Completion of $mathbb{E}$ by $mathbb{N}_mu$ regarding Lebesgue measure - are my arguments valid












0












$begingroup$


I have the measure space $(X,mathbb{E},mu)$ where $mathbb{E}$ is a $sigma$-algebra on $X$ and $mu$ is the Lebesgue measure.



I have that
$$mathbb{N}_mu={Esubseteq X; |; text{there exist }Einmathbb{E};text{so};Nsubseteq E;;text{and};mu(E)=0}inmathbb{N}_mu},$$
so is the following argumentation valid:



The enlargement of $mathbb{E}$ with respect to $mu$ is:
$$mathbb{E}_mu=mathbb{E}cupmathbb{N}_mu,$$
so
$$mathbb{E}subseteq mathbb{E}_muquadquadtext{and}quadquadmathbb{N}_musubseteq mathbb{E}_mu$$
as
$mathbb{E}=mathbb{E}_mu$ ${mathbb{N}_mu}$ and $mathbb{N}_mu=mathbb{E}_mu$ ${mathbb{E}}$.






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I misunderstood what you did, yes this is basically correct.
    $endgroup$
    – Ian
    Dec 1 '18 at 18:58












  • $begingroup$
    Except there are some places where you are sloppy...for example you really want to add sets $B$ such that there is a measurable set $A$ and a measurable set $C$ with $mu(C)=0$ and $B setminus A subseteq C$.
    $endgroup$
    – Ian
    Dec 1 '18 at 19:11










  • $begingroup$
    your definition of $Bbb N_mu$ seems wrong. I guess that $Bbb N_mu$ want to mean the collection of subsets of any subset of the original $sigma$-algebra of measure zero. If this is the case you must write in the definition $Esubseteq N$ instead of $Nsubseteq E$
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:28












  • $begingroup$
    there are some more typos, but the assertions are fine besides that. You want to write $Bbb E_musetminusBbb N_mu$ and $Bbb E_musetminusBbb E$ instead of what you wrote
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:35












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Frederik
    Dec 2 '18 at 9:30
















0












$begingroup$


I have the measure space $(X,mathbb{E},mu)$ where $mathbb{E}$ is a $sigma$-algebra on $X$ and $mu$ is the Lebesgue measure.



I have that
$$mathbb{N}_mu={Esubseteq X; |; text{there exist }Einmathbb{E};text{so};Nsubseteq E;;text{and};mu(E)=0}inmathbb{N}_mu},$$
so is the following argumentation valid:



The enlargement of $mathbb{E}$ with respect to $mu$ is:
$$mathbb{E}_mu=mathbb{E}cupmathbb{N}_mu,$$
so
$$mathbb{E}subseteq mathbb{E}_muquadquadtext{and}quadquadmathbb{N}_musubseteq mathbb{E}_mu$$
as
$mathbb{E}=mathbb{E}_mu$ ${mathbb{N}_mu}$ and $mathbb{N}_mu=mathbb{E}_mu$ ${mathbb{E}}$.






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I misunderstood what you did, yes this is basically correct.
    $endgroup$
    – Ian
    Dec 1 '18 at 18:58












  • $begingroup$
    Except there are some places where you are sloppy...for example you really want to add sets $B$ such that there is a measurable set $A$ and a measurable set $C$ with $mu(C)=0$ and $B setminus A subseteq C$.
    $endgroup$
    – Ian
    Dec 1 '18 at 19:11










  • $begingroup$
    your definition of $Bbb N_mu$ seems wrong. I guess that $Bbb N_mu$ want to mean the collection of subsets of any subset of the original $sigma$-algebra of measure zero. If this is the case you must write in the definition $Esubseteq N$ instead of $Nsubseteq E$
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:28












  • $begingroup$
    there are some more typos, but the assertions are fine besides that. You want to write $Bbb E_musetminusBbb N_mu$ and $Bbb E_musetminusBbb E$ instead of what you wrote
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:35












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Frederik
    Dec 2 '18 at 9:30














0












0








0





$begingroup$


I have the measure space $(X,mathbb{E},mu)$ where $mathbb{E}$ is a $sigma$-algebra on $X$ and $mu$ is the Lebesgue measure.



