Why is $mathscr{A}_{infty}=bigcup_{ninmathbb{N}} mathscr{A}_n$ never a $sigma$-algebra?
Problem: Let $(X, mathscr{A})$ be a measurable space and $(mathscr{A}_n)_{ninmathbb{N}}$ be a strictly increasing sequence of $sigma$-algebras. Show that
$$ mathscr{A}_{infty} := bigcup_{ninmathbb{N}} mathscr{A}_n $$
is never a $sigma$-algebra.
This is a problem of a book of Rene Schilling. He shows an answer of this problem on his homepage. However I don't understand the last part, Step 5. I don't know there always exists the smallest set $B_n$.
Here is his solutions. This problem is Problem 3.8. [pp.24-26]
measure-theory elementary-set-theory
edited 2 days ago
Davide Giraudo
125k16150260
asked Dec 30 '18 at 11:16
taku
111
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taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
Problem: Let $(X, mathscr{A})$ be a measurable space and $(mathscr{A}_n)_{ninmathbb{N}}$ be a strictly increasing sequence of $sigma$-algebras. Show that
$$ mathscr{A}_{infty} := bigcup_{ninmathbb{N}} mathscr{A}_n $$
is never a $sigma$-algebra.
This is a problem of a book of Rene Schilling. He shows an answer of this problem on his homepage. However I don't understand the last part, Step 5. I don't know there always exists the smallest set $B_n$.
Here is his solutions. This problem is Problem 3.8. [pp.24-26]
measure-theory elementary-set-theory
edited 2 days ago
Davide Giraudo
125k16150260
asked Dec 30 '18 at 11:16
taku
111
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
– Sangchul Lee
Dec 30 '18 at 11:21
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
– Sangchul Lee
Dec 30 '18 at 11:28
Thank you for your editing.
– taku
Dec 30 '18 at 11:51
1
$nin mathbb{N}in mathscr{A}_n$
– d.k.o.
Dec 30 '18 at 19:16
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
– dpb492
2 days ago
add a comment |
Problem: Let $(X, mathscr{A})$ be a measurable space and $(mathscr{A}_n)_{ninmathbb{N}}$ be a strictly increasing sequence of $sigma$-algebras. Show that
$$ mathscr{A}_{infty} := bigcup_{ninmathbb{N}} mathscr{A}_n $$
is never a $sigma$-algebra.
This is a problem of a book of Rene Schilling. He shows an answer of this problem on his homepage. However I don't understand the last part, Step 5. I don't know there always exists the smallest set $B_n$.
Here is his solutions. This problem is Problem 3.8. [pp.24-26]
measure-theory elementary-set-theory
edited 2 days ago
Davide Giraudo
125k16150260
asked Dec 30 '18 at 11:16
taku
111
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Problem: Let $(X, mathscr{A})$ be a measurable space and $(mathscr{A}_n)_{ninmathbb{N}}$ be a strictly increasing sequence of $sigma$-algebras. Show that
$$ mathscr{A}_{infty} := bigcup_{ninmathbb{N}} mathscr{A}_n $$
is never a $sigma$-algebra.
This is a problem of a book of Rene Schilling. He shows an answer of this problem on his homepage. However I don't understand the last part, Step 5. I don't know there always exists the smallest set $B_n$.
Here is his solutions. This problem is Problem 3.8. [pp.24-26]
measure-theory elementary-set-theory
measure-theory elementary-set-theory
edited 2 days ago
Davide Giraudo
125k16150260
asked Dec 30 '18 at 11:16
taku
111
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Davide Giraudo
125k16150260
asked Dec 30 '18 at 11:16
taku
111
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Davide Giraudo
125k16150260
edited 2 days ago
Davide Giraudo
125k16150260
edited 2 days ago
Davide Giraudo
125k16150260
125k16150260
asked Dec 30 '18 at 11:16
taku
111
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 30 '18 at 11:16
taku
111
asked Dec 30 '18 at 11:16
taku
111
111
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
taku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
– Sangchul Lee
Dec 30 '18 at 11:21
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
– Sangchul Lee
Dec 30 '18 at 11:28
Thank you for your editing.
– taku
Dec 30 '18 at 11:51
1
$nin mathbb{N}in mathscr{A}_n$
– d.k.o.
