Show that the set $D_b$ is a Dynkin-System
$begingroup$
let $X neq emptyset$, ${D_B} := {Q subseteq X: Q cap B in sigma(mathcal{E}) }$, where $sigma(mathcal{E}):= bigcup_{mathcal{E} subseteq mathcal{D_{i}}} D_i$ and $D_i$ is a family of Dynkin systems $ forall i in I$.
That is, $emptyset neq X, D subseteq mathcal{P}(X)$ is a Dynkin system when:
$emptyset, X in D$.
$A in D to Xsetminus A in D$.- For every sequence $(A_{n})_{ninmathbb{N}}$ of pairwise disjoint sets $A_i$ in $D$, the union $bigcup_{ninmathbb{N}} A_n in D$.
Moreover, $B in sigma(mathcal{E})$ and $ mathcal{E} subseteq mathcal{P}(X)$. What I want to prove is that $D_B$ is indeed a Dynkin System. I struggle with the second and third condition and would appreciate some help.
measure-theory elementary-set-theory
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
$endgroup$
add a comment |
$begingroup$
let $X neq emptyset$, ${D_B} := {Q subseteq X: Q cap B in sigma(mathcal{E}) }$, where $sigma(mathcal{E}):= bigcup_{mathcal{E} subseteq mathcal{D_{i}}} D_i$ and $D_i$ is a family of Dynkin systems $ forall i in I$.
That is, $emptyset neq X, D subseteq mathcal{P}(X)$ is a Dynkin system when:
$emptyset, X in D$.
$A in D to Xsetminus A in D$.- For every sequence $(A_{n})_{ninmathbb{N}}$ of pairwise disjoint sets $A_i$ in $D$, the union $bigcup_{ninmathbb{N}} A_n in D$.
Moreover, $B in sigma(mathcal{E})$ and $ mathcal{E} subseteq mathcal{P}(X)$. What I want to prove is that $D_B$ is indeed a Dynkin System. I struggle with the second and third condition and would appreciate some help.
measure-theory elementary-set-theory
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
$endgroup$
1
$begingroup$
Are you sure, that $sigma(mathcal E)$ isn't ment to be the Dynkin system generated by $mathcal E$? This would be the intersection over all Dynkin systems containing $mathcal E$. Do you need this for Dynkin's $pi$-$lambda$ lemma or sth. similar? Cause it looks a little bit like a part of the proof.
$endgroup$
– N.Beck
Dec 4 '18 at 18:39
$begingroup$
Yes, it is meant to be the Dynkin system genereated by $mathcal{E}$. Isn't that what my writings are saying?. I need this to prove that $sigma(mathcal{E}) = delta(mathcal{E})$, where $delta(mathcal{E}) := bigcup_{mathcal{E} subseteq mathcal{A_{i}}} A_i$, with $mathcal{A}$ being the $sigma$ -Algebra on X
$endgroup$
– Michael Maier
Dec 4 '18 at 18:57
add a comment |
$begingroup$
let $X neq emptyset$, ${D_B} := {Q subseteq X: Q cap B in sigma(mathcal{E}) }$, where $sigma(mathcal{E}):= bigcup_{mathcal{E} subseteq mathcal{D_{i}}} D_i$ and $D_i$ is a family of Dynkin systems $ forall i in I$.
That is, $emptyset neq X, D subseteq mathcal{P}(X)$ is a Dynkin system when:
$emptyset, X in D$.
$A in D to Xsetminus A in D$.- For every sequence $(A_{n})_{ninmathbb{N}}$ of pairwise disjoint sets $A_i$ in $D$, the union $bigcup_{ninmathbb{N}} A_n in D$.
