty}chi_{E_{n_k}}pmod{mu}$. The latter is in $mathcal G$ and we are done.
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Let $mu (E_{n_{k}} Delta E)<frac 1 {2^{k}}$ for each $k$. Then $sum_k mu (E_{n_{k}} Delta E)<infty $ so $mu ( lim sup_k (E_{n_{k}} Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_{n_{k}}$.
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
add a comment |
$begingroup$
Let $mu (E_{n_{k}} Delta E)<frac 1 {2^{k}}$ for each $k$. Then $sum_k mu (E_{n_{k}} Delta E)<infty $ so $mu ( lim sup_k (E_{n_{k}} Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_{n_{k}}$.
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
add a comment |
$begingroup$
Let $mu (E_{n_{k}} Delta E)<frac 1 {2^{k}}$ for each $k$. Then $sum_k mu (E_{n_{k}} Delta E)<infty $ so $mu ( lim sup_k (E_{n_{k}} Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_{n_{k}}$.
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
$endgroup$
Let $mu (E_{n_{k}} Delta E)<frac 1 {2^{k}}$ for each $k$. Then $sum_k mu (E_{n_{k}} Delta E)<infty $ so $mu ( lim sup_k (E_{n_{k}} Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_{n_{k}}$.
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
answered Jul 23 '18 at 7:26
Kavi Rama MurthyKavi Rama Murthy
69.9k53170
69.9k53170
add a comment |
add a comment |
$begingroup$
Another way to see the result is the following:
Let $Ein mathcal F$ be fixed.
By hypothesis, we can find a sequence $n_1<n_2< n_3<cdots$ such that $chi_{E_{n_k}}to chi_E$ in $L^1$ as $kto infty$.
As is well-known, we may pass to a subsequence and assume that the convergence occurs pointwise a.e.
But then we have $lim_{kto infty} chi_{E_{n_k}} = limsup_{kto infty}chi_{E_{n_k}}pmod{mu}$. The latter is in $mathcal G$ and we are done.
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
add a comment |
$begingroup$
Another way to see the result is the following:
Let $Ein mathcal F$ be fixed.
By hypothesis, we can find a sequence $n_1<n_2< n_3<cdots$ such that $chi_{E_{n_k}}to chi_E$ in $L^1$ as $kto infty$.
As is well-known, we may pass to a subsequence and assume that the convergence occurs pointwise a.e.
But then we have $lim_{kto infty} chi_{E_{n_k}} = limsup_{kto infty}chi_{E_{n_k}}pmod{mu}$. The latter is in $mathcal G$ and we are done.
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
add a comment |
$begingroup$
Another way to see the result is the following:
Let $Ein mathcal F$ be fixed.
By hypothesis, we can find a sequence $n_1<n_2< n_3<cdots$ such that $chi_{E_{n_k}}to chi_E$ in $L^1$ as $kto infty$.
As is well-known, we may pass to a subsequence and assume that the convergence occurs pointwise a.e.
But then we have $lim_{kto infty} chi_{E_{n_k}} = limsup_{kto infty}chi_{E_{n_k}}pmod{mu}$. The latter is in $mathcal G$ and we are done.
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
$endgroup$
Another way to see the result is the following:
Let $Ein mathcal F$ be fixed.
By hypothesis, we can find a sequence $n_1<n_2< n_3<cdots$ such that $chi_{E_{n_k}}to chi_E$ in $L^1$ as $kto infty$.
As is well-known, we may pass to a subsequence and assume that the convergence occurs pointwise a.e.
But then we have $lim_{kto infty} chi_{E_{n_k}} = limsup_{kto infty}chi_{E_{n_k}}pmod{mu}$. The latter is in $mathcal G$ and we are done.
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
answered Dec 10 '18 at 13:34
caffeinemachinecaffeinemachine
6,67121455
6,67121455
add a comment |
add a comment |
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