Proof that $P(A_{n} cap B)rightarrow P(B)$ given $A_{n} nearrow A$ and $P(A)=1$












1












$begingroup$


Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.



Proof that $P(A_{n} cap B)rightarrow P(B)$



This is how I am starting the answer:



Let $B = cup (A_{n}cap B)$



Is it the right way?



Any help. I am really stuck in here.






probability measure-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How is the title of your question related to the question itself...?
    $endgroup$
    – saz
    Jan 13 at 11:54
















1












$begingroup$


Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.



Proof that $P(A_{n} cap B)rightarrow P(B)$



This is how I am starting the answer:



Let $B = cup (A_{n}cap B)$



Is it the right way?



Any help. I am really stuck in here.






probability measure-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How is the title of your question related to the question itself...?
    $endgroup$
    – saz
    Jan 13 at 11:54














1












1








1


2



$begingroup$


Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.



Proof that $P(A_{n} cap B)rightarrow P(B)$



This is how I am starting the answer:



Let $B = cup (A_{n}cap B)$



Is it the right way?



Any help. I am really stuck in here.






probability measure-theory







share|cite|improve this question











$endgroup$




Let $A$, $B$ and $A_{1},A_{2},ldots,A_{n}$ aleatory events in a sample space, with $A_{n} nearrow A$, and $P(A)=1$.



Proof that $P(A_{n} cap B)rightarrow P(B)$



This is how I am starting the answer:



Let $B = cup (A_{n}cap B)$



Is it the right way?



Any help. I am really stuck in here.






probability measure-theory




probability measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 10:42









Lee David Chung Lin

4,26031141




4,26031141










asked Jan 9 at 1:10









LauraLaura

2408




2408








  • 1




    $begingroup$
    How is the title of your question related to the question itself...?
    $endgroup$
    – saz
    Jan 13 at 11:54














  • 1




    $begingroup$
    How is the title of your question related to the question itself...?
    $endgroup$
    – saz
    Jan 13 at 11:54








1




1




$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54




$begingroup$
How is the title of your question related to the question itself...?
$endgroup$
– saz
Jan 13 at 11:54










2 Answers
2






active

oldest

votes


















2





+50







$begingroup$

Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).



Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
      $endgroup$
      – Laura
      Jan 9 at 1:28













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2





    +50







    $begingroup$

    Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).



    Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}






    share|cite|improve this answer









    $endgroup$


















      2





      +50







      $begingroup$

      Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).



      Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}






      share|cite|improve this answer









      $endgroup$
















        2





        +50







        2





        +50



        2




        +50



        $begingroup$

        Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).



        Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}






        share|cite|improve this answer









        $endgroup$



        Elaborating on the answer above, note that since $A_n nearrow A$, then $(A_n cap B) nearrow (A cap B)$. Proving this should be straightforward (e.g. from elementary set theory).



        Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c cap B) leq P(A^c) = 0$, we see that begin{align}P(B) &= P(A cap B) + P(A^c cap B) \&= P(A cap B).end{align} Putting it all together, we see that begin{align} Pbigg( bigcup_i A_i cap B bigg) &= lim_i P(A_i cap B) \ &= P( A cap B)\&= P(B).end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 17:26









        OldGodzillaOldGodzilla

        58227




        58227























            3












            $begingroup$

            Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
              $endgroup$
              – Laura
              Jan 9 at 1:28


















            3












            $begingroup$

            Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
              $endgroup$
              – Laura
              Jan 9 at 1:28
















            3












            3








            3





            $begingroup$

            Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.






            share|cite|improve this answer









            $endgroup$



            Just show $(A_n cap B) nearrow (A cap B)$ and note $P(A cap B) = P(B)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 1:17









            angryavianangryavian

            40.9k23380




            40.9k23380












            • $begingroup$
              Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
              $endgroup$
              – Laura
              Jan 9 at 1:28




















            • $begingroup$
              Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
              $endgroup$
              – Laura
              Jan 9 at 1:28


















            $begingroup$
            Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
            $endgroup$
            – Laura
            Jan 9 at 1:28






            $begingroup$
            Thanks @angryavian This was my Idea $B=∪(An∩B)$ Could give me a formaly answer?
            $endgroup$
            – Laura
            Jan 9 at 1:28




















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