Borel-Cantelli-like inequality

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I’m wondering if the following is true:



If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.



Here $m$ is the Lebesgue measure.







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    up vote
    0
    down vote

    favorite












    I’m wondering if the following is true:



    If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.



    Here $m$ is the Lebesgue measure.







    share|cite|improve this question






















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I’m wondering if the following is true:



      If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.



      Here $m$ is the Lebesgue measure.







      share|cite|improve this question












      I’m wondering if the following is true:



      If $A_n subseteq [0,1]$ satisfies $m(A_n) geq epsilon > 0$, then $m(limsup A_n) > 0$.



      Here $m$ is the Lebesgue measure.









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 17 at 2:28









      Seh-kai

      1027




      1027




















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          Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$






            share|cite|improve this answer
























              up vote
              3
              down vote



              accepted










              Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$






              share|cite|improve this answer






















                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$






                share|cite|improve this answer












                Yes. Let $B_n = bigcup_kge n A_k.$ Then we have $B_1 supseteq B_2supseteqldots$ and $$m(B_n)ge m(A_n)ge epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ mleft(limsup_j A_j right)=mleft(bigcap_nge 1 B_nright) = lim_n m(B_n)ge epsilon$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 17 at 3:51









                spaceisdarkgreen

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