Can we approximate a sigma-algebra by an increasing sequence of finite sigma-algebras?

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Definition.
We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $setE_n_nin N$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.




Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $setE_n_nin N$, then for all $Fin mc F$, there if $Gin mc G$ such that $mu(FDelta G)=0$?





measure-theory




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    up vote
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    $newcommandNmathbb N$
    $newcommandmcmathcal$
    $newcommandset[1]#1$
    Definition.
    We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $setE_n_nin N$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.




    Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $setE_n_nin N$, then for all $Fin mc F$, there if $Gin mc G$ such that $mu(FDelta G)=0$?





    measure-theory




    share|cite|improve this question





















      up vote
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      up vote
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      down vote

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      $newcommandNmathbb N$
      $newcommandmcmathcal$
      $newcommandset[1]#1$
      Definition.
      We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $setE_n_nin N$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.




      Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $setE_n_nin N$, then for all $Fin mc F$, there if $Gin mc G$ such that $mu(FDelta G)=0$?





      measure-theory




      share|cite|improve this question











      $newcommandNmathbb N$
      $newcommandmcmathcal$
      $newcommandset[1]#1$
      Definition.
      We say that a measure space $(X, mc F, mu)$ has a countable basis if there is a collection $setE_n_nin N$ of measurable subsets of $X$ such that for all $varepsilon>0$ and for all $Ein mc F$, there is $n$ such that $mu(E_nDelta E)< varepsilon$.




      Question. With notation as in the above definition, is it true that if $mc G$ denotes the $sigma$-algebra generated by $setE_n_nin N$, then for all $Fin mc F$, there if $Gin mc G$ such that $mu(FDelta G)=0$?






      measure-theory





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      asked Jul 23 at 5:38









      caffeinemachine

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          Let $mu (E_n_k Delta E)<frac 1 2^k$ for each $k$. Then $sum_k mu (E_n_k Delta E)<infty $ so $mu ( lim sup_k (E_n_k Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_n_k$.






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            accepted










            Let $mu (E_n_k Delta E)<frac 1 2^k$ for each $k$. Then $sum_k mu (E_n_k Delta E)<infty $ so $mu ( lim sup_k (E_n_k Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_n_k$.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              Let $mu (E_n_k Delta E)<frac 1 2^k$ for each $k$. Then $sum_k mu (E_n_k Delta E)<infty $ so $mu ( lim sup_k (E_n_k Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_n_k$.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Let $mu (E_n_k Delta E)<frac 1 2^k$ for each $k$. Then $sum_k mu (E_n_k Delta E)<infty $ so $mu ( lim sup_k (E_n_k Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_n_k$.






                share|cite|improve this answer













                Let $mu (E_n_k Delta E)<frac 1 2^k$ for each $k$. Then $sum_k mu (E_n_k Delta E)<infty $ so $mu ( lim sup_k (E_n_k Delta E))=0$. This implies $mu (FDelta E)=0$ where $F=lim sup E_n_k$.







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                answered Jul 23 at 7:26









                Kavi Rama Murthy

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