Borel-Cantelli Theorem of a Finite Series of Independent Events











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Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.



We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.



However part of most of the proofs I have read do additional step where
$bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.



I don't understand why those two values are equal.










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    Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.



    We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.



    However part of most of the proofs I have read do additional step where
    $bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.



    I don't understand why those two values are equal.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.



      We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.



      However part of most of the proofs I have read do additional step where
      $bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.



      I don't understand why those two values are equal.










      share|cite|improve this question













      Let ${A_n}_{n=1}^{infty}$ be a independent sequence of events such that $sum_{n=1}^{infty}P(A_n) <infty$, then $P(A_n i.o.) =0 $.



      We have that $P(A_n i.o.) =bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)$ which I understand, its the definition.



      However part of most of the proofs I have read do additional step where
      $bigcap_{m=1}^{infty} bigcup_{n=m}^{infty} P(A_n)= lim_{mrightarrow infty} P(bigcup_{n=m}^{infty} A_n) $.



      I don't understand why those two values are equal.







      probability-theory statistics proof-explanation






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      asked Nov 15 at 16:28









      kpr62

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          Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$



          This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$



          Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.



          Can you figure out why yourself?






          share|cite|improve this answer





















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            Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$



            This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$



            Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.



            Can you figure out why yourself?






            share|cite|improve this answer

























              up vote
              0
              down vote













              Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$



              This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$



              Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.



              Can you figure out why yourself?






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$



                This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$



                Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.



                Can you figure out why yourself?






                share|cite|improve this answer












                Let $B_m=bigcup_{n=m}^{infty}A_n$ and $$B:=bigcap_{m=1}^{infty}B_mtag1$$



                This with: $$B_1supseteq B_2supseteq B_3supseteqcdotstag2$$



                Based on $(1)$ and $(2)$ that in can be shown that $P(B_n)downarrow P(B)$.



                Can you figure out why yourself?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 15 at 16:42









                drhab

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                94.8k543125






























                     

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