Completion of a $sigma$-algebra is a $sigma$-algebra












2












$begingroup$



Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
begin{align*}
mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
Prove that $mathcal{G}$ is a $sigma$-algebra.




I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?



I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.






probability-theory measure-theory elementary-set-theory







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
    begin{align*}
    mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
    Prove that $mathcal{G}$ is a $sigma$-algebra.




    I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?



    I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.






    probability-theory measure-theory elementary-set-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
      begin{align*}
      mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
      Prove that $mathcal{G}$ is a $sigma$-algebra.




      I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?



      I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.






      probability-theory measure-theory elementary-set-theory







      share|cite|improve this question











      $endgroup$





      Let $(Omega,mathcal{F},mathbb{P})$ be a probability space. Let
      begin{align*}
      mathcal{N}&= left{NsubseteqOmega:exists Finmathcal{F}, Nsubseteq F,mathbb{P}(F)=0right} \ mathcal{G}&=left{Acup N:Ainmathcal{F},Ninmathcal{N}right} end{align*}
      Prove that $mathcal{G}$ is a $sigma$-algebra.




      I've shown that $Omegainmathcal{G}$ and that if $(A_n)_nsubseteqmathcal{G}$ then $cup_n A_ninmathcal{G}$. How can I show that if $Ainmathcal{G}$ then $A^cinmathcal{G}$?



      I also noticed $mathcal{F}subseteqmathcal{G}$ if it helps somehow.






      probability-theory measure-theory elementary-set-theory




      probability-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 Jan 22 at 12:35









      Davide Giraudo

      128k17156268




      128k17156268










      asked Jan 18 at 17:42









      J. DoeJ. Doe

      16613




      16613






















          2 Answers
          2






          active

          oldest

          votes


















          3





          +50







          $begingroup$

          Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,



          $$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$



          As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that



          $$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$



          here we have used the general fact that



          $$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that



          $$G^c = tilde{A} cup tilde{N} tag{2}$$



          for



          $$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$



          As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is $(1)$ true? @saz
            $endgroup$
            – J. Doe
            Jan 22 at 10:26








          • 1




            $begingroup$
            @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
            $endgroup$
            – saz
            Jan 23 at 7:10



















          2












          $begingroup$

          Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.



          Then
          $$
          begin{aligned}
          A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
          A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
          end{aligned}
          $$
          so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
          $$
          0le
          Bbb P(A^c-(A^ccap F^c))
          =
          Bbb P(A^ccap F^{cc})
          =
          Bbb P(A^ccap F)
          le
          Bbb P(F)
          =0
          .
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:07








          • 1




            $begingroup$
            $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
            $endgroup$
            – dan_fulea
            Jan 18 at 18:11










          • $begingroup$
            Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:23












          • $begingroup$
            $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
            $endgroup$
            – dan_fulea
            Jan 18 at 18:27












          • $begingroup$
            We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
            $endgroup$
            – J. Doe
            Jan 19 at 9:59














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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3





          +50







          $begingroup$

          Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,



          $$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$



          As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that



          $$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$



          here we have used the general fact that



          $$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that



          $$G^c = tilde{A} cup tilde{N} tag{2}$$



          for



          $$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$



          As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is $(1)$ true? @saz
            $endgroup$
            – J. Doe
            Jan 22 at 10:26








          • 1




            $begingroup$
            @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
            $endgroup$
            – saz
            Jan 23 at 7:10
















          3





          +50







          $begingroup$

          Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,



          $$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$



          As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that



          $$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$



          here we have used the general fact that



          $$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that



          $$G^c = tilde{A} cup tilde{N} tag{2}$$



          for



          $$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$



          As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why is $(1)$ true? @saz
            $endgroup$
            – J. Doe
            Jan 22 at 10:26








          • 1




            $begingroup$
            @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
            $endgroup$
            – saz
            Jan 23 at 7:10














          3





          +50







          3





          +50



          3




          +50



          $begingroup$

          Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,



          $$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$



          As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that



          $$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$



          here we have used the general fact that



          $$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that



          $$G^c = tilde{A} cup tilde{N} tag{2}$$



          for



          $$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$



          As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.