I have that
$$mathbb{N}_mu={Esubseteq X; |; text{there exist }Einmathbb{E};text{so};Nsubseteq E;;text{and};mu(E)=0}inmathbb{N}_mu},$$
so is the following argumentation valid:



The enlargement of $mathbb{E}$ with respect to $mu$ is:
$$mathbb{E}_mu=mathbb{E}cupmathbb{N}_mu,$$
so
$$mathbb{E}subseteq mathbb{E}_muquadquadtext{and}quadquadmathbb{N}_musubseteq mathbb{E}_mu$$
as
$mathbb{E}=mathbb{E}_mu$ ${mathbb{N}_mu}$ and $mathbb{N}_mu=mathbb{E}_mu$ ${mathbb{E}}$.






measure-theory lebesgue-measure







share|cite|improve this question











$endgroup$




I have the measure space $(X,mathbb{E},mu)$ where $mathbb{E}$ is a $sigma$-algebra on $X$ and $mu$ is the Lebesgue measure.



I have that
$$mathbb{N}_mu={Esubseteq X; |; text{there exist }Einmathbb{E};text{so};Nsubseteq E;;text{and};mu(E)=0}inmathbb{N}_mu},$$
so is the following argumentation valid:



The enlargement of $mathbb{E}$ with respect to $mu$ is:
$$mathbb{E}_mu=mathbb{E}cupmathbb{N}_mu,$$
so
$$mathbb{E}subseteq mathbb{E}_muquadquadtext{and}quadquadmathbb{N}_musubseteq mathbb{E}_mu$$
as
$mathbb{E}=mathbb{E}_mu$ ${mathbb{N}_mu}$ and $mathbb{N}_mu=mathbb{E}_mu$ ${mathbb{E}}$.






measure-theory lebesgue-measure




measure-theory lebesgue-measure






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 9:26







Frederik

















asked Dec 1 '18 at 18:40









FrederikFrederik

1009




1009








  • 1




    $begingroup$
    I misunderstood what you did, yes this is basically correct.
    $endgroup$
    – Ian
    Dec 1 '18 at 18:58












  • $begingroup$
    Except there are some places where you are sloppy...for example you really want to add sets $B$ such that there is a measurable set $A$ and a measurable set $C$ with $mu(C)=0$ and $B setminus A subseteq C$.
    $endgroup$
    – Ian
    Dec 1 '18 at 19:11










  • $begingroup$
    your definition of $Bbb N_mu$ seems wrong. I guess that $Bbb N_mu$ want to mean the collection of subsets of any subset of the original $sigma$-algebra of measure zero. If this is the case you must write in the definition $Esubseteq N$ instead of $Nsubseteq E$
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:28












  • $begingroup$
    there are some more typos, but the assertions are fine besides that. You want to write $Bbb E_musetminusBbb N_mu$ and $Bbb E_musetminusBbb E$ instead of what you wrote
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:35












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Frederik
    Dec 2 '18 at 9:30














  • 1




    $begingroup$
    I misunderstood what you did, yes this is basically correct.
    $endgroup$
    – Ian
    Dec 1 '18 at 18:58












  • $begingroup$
    Except there are some places where you are sloppy...for example you really want to add sets $B$ such that there is a measurable set $A$ and a measurable set $C$ with $mu(C)=0$ and $B setminus A subseteq C$.
    $endgroup$
    – Ian
    Dec 1 '18 at 19:11










  • $begingroup$
    your definition of $Bbb N_mu$ seems wrong. I guess that $Bbb N_mu$ want to mean the collection of subsets of any subset of the original $sigma$-algebra of measure zero. If this is the case you must write in the definition $Esubseteq N$ instead of $Nsubseteq E$
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:28












  • $begingroup$
    there are some more typos, but the assertions are fine besides that. You want to write $Bbb E_musetminusBbb N_mu$ and $Bbb E_musetminusBbb E$ instead of what you wrote
    $endgroup$
    – Masacroso
    Dec 1 '18 at 19:35












  • $begingroup$
    Thank you for your help!
    $endgroup$
    – Frederik
    Dec 2 '18 at 9:30








1




1




$begingroup$
I misunderstood what you did, yes this is basically correct.
$endgroup$
– Ian
Dec 1 '18 at 18:58






$begingroup$
I misunderstood what you did, yes this is basically correct.
$endgroup$
– Ian
Dec 1 '18 at 18:58














$begingroup$
Except there are some places where you are sloppy...for example you really want to add sets $B$ such that there is a measurable set $A$ and a measurable set $C$ with $mu(C)=0$ and $B setminus A subseteq C$.
$endgroup$
– Ian
Dec 1 '18 at 19:11




$begingroup$
Except there are some places where you are sloppy...for example you really want to add sets $B$ such that there is a measurable set $A$ and a measurable set $C$ with $mu(C)=0$ and $B setminus A subseteq C$.
$endgroup$
– Ian
Dec 1 '18 at 19:11