Moreover, $B in sigma(mathcal{E})$ and $ mathcal{E} subseteq mathcal{P}(X)$. What I want to prove is that $D_B$ is indeed a Dynkin System. I struggle with the second and third condition and would appreciate some help.
measure-theory elementary-set-theory
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
$endgroup$
let $X neq emptyset$, ${D_B} := {Q subseteq X: Q cap B in sigma(mathcal{E}) }$, where $sigma(mathcal{E}):= bigcup_{mathcal{E} subseteq mathcal{D_{i}}} D_i$ and $D_i$ is a family of Dynkin systems $ forall i in I$.
That is, $emptyset neq X, D subseteq mathcal{P}(X)$ is a Dynkin system when:
$emptyset, X in D$.
$A in D to Xsetminus A in D$.- For every sequence $(A_{n})_{ninmathbb{N}}$ of pairwise disjoint sets $A_i$ in $D$, the union $bigcup_{ninmathbb{N}} A_n in D$.
Moreover, $B in sigma(mathcal{E})$ and $ mathcal{E} subseteq mathcal{P}(X)$. What I want to prove is that $D_B$ is indeed a Dynkin System. I struggle with the second and third condition and would appreciate some help.
measure-theory elementary-set-theory
measure-theory elementary-set-theory
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
edited Dec 4 '18 at 17:54
Henno Brandsma
108k347114
108k347114
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
asked Dec 4 '18 at 17:17
Michael MaierMichael Maier
859
859
1
$begingroup$
Are you sure, that $sigma(mathcal E)$ isn't ment to be the Dynkin system generated by $mathcal E$? This would be the intersection over all Dynkin systems containing $mathcal E$. Do you need this for Dynkin's $pi$-$lambda$ lemma or sth. similar? Cause it looks a little bit like a part of the proof.
$endgroup$
– N.Beck
Dec 4 '18 at 18:39
$begingroup$
Yes, it is meant to be the Dynkin system genereated by $mathcal{E}$. Isn't that what my writings are saying?. I need this to prove that $sigma(mathcal{E}) = delta(mathcal{E})$, where $delta(mathcal{E}) := bigcup_{mathcal{E} subseteq mathcal{A_{i}}} A_i$, with $mathcal{A}$ being the $sigma$ -Algebra on X
$endgroup$
– Michael Maier
Dec 4 '18 at 18:57
add a comment |
1
$begingroup$
Are you sure, that $sigma(mathcal E)$ isn't ment to be the Dynkin system generated by $mathcal E$? This would be the intersection over all Dynkin systems containing $mathcal E$. Do you need this for Dynkin's $pi$-$lambda$ lemma or sth. similar? Cause it looks a little bit like a part of the proof.
$endgroup$
– N.Beck
Dec 4 '18 at 18:39
$begingroup$
Yes, it is meant to be the Dynkin system genereated by $mathcal{E}$. Isn't that what my writings are saying?. I need this to prove that $sigma(mathcal{E}) = delta(mathcal{E})$, where $delta(mathcal{E}) := bigcup_{mathcal{E} subseteq mathcal{A_{i}}} A_i$, with $mathcal{A}$ being the $sigma$ -Algebra on X
$endgroup$
– Michael Maier
Dec 4 '18 at 18:57
1
1
$begingroup$
Are you sure, that $sigma(mathcal E)$ isn't ment to be the Dynkin system generated by $mathcal E$? This would be the intersection over all Dynkin systems containing $mathcal E$. Do you need this for Dynkin's $pi$-$lambda$ lemma or sth. similar? Cause it looks a little bit like a part of the proof.
$endgroup$
– N.Beck
Dec 4 '18 at 18:39
$begingroup$
Are you sure, that $sigma(mathcal E)$ isn't ment to be the Dynkin system generated by $mathcal E$? This would be the intersection over all Dynkin systems containing $mathcal E$. Do you need this for Dynkin's $pi$-$lambda$ lemma or sth. similar? Cause it looks a little bit like a part of the proof.