          share|cite|improve this answer











          $endgroup$



          Let $G=A cup N$ for $A in mathcal{F}$ and $N in mathcal{N}$. By definition, there exists $F in mathcal{F}$ such that $N subseteq F$ and $mathbb{P}(F)=0$. Clearly,



          $$G^c = A^c cap N^c = underbrace{A^c}_W backslash underbrace{(N cap A^c)}_{U}$$



          As $$underbrace{N cap A^c}_{U} subseteq underbrace{F cap A^c}_{V} subseteq underbrace{A^c}_{W}$$ it follows that



          $$G^c = underbrace{big(A^c backslash (F cap A^c)big)}_{W backslash V} cup underbrace{big( (F cap A^c) backslash (N cap A^c) big)}_{V backslash U}; tag{1}$$



          here we have used the general fact that



          $$U subseteq V subseteq W implies W backslash U = (W backslash V) cup (V backslash U).$$ Consequently, we have shown that



          $$G^c = tilde{A} cup tilde{N} tag{2}$$



          for



          $$tilde{A} := A^c backslash (F cap A^c)quad text{and} quad tilde{N}:= (F cap A^c) backslash (N cap A^c).$$



          As $tilde{N} subseteq F in mathcal{F}$ and $mathbb{P}(F)=0$, we have $tilde{N} in mathcal{N}$. Morover, $A in mathcal{F}$ and $F in mathcal{F}$ implies $tilde{A} in mathcal{F}$. Consequently, $(2)$ shows that $G^c in mathcal{G}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 23 at 7:53

























          answered Jan 21 at 18:51









          sazsaz

          82.7k863132




          82.7k863132












          • $begingroup$
            Why is $(1)$ true? @saz
            $endgroup$
            – J. Doe
            Jan 22 at 10:26








          • 1




            $begingroup$
            @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
            $endgroup$
            – saz
            Jan 23 at 7:10


















          • $begingroup$
            Why is $(1)$ true? @saz
            $endgroup$
            – J. Doe
            Jan 22 at 10:26








          • 1




            $begingroup$
            @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
            $endgroup$
            – saz
            Jan 23 at 7:10
















          $begingroup$
          Why is $(1)$ true? @saz
          $endgroup$
          – J. Doe
          Jan 22 at 10:26






          $begingroup$
          Why is $(1)$ true? @saz
          $endgroup$
          – J. Doe
          Jan 22 at 10:26






          1




          1




          $begingroup$
          @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
          $endgroup$
          – saz
          Jan 23 at 7:10




          $begingroup$
          @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now.
          $endgroup$
          – saz
          Jan 23 at 7:10











          2












          $begingroup$

          Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.



          Then
          $$
          begin{aligned}
          A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
          A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
          end{aligned}
          $$
          so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
          $$
          0le
          Bbb P(A^c-(A^ccap F^c))
          =
          Bbb P(A^ccap F^{cc})
          =
          Bbb P(A^ccap F)
          le
          Bbb P(F)
          =0
          .
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:07








          • 1




            $begingroup$
            $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
            $endgroup$
            – dan_fulea
            Jan 18 at 18:11










          • $begingroup$
            Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:23












          • $begingroup$
            $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
            $endgroup$
            – dan_fulea
            Jan 18 at 18:27












          • $begingroup$
            We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
            $endgroup$
            – J. Doe
            Jan 19 at 9:59


















          2












          $begingroup$

          Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.



          Then
          $$
          begin{aligned}
          A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
          A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
          end{aligned}
          $$
          so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
          $$
          0le
          Bbb P(A^c-(A^ccap F^c))
          =
          Bbb P(A^ccap F^{cc})
          =
          Bbb P(A^ccap F)
          le
          Bbb P(F)
          =0
          .
          $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:07








          • 1




            $begingroup$
            $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
            $endgroup$
            – dan_fulea
            Jan 18 at 18:11










          • $begingroup$
            Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:23












          • $begingroup$
            $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
            $endgroup$
            – dan_fulea
            Jan 18 at 18:27












          • $begingroup$
            We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
            $endgroup$
            – J. Doe
            Jan 19 at 9:59
















          2












          2








          2





          $begingroup$

          Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.