Dec 30 '18 at 19:16
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
– dpb492
2 days ago
add a comment |
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
– Sangchul Lee
Dec 30 '18 at 11:21
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
– Sangchul Lee
Dec 30 '18 at 11:28
Thank you for your editing.
– taku
Dec 30 '18 at 11:51
1
$nin mathbb{N}in mathscr{A}_n$
– d.k.o.
Dec 30 '18 at 19:16
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
– dpb492
2 days ago
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
– Sangchul Lee
Dec 30 '18 at 11:21
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
– Sangchul Lee
Dec 30 '18 at 11:21
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
– Sangchul Lee
Dec 30 '18 at 11:28
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
– Sangchul Lee
Dec 30 '18 at 11:28
Thank you for your editing.
– taku
Dec 30 '18 at 11:51
Thank you for your editing.
– taku
Dec 30 '18 at 11:51
1
1
$nin mathbb{N}in mathscr{A}_n$
– d.k.o.
Dec 30 '18 at 19:16
$nin mathbb{N}in mathscr{A}_n$
– d.k.o.
Dec 30 '18 at 19:16
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
– dpb492
2 days ago
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
– dpb492
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
The problem reduces to the following.
Let $mathcal A$ be a $sigma$-algebra on $mathbb N$. For all $kinmathbb N$, there exists a minimal element (for the inclusion) $E_k$ of $mathcal A$ containing $k$.
Indeed, define
$$
I:=left{iinmathbb Nmid exists A_iinmathcal A,kin A_i,inotin A_iright}.
$$
For each $iin I$, choose $A_iinmathcal A$ such that $kin A_i$ and $inotin A_i$. Define
$$
E_k:= bigcap_{iin I}A_i.
$$
Then $E_kinmathcal A$ and $kin E_k$. Let $Binmathcal A$ be such that $kin B$. We have to check that $E_ksubset B$. Observe that $inotin A_i$ hence
$$
E_k= bigcap_{iin I}A_isetminus {i}=bigcap_{iin I}A_icap left(mathbb Nsetminus {i}right)=E_ksetminus I.
$$
Let $jin E_k=E_ksetminus I$. We know that for all $Sinmathcal A$, we either have $knotin S$ or $jin S$. Apply this to $B$ to get that $knotin B$ or $jin B$. Since $kin B$, the only possibility is that $jin B$, which proves that $E_ksubset B$.
answered yesterday
Davide Giraudo
125k16150260
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The problem reduces to the following.
Let $mathcal A$ be a $sigma$-algebra on $mathbb N$. For all $kinmathbb N$, there exists a minimal element (for the inclusion) $E_k$ of $mathcal A$ containing $k$.
Indeed, define
$$
I:=left{iinmathbb Nmid exists A_iinmathcal A,kin A_i,inotin A_iright}.
$$
For each $iin I$, choose $A_iinmathcal A$ such that $kin A_i$ and $inotin A_i$. Define
$$
E_k:= bigcap_{iin I}A_i.
$$
Then $E_kinmathcal A$ and $kin E_k$. Let $Binmathcal A$ be such that $kin B$. We have to check that $E_ksubset B$. Observe that $inotin A_i$ hence
$$
E_k= bigcap_{iin I}A_isetminus {i}=bigcap_{iin I}A_icap left(mathbb Nsetminus {i}right)=E_ksetminus I.
$$
Let $jin E_k=E_ksetminus I$. We know that for all $Sinmathcal A$, we either have $knotin S$ or $jin S$. Apply this to $B$ to get that $knotin B$ or $jin B$. Since $kin B$, the only possibility is that $jin B$, which proves that $E_ksubset B$.
answered yesterday
Davide Giraudo
125k16150260
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
add a comment |
The problem reduces to the following.
Let $mathcal A$ be a $sigma$-algebra on $mathbb N$. For all $kinmathbb N$, there exists a minimal element (for the inclusion) $E_k$ of $mathcal A$ containing $k$.
