Show that the set $D_b$ is a Dynkin-System












0












$begingroup$


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:





  1. $emptyset, X in D$.


  2. $A in D to Xsetminus A in D$.

  3. 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







share|cite|improve this question











$endgroup$








  • 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


















0












$begingroup$


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:





  1. $emptyset, X in D$.


  2. $A in D to Xsetminus A in D$.

  3. 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







share|cite|improve this question











$endgroup$








  • 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
















0












0








0





$begingroup$


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:





  1. $emptyset, X in D$.


  2. $A in D to Xsetminus A in D$.

  3. 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







share|cite|improve this question











$endgroup$




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:





  1. $emptyset, X in D$.


  2. $A in D to Xsetminus A in D$.

  3. 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 17:54









Henno Brandsma

108k347114




108k347114










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
















  • 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












1 Answer
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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:




  1. An arbitrary intersection of Dynkin systems is a Dynkin system.

  2. 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 $bigcup_{ngeqslant 1}Q_n$.






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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:




    1. An arbitrary intersection of Dynkin systems is a Dynkin system.

    2. 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 $bigcup_{ngeqslant 1}Q_n$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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:




      1. An arbitrary intersection of Dynkin systems is a Dynkin system.

      2. 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 $bigcup_{ngeqslant 1}Q_n$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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:




        1. An arbitrary intersection of Dynkin systems is a Dynkin system.

        2. 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 $bigcup_{ngeqslant 1}Q_n$.






        share|cite|improve this answer









        $endgroup$



        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:




        1. An arbitrary intersection of Dynkin systems is a Dynkin system.

        2. 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 $bigcup_{ngeqslant 1}Q_n$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 23:36









        Davide GiraudoDavide Giraudo

        126k16150261




        126k16150261






























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