$begingroup$
your definition of $Bbb N_mu$ seems wrong. I guess that $Bbb N_mu$ want to mean the collection of subsets of any subset of the original $sigma$-algebra of measure zero. If this is the case you must write in the definition $Esubseteq N$ instead of $Nsubseteq E$
$endgroup$
– Masacroso
Dec 1 '18 at 19:28






$begingroup$
your definition of $Bbb N_mu$ seems wrong. I guess that $Bbb N_mu$ want to mean the collection of subsets of any subset of the original $sigma$-algebra of measure zero. If this is the case you must write in the definition $Esubseteq N$ instead of $Nsubseteq E$
$endgroup$
– Masacroso
Dec 1 '18 at 19:28














$begingroup$
there are some more typos, but the assertions are fine besides that. You want to write $Bbb E_musetminusBbb N_mu$ and $Bbb E_musetminusBbb E$ instead of what you wrote
$endgroup$
– Masacroso
Dec 1 '18 at 19:35






$begingroup$
there are some more typos, but the assertions are fine besides that. You want to write $Bbb E_musetminusBbb N_mu$ and $Bbb E_musetminusBbb E$ instead of what you wrote
$endgroup$
– Masacroso
Dec 1 '18 at 19:35














$begingroup$
Thank you for your help!
$endgroup$
– Frederik
Dec 2 '18 at 9:30




$begingroup$
Thank you for your help!
$endgroup$
– Frederik
Dec 2 '18 at 9:30










1 Answer
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If you start with $(X,mathcal{M},mu)$, the completed $sigma$-algebra is



$$mathcal{M}^*={ B subseteq X : text{ there exists } A,C in mathcal{M} : mu(C)=0 wedge B setminus A subseteq C }.$$



Note that fixing $C=emptyset$ inside that set builder reproduces $mathcal{M}$ itself, so no union is required. Similarly fixing $A=emptyset$ reproduces all of the actual null sets in the completed space.



This is not quite what you wrote, what you wrote has some minor errors that make it rather nonsensical (for example $E$ being bound under two different quantifiers in the same expression).






share|cite|improve this answer











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    $begingroup$

    If you start with $(X,mathcal{M},mu)$, the completed $sigma$-algebra is



    $$mathcal{M}^*={ B subseteq X : text{ there exists } A,C in mathcal{M} : mu(C)=0 wedge B setminus A subseteq C }.$$



    Note that fixing $C=emptyset$ inside that set builder reproduces $mathcal{M}$ itself, so no union is required. Similarly fixing $A=emptyset$ reproduces all of the actual null sets in the completed space.



    This is not quite what you wrote, what you wrote has some minor errors that make it rather nonsensical (for example $E$ being bound under two different quantifiers in the same expression).






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If you start with $(X,mathcal{M},mu)$, the completed $sigma$-algebra is



      $$mathcal{M}^*={ B subseteq X : text{ there exists } A,C in mathcal{M} : mu(C)=0 wedge B setminus A subseteq C }.$$



      Note that fixing $C=emptyset$ inside that set builder reproduces $mathcal{M}$ itself, so no union is required. Similarly fixing $A=emptyset$ reproduces all of the actual null sets in the completed space.



      This is not quite what you wrote, what you wrote has some minor errors that make it rather nonsensical (for example $E$ being bound under two different quantifiers in the same expression).






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If you start with $(X,mathcal{M},mu)$, the completed $sigma$-algebra is



        $$mathcal{M}^*={ B subseteq X : text{ there exists } A,C in mathcal{M} : mu(C)=0 wedge B setminus A subseteq C }.$$



        Note that fixing $C=emptyset$ inside that set builder reproduces $mathcal{M}$ itself, so no union is required. Similarly fixing $A=emptyset$ reproduces all of the actual null sets in the completed space.



        This is not quite what you wrote, what you wrote has some minor errors that make it rather nonsensical (for example $E$ being bound under two different quantifiers in the same expression).






        share|cite|improve this answer











        $endgroup$



        If you start with $(X,mathcal{M},mu)$, the completed $sigma$-algebra is



        $$mathcal{M}^*={ B subseteq X : text{ there exists } A,C in mathcal{M} : mu(C)=0 wedge B setminus A subseteq C }.$$



        Note that fixing $C=emptyset$ inside that set builder reproduces $mathcal{M}$ itself, so no union is required. Similarly fixing $A=emptyset$ reproduces all of the actual null sets in the completed space.



        This is not quite what you wrote, what you wrote has some minor errors that make it rather nonsensical (for example $E$ being bound under two different quantifiers in the same expression).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 19:40

























        answered Dec 1 '18 at 18:46









        IanIan

        67.6k25387




        67.6k25387






























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