$endgroup$
– N.Beck
Dec 4 '18 at 18:39
$begingroup$
Yes, it is meant to be the Dynkin system genereated by $mathcal{E}$. Isn't that what my writings are saying?. I need this to prove that $sigma(mathcal{E}) = delta(mathcal{E})$, where $delta(mathcal{E}) := bigcup_{mathcal{E} subseteq mathcal{A_{i}}} A_i$, with $mathcal{A}$ being the $sigma$ -Algebra on X
$endgroup$
– Michael Maier
Dec 4 '18 at 18:57
$begingroup$
Yes, it is meant to be the Dynkin system genereated by $mathcal{E}$. Isn't that what my writings are saying?. I need this to prove that $sigma(mathcal{E}) = delta(mathcal{E})$, where $delta(mathcal{E}) := bigcup_{mathcal{E} subseteq mathcal{A_{i}}} A_i$, with $mathcal{A}$ being the $sigma$ -Algebra on X
$endgroup$
– Michael Maier
Dec 4 '18 at 18:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It seems that the Dynkin system generated by $mathcal E$ is the intersection of all the Dynkin systems containing $mathcal E$. Denoting by $mathcal D$ the collection of the Dynkin systems containing $mathcal E$, this means that $sigmaleft(mathcal Eright)=bigcap_{Dinmathcal D}D$. With these notations,
$$
D_B=left{Qsubseteq X: Qcap Bin bigcap_{Dinmathcal D}Dright}
=bigcap_{Dinmathcal D} left{Qsubseteq X: Qcap Bin Dright}.
$$
To conclude, we need the following two facts:
- An arbitrary intersection of Dynkin systems is a Dynkin system.
- For each Dynkin system $D$, the collection $mathcal A_D=left{Qsubseteq X: Qcap Bin Dright}$ is a Dynkin system.
Let us show the second fact. Observe that for all $Dinmathcal D$, $Binmathcal D$ hence $Xin mathcal A_D$. Let $Qinmathcal A_D$. Write
$$
left(Xsetminus Qright)cap B=Bcap Q^c=left(B^ccup left(Qcap Bright) right)^c.
$$
Since $D$ is Dynkin system, $B^cin D$ and by assumption $Qcap Bin D$ hence by the third point of the definition, the disjoint union $B^ccup left(Qcap Bright) $ belongs to $D$. Using again the second point of the definition shows that $Xsetminus Qin mathcal A_D$.
For the third condition, let $left(Q_nright)_{ngeqslant 1}$ be a pairwise disjoint sequence of elements of $mathcal A_D$. We know that $Q_ncap Bin D$ for all $n$; since the sequence $left(Q_ncap Bright)_{ngeqslant 1}$ is disjoint and consists of elements of $D$, we know that $bigcup_{ngeqslant 1}left(Q_ncap Bright)=
left(bigcup_{ngeqslant 1}Q_nright)cap B$ is an element of $D$ hence so is $bigcup_{ngeqslant 1}Q_n$.
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
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$begingroup$
It seems that the Dynkin system generated by $mathcal E$ is the intersection of all the Dynkin systems containing $mathcal E$. Denoting by $mathcal D$ the collection of the Dynkin systems containing $mathcal E$, this means that $sigmaleft(mathcal Eright)=bigcap_{Dinmathcal D}D$. With these notations,
$$
D_B=left{Qsubseteq X: Qcap Bin bigcap_{Dinmathcal D}Dright}
=bigcap_{Dinmathcal D} left{Qsubseteq X: Qcap Bin Dright}.
$$
To conclude, we need the following two facts:
- An arbitrary intersection of Dynkin systems is a Dynkin system.
- For each Dynkin system $D$, the collection $mathcal A_D=left{Qsubseteq X: Qcap Bin Dright}$ is a Dynkin system.