Indeed, define
$$
I:=left{iinmathbb Nmid exists A_iinmathcal A,kin A_i,inotin A_iright}.
$$
For each $iin I$, choose $A_iinmathcal A$ such that $kin A_i$ and $inotin A_i$. Define
$$
E_k:= bigcap_{iin I}A_i.
$$
Then $E_kinmathcal A$ and $kin E_k$. Let $Binmathcal A$ be such that $kin B$. We have to check that $E_ksubset B$. Observe that $inotin A_i$ hence
$$
E_k= bigcap_{iin I}A_isetminus {i}=bigcap_{iin I}A_icap left(mathbb Nsetminus {i}right)=E_ksetminus I.
$$
Let $jin E_k=E_ksetminus I$. We know that for all $Sinmathcal A$, we either have $knotin S$ or $jin S$. Apply this to $B$ to get that $knotin B$ or $jin B$. Since $kin B$, the only possibility is that $jin B$, which proves that $E_ksubset B$.
answered yesterday
Davide Giraudo
125k16150260
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
add a comment |
The problem reduces to the following.
Let $mathcal A$ be a $sigma$-algebra on $mathbb N$. For all $kinmathbb N$, there exists a minimal element (for the inclusion) $E_k$ of $mathcal A$ containing $k$.
Indeed, define
$$
I:=left{iinmathbb Nmid exists A_iinmathcal A,kin A_i,inotin A_iright}.
$$
For each $iin I$, choose $A_iinmathcal A$ such that $kin A_i$ and $inotin A_i$. Define
$$
E_k:= bigcap_{iin I}A_i.
$$
Then $E_kinmathcal A$ and $kin E_k$. Let $Binmathcal A$ be such that $kin B$. We have to check that $E_ksubset B$. Observe that $inotin A_i$ hence
$$
E_k= bigcap_{iin I}A_isetminus {i}=bigcap_{iin I}A_icap left(mathbb Nsetminus {i}right)=E_ksetminus I.
$$
Let $jin E_k=E_ksetminus I$. We know that for all $Sinmathcal A$, we either have $knotin S$ or $jin S$. Apply this to $B$ to get that $knotin B$ or $jin B$. Since $kin B$, the only possibility is that $jin B$, which proves that $E_ksubset B$.
answered yesterday
Davide Giraudo
125k16150260
The problem reduces to the following.
Let $mathcal A$ be a $sigma$-algebra on $mathbb N$. For all $kinmathbb N$, there exists a minimal element (for the inclusion) $E_k$ of $mathcal A$ containing $k$.
Indeed, define
$$
I:=left{iinmathbb Nmid exists A_iinmathcal A,kin A_i,inotin A_iright}.
$$
For each $iin I$, choose $A_iinmathcal A$ such that $kin A_i$ and $inotin A_i$. Define
$$
E_k:= bigcap_{iin I}A_i.
$$
Then $E_kinmathcal A$ and $kin E_k$. Let $Binmathcal A$ be such that $kin B$. We have to check that $E_ksubset B$. Observe that $inotin A_i$ hence
$$
E_k= bigcap_{iin I}A_isetminus {i}=bigcap_{iin I}A_icap left(mathbb Nsetminus {i}right)=E_ksetminus I.
$$
Let $jin E_k=E_ksetminus I$. We know that for all $Sinmathcal A$, we either have $knotin S$ or $jin S$. Apply this to $B$ to get that $knotin B$ or $jin B$. Since $kin B$, the only possibility is that $jin B$, which proves that $E_ksubset B$.
answered yesterday
Davide Giraudo
125k16150260
answered yesterday
Davide Giraudo
125k16150260
answered yesterday
Davide Giraudo
125k16150260
answered yesterday
Davide Giraudo
125k16150260
125k16150260
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
add a comment |
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
– taku
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
– Davide Giraudo
yesterday
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
I see there is no problem,Sorry,but I don’t understand why for all S∈A k is not an element of S or j∈S.
– taku
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
We know that $j$ is not an element of $I$: this means that for each element $S$ of $mathcal A$, the assertion ($kin S$ and $jnotin S$) is not true.
– Davide Giraudo
23 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
I have completely understood! Thank you for your answer.
– taku
21 hours ago
add a comment |
taku is a new contributor. Be nice, and check out our Code of Conduct.
taku is a new contributor. Be nice, and check out our Code of Conduct.
taku is a new contributor. Be nice, and check out our Code of Conduct.
taku is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
– Sangchul Lee
Dec 30 '18 at 11:21
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
– Sangchul Lee
Dec 30 '18 at 11:28
Thank you for your editing.
– taku
Dec 30 '18 at 11:51
1
$nin mathbb{N}in mathscr{A}_n$
– d.k.o.
Dec 30 '18 at 19:16
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
– dpb492
2 days ago