          Then
          $$
          begin{aligned}
          A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
          A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
          end{aligned}
          $$
          so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
          $$
          0le
          Bbb P(A^c-(A^ccap F^c))
          =
          Bbb P(A^ccap F^{cc})
          =
          Bbb P(A^ccap F)
          le
          Bbb P(F)
          =0
          .
          $$






          share|cite|improve this answer









          $endgroup$



          Let $G$ be an event in $mathcal G$, so we can write it in the form $G=Acup N$ with $Ainmathcal F$, $Ninmathcal N$, so $Nsubseteq F$, for some suitable $Fin mathcal F$, $Bbb P(F)=0$.



          Then
          $$
          begin{aligned}
          A &subseteq {color{red}{G}}=Acup Nsubseteq Acup Ftext{ leads to}\
          A^c &supseteq {color{red}{G^c}}=A^ccap N^csupseteq A^ccap F^c ,
          end{aligned}
          $$
          so $G^c$ lies between two events in $mathcal G$ that differ by a null set,
          $$
          0le
          Bbb P(A^c-(A^ccap F^c))
          =
          Bbb P(A^ccap F^{cc})
          =
          Bbb P(A^ccap F)
          le
          Bbb P(F)
          =0
          .
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 17:54









          dan_fuleadan_fulea

          7,1981513




          7,1981513












          • $begingroup$
            I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:07








          • 1




            $begingroup$
            $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
            $endgroup$
            – dan_fulea
            Jan 18 at 18:11










          • $begingroup$
            Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:23












          • $begingroup$
            $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
            $endgroup$
            – dan_fulea
            Jan 18 at 18:27












          • $begingroup$
            We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
            $endgroup$
            – J. Doe
            Jan 19 at 9:59




















          • $begingroup$
            I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:07








          • 1




            $begingroup$
            $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
            $endgroup$
            – dan_fulea
            Jan 18 at 18:11










          • $begingroup$
            Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
            $endgroup$
            – J. Doe
            Jan 18 at 18:23












          • $begingroup$
            $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
            $endgroup$
            – dan_fulea
            Jan 18 at 18:27












          • $begingroup$
            We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
            $endgroup$
            – J. Doe
            Jan 19 at 9:59


















          $begingroup$
          I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
          $endgroup$
          – J. Doe
          Jan 18 at 18:07






          $begingroup$
          I got that $A^ccap F^csubseteq G^csubseteq A^c$, but how does $mathbb{P}(A^ccap F)=0$ leads us to $G^cinmathcal{G}$? @dan_fulea
          $endgroup$
          – J. Doe
          Jan 18 at 18:07






          1




          1




          $begingroup$
          $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
          $endgroup$
          – dan_fulea
          Jan 18 at 18:11




          $begingroup$
          $G^c$ is then between $A_1:=A^ccap F^c$ and $A_1cup N_1$ for that $N_1$ that makes $A_1cup N_1=A^c$.
          $endgroup$
          – dan_fulea
          Jan 18 at 18:11












          $begingroup$
          Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
          $endgroup$
          – J. Doe
          Jan 18 at 18:23






          $begingroup$
          Sorry for stubborning but I still don't get it. We have $A_1=A^ccap F^csubseteq G^csubseteq A^c=A_1cup N_1$, but what does $G^c$ equal to for getting into the set $mathcal{G}$? @dan_fulea
          $endgroup$
          – J. Doe
          Jan 18 at 18:23














          $begingroup$
          $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
          $endgroup$
          – dan_fulea
          Jan 18 at 18:27






          $begingroup$
          $G^c=A_1cupcolor{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point...
          $endgroup$
          – dan_fulea
          Jan 18 at 18:27














          $begingroup$
          We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
          $endgroup$
          – J. Doe
          Jan 19 at 9:59






          $begingroup$
          We need that $G^c-A_1=(A^ccap N^c)-(A^ccap F^c)inmathcal{N}$. Why is that true? I observe that $G^c-A_1subseteq A^ccap N^c$ but not sure if it helps. @dan_fulea
          $endgroup$
          – J. Doe
          Jan 19 at 9:59




















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