Let us show the second fact. Observe that for all $Dinmathcal D$, $Binmathcal D$ hence $Xin mathcal A_D$. Let $Qinmathcal A_D$. Write
$$
left(Xsetminus Qright)cap B=Bcap Q^c=left(B^ccup left(Qcap Bright) right)^c.
$$
Since $D$ is Dynkin system, $B^cin D$ and by assumption $Qcap Bin D$ hence by the third point of the definition, the disjoint union $B^ccup left(Qcap Bright) $ belongs to $D$. Using again the second point of the definition shows that $Xsetminus Qin mathcal A_D$.
For the third condition, let $left(Q_nright)_{ngeqslant 1}$ be a pairwise disjoint sequence of elements of $mathcal A_D$. We know that $Q_ncap Bin D$ for all $n$; since the sequence $left(Q_ncap Bright)_{ngeqslant 1}$ is disjoint and consists of elements of $D$, we know that $bigcup_{ngeqslant 1}left(Q_ncap Bright)=
left(bigcup_{ngeqslant 1}Q_nright)cap B$ is an element of $D$ hence so is $bigcup_{ngeqslant 1}Q_n$.
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
It seems that the Dynkin system generated by $mathcal E$ is the intersection of all the Dynkin systems containing $mathcal E$. Denoting by $mathcal D$ the collection of the Dynkin systems containing $mathcal E$, this means that $sigmaleft(mathcal Eright)=bigcap_{Dinmathcal D}D$. With these notations,
$$
D_B=left{Qsubseteq X: Qcap Bin bigcap_{Dinmathcal D}Dright}
=bigcap_{Dinmathcal D} left{Qsubseteq X: Qcap Bin Dright}.
$$
To conclude, we need the following two facts:
- An arbitrary intersection of Dynkin systems is a Dynkin system.
- For each Dynkin system $D$, the collection $mathcal A_D=left{Qsubseteq X: Qcap Bin Dright}$ is a Dynkin system.
Let us show the second fact. Observe that for all $Dinmathcal D$, $Binmathcal D$ hence $Xin mathcal A_D$. Let $Qinmathcal A_D$. Write
$$
left(Xsetminus Qright)cap B=Bcap Q^c=left(B^ccup left(Qcap Bright) right)^c.
$$
Since $D$ is Dynkin system, $B^cin D$ and by assumption $Qcap Bin D$ hence by the third point of the definition, the disjoint union $B^ccup left(Qcap Bright) $ belongs to $D$. Using again the second point of the definition shows that $Xsetminus Qin mathcal A_D$.
For the third condition, let $left(Q_nright)_{ngeqslant 1}$ be a pairwise disjoint sequence of elements of $mathcal A_D$. We know that $Q_ncap Bin D$ for all $n$; since the sequence $left(Q_ncap Bright)_{ngeqslant 1}$ is disjoint and consists of elements of $D$, we know that $bigcup_{ngeqslant 1}left(Q_ncap Bright)=
left(bigcup_{ngeqslant 1}Q_nright)cap B$ is an element of $D$ hence so is $bigcup_{ngeqslant 1}Q_n$.
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
It seems that the Dynkin system generated by $mathcal E$ is the intersection of all the Dynkin systems containing $mathcal E$. Denoting by $mathcal D$ the collection of the Dynkin systems containing $mathcal E$, this means that $sigmaleft(mathcal Eright)=bigcap_{Dinmathcal D}D$. With these notations,
$$
D_B=left{Qsubseteq X: Qcap Bin bigcap_{Dinmathcal D}Dright}
=bigcap_{Dinmathcal D} left{Qsubseteq X: Qcap Bin Dright}.
$$
To conclude, we need the following two facts:
- An arbitrary intersection of Dynkin systems is a Dynkin system.
- For each Dynkin system $D$, the collection $mathcal A_D=left{Qsubseteq X: Qcap Bin Dright}$ is a Dynkin system.
