If $m^*(E)=0$, prove that $m^*(E_n)=0$ for each $n$












0














Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
.



Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.



My attempt:



Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.



QED





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  • 3




    Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
    – T. Bongers
    Dec 3 '18 at 0:10
















0














Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
.



Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.



My attempt:



Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.



QED





Is this correct?










share|cite|improve this question


















  • 3




    Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
    – T. Bongers
    Dec 3 '18 at 0:10














0












0








0







Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
.



Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.



My attempt:



Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.



QED





Is this correct?










share|cite|improve this question













Define $displaystyle m^*(E)=infleft{sum_{n=1}^inftyell(I_n):Esubsetbigcup
_{n=1}^infty I_nright}$
.



Let $E=bigcup_{n=1}^infty E_n$. Suppose $m^*(E)=0$. Prove that $m^*(E_n)=0$ for every $n$.



My attempt:



Since $m^*(E)=0$, $E$ is a countable set. This means each of the $E_n$ is also countable and so $m^*(E_n)=0$ for all $n$.



QED





Is this correct?







real-analysis proof-verification






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asked Dec 3 '18 at 0:04









Thomas

713416




713416








  • 3




    Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
    – T. Bongers
    Dec 3 '18 at 0:10














  • 3




    Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
    – T. Bongers
    Dec 3 '18 at 0:10








3




3




Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10




Not at all correct. Many uncountable sets have outer measure zero, e.g. many Cantor sets.
– T. Bongers
Dec 3 '18 at 0:10










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$E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.






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    1 Answer
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    1 Answer
    1






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    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2














    $E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.






    share|cite|improve this answer


























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      $E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.






      share|cite|improve this answer
























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        2








        2






        $E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.






        share|cite|improve this answer












        $E_n subseteq E$ for all $n$. By monotonicity, $m^*(E_n) = 0$.







        share|cite|improve this answer












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        answered Dec 3 '18 at 0:11









        Lukas Kofler

        1,3552519




        1,3552519






























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