Why is $mathscr{A}_{infty}=bigcup_{ninmathbb{N}} mathscr{A}_n$ never a $sigma$-algebra?
$begingroup$
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 Jan 3 at 0:08
Davide Giraudo
125k16150261
asked Dec 30 '18 at 11:16
takutaku
111
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 0:08
Davide Giraudo
125k16150261
asked Dec 30 '18 at 11:16
takutaku
111
$endgroup$
$begingroup$
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:21
$begingroup$
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:28
$begingroup$
Thank you for your editing.
$endgroup$
– taku
Dec 30 '18 at 11:51
1
$begingroup$
$nin mathbb{N}in mathscr{A}_n$
$endgroup$
– d.k.o.
Dec 30 '18 at 19:16
$begingroup$
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
$endgroup$
– dpb492
Jan 3 at 1:59
add a comment |
$begingroup$
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 Jan 3 at 0:08
Davide Giraudo
125k16150261
asked Dec 30 '18 at 11:16
takutaku
111
$endgroup$
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 Jan 3 at 0:08
Davide Giraudo
125k16150261
asked Dec 30 '18 at 11:16
takutaku
111
edited Jan 3 at 0:08
Davide Giraudo
125k16150261
asked Dec 30 '18 at 11:16
takutaku
111
edited Jan 3 at 0:08
Davide Giraudo
125k16150261
edited Jan 3 at 0:08
Davide Giraudo
125k16150261
edited Jan 3 at 0:08
Davide Giraudo
125k16150261
125k16150261
asked Dec 30 '18 at 11:16
takutaku
111
asked Dec 30 '18 at 11:16
takutaku
111
asked Dec 30 '18 at 11:16
takutaku
111
111
$begingroup$
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:21
$begingroup$
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:28
$begingroup$
Thank you for your editing.
$endgroup$
– taku
Dec 30 '18 at 11:51
1
$begingroup$
$nin mathbb{N}in mathscr{A}_n$
$endgroup$
– d.k.o.
Dec 30 '18 at 19:16
$begingroup$
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
$endgroup$
– dpb492
Jan 3 at 1:59
add a comment |
$begingroup$
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:21
$begingroup$
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:28
$begingroup$
Thank you for your editing.
$endgroup$
– taku
Dec 30 '18 at 11:51
1
$begingroup$
$nin mathbb{N}in mathscr{A}_n$
$endgroup$
– d.k.o.
Dec 30 '18 at 19:16
$begingroup$
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
$endgroup$
– dpb492
Jan 3 at 1:59
$begingroup$
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:21
$begingroup$
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:21
$begingroup$
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:28
$begingroup$
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:28
$begingroup$
Thank you for your editing.
$endgroup$
– taku
Dec 30 '18 at 11:51
$begingroup$
Thank you for your editing.
$endgroup$
– taku
Dec 30 '18 at 11:51
1
1
$begingroup$
$nin mathbb{N}in mathscr{A}_n$
$endgroup$
– d.k.o.
Dec 30 '18 at 19:16
$begingroup$
$nin mathbb{N}in mathscr{A}_n$
$endgroup$
– d.k.o.
Dec 30 '18 at 19:16
$begingroup$
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
$endgroup$
– dpb492
Jan 3 at 1:59
$begingroup$
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
$endgroup$
– dpb492
Jan 3 at 1:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
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 Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
add a comment |
$begingroup$
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 Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
add a comment |
$begingroup$
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 Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
$endgroup$
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 Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
answered Jan 3 at 15:04
Davide GiraudoDavide Giraudo
125k16150261
125k16150261
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
add a comment |
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
I think the cardinality of σ-algebra is’t necessarily countable.So,I think “I” may be not well-defined.
$endgroup$
– taku
Jan 4 at 0:07
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
Actually, a $sigma$-algebra is either finite or uncountable but why would this be a problem in the definition of $I$?
$endgroup$
– Davide Giraudo
Jan 4 at 9:59
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– taku
Jan 4 at 14:09
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
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.
$endgroup$
– Davide Giraudo
Jan 4 at 14:12
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
$begingroup$
I have completely understood! Thank you for your answer.
$endgroup$
– taku
Jan 4 at 16:16
add a comment |
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$begingroup$
Welcome to MSE! Please take time to read how to format your question using MathJax, see this for instance.
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:21
$begingroup$
I also slightly changed the title and the formatting of the question. Hope this is to your liking!
$endgroup$
– Sangchul Lee
Dec 30 '18 at 11:28
$begingroup$
Thank you for your editing.
$endgroup$
– taku
Dec 30 '18 at 11:51
1
$begingroup$
$nin mathbb{N}in mathscr{A}_n$
$endgroup$
– d.k.o.
Dec 30 '18 at 19:16
$begingroup$
To echo what user d.k.o mentioned, I think there is an implicit use of the well-ordering theorem.
$endgroup$
– dpb492
Jan 3 at 1:59