Let us show the second fact. Observe that for all $Dinmathcal D$, $Binmathcal D$ hence $Xin mathcal A_D$. Let $Qinmathcal A_D$. Write
$$
left(Xsetminus Qright)cap B=Bcap Q^c=left(B^ccup left(Qcap Bright) right)^c.
$$
Since $D$ is Dynkin system, $B^cin D$ and by assumption $Qcap Bin D$ hence by the third point of the definition, the disjoint union $B^ccup left(Qcap Bright) $ belongs to $D$. Using again the second point of the definition shows that $Xsetminus Qin mathcal A_D$.
For the third condition, let $left(Q_nright)_{ngeqslant 1}$ be a pairwise disjoint sequence of elements of $mathcal A_D$. We know that $Q_ncap Bin D$ for all $n$; since the sequence $left(Q_ncap Bright)_{ngeqslant 1}$ is disjoint and consists of elements of $D$, we know that $bigcup_{ngeqslant 1}left(Q_ncap Bright)=
left(bigcup_{ngeqslant 1}Q_nright)cap B$ is an element of $D$ hence so is $bigcup_{ngeqslant 1}Q_n$.
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
It seems that the Dynkin system generated by $mathcal E$ is the intersection of all the Dynkin systems containing $mathcal E$. Denoting by $mathcal D$ the collection of the Dynkin systems containing $mathcal E$, this means that $sigmaleft(mathcal Eright)=bigcap_{Dinmathcal D}D$. With these notations,
$$
D_B=left{Qsubseteq X: Qcap Bin bigcap_{Dinmathcal D}Dright}
=bigcap_{Dinmathcal D} left{Qsubseteq X: Qcap Bin Dright}.
$$
To conclude, we need the following two facts:
- An arbitrary intersection of Dynkin systems is a Dynkin system.
- For each Dynkin system $D$, the collection $mathcal A_D=left{Qsubseteq X: Qcap Bin Dright}$ is a Dynkin system.
Let us show the second fact. Observe that for all $Dinmathcal D$, $Binmathcal D$ hence $Xin mathcal A_D$. Let $Qinmathcal A_D$. Write
$$
left(Xsetminus Qright)cap B=Bcap Q^c=left(B^ccup left(Qcap Bright) right)^c.
$$
Since $D$ is Dynkin system, $B^cin D$ and by assumption $Qcap Bin D$ hence by the third point of the definition, the disjoint union $B^ccup left(Qcap Bright) $ belongs to $D$. Using again the second point of the definition shows that $Xsetminus Qin mathcal A_D$.
For the third condition, let $left(Q_nright)_{ngeqslant 1}$ be a pairwise disjoint sequence of elements of $mathcal A_D$. We know that $Q_ncap Bin D$ for all $n$; since the sequence $left(Q_ncap Bright)_{ngeqslant 1}$ is disjoint and consists of elements of $D$, we know that $bigcup_{ngeqslant 1}left(Q_ncap Bright)=
left(bigcup_{ngeqslant 1}Q_nright)cap B$ is an element of $D$ hence so is $bigcup_{ngeqslant 1}Q_n$.
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 4 at 23:36
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
add a comment |
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$begingroup$
Are you sure, that $sigma(mathcal E)$ isn't ment to be the Dynkin system generated by $mathcal E$? This would be the intersection over all Dynkin systems containing $mathcal E$. Do you need this for Dynkin's $pi$-$lambda$ lemma or sth. similar? Cause it looks a little bit like a part of the proof.
$endgroup$
– N.Beck
Dec 4 '18 at 18:39
$begingroup$
Yes, it is meant to be the Dynkin system genereated by $mathcal{E}$. Isn't that what my writings are saying?. I need this to prove that $sigma(mathcal{E}) = delta(mathcal{E})$, where $delta(mathcal{E}) := bigcup_{mathcal{E} subseteq mathcal{A_{i}}} A_i$, with $mathcal{A}$ being the $sigma$ -Algebra on X
$endgroup$
– Michael Maier
Dec 4 '18 at 18:57