Hausdorff measure (open subset containing x)











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Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.



Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.



How to show this implication?



I tried to use:



Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.



So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.



Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?



Is this way correct or can it be proved differently?






measure-theory







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    up vote
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    favorite












    Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.



    Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.



    How to show this implication?



    I tried to use:



    Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.



    So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.



    Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?



    Is this way correct or can it be proved differently?






    measure-theory







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.



      Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.



      How to show this implication?



      I tried to use:



      Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.



      So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.



      Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?



      Is this way correct or can it be proved differently?






      measure-theory







      share|cite|improve this question













      Let $s geq 0$ be a real number and $E subset mathbb{R^n}$.



      Suppose that for all $x in E$ there is an open subset $U subset mathbb{R^n}$ which contains $x$, and $mathcal{H^s}(E cap U)=0 Rightarrow mathcal{H^s}(E)=0$.



      How to show this implication?



      I tried to use:



      Since $E subset mathbb{R^n}$, it exists a number $s_0=$inf$lbrace s in [0, infty):mathcal{H^s}(E)=0 rbrace$.



      So $s_0(E)leq s$ and for all $x in E$ there is an open subset $U Rightarrow s_0(E) geq s$.



      Here I don't see how to conclude. It should be possible to use that every open cover of $E subset mathbb{R^n}$ has a countable subcover, but how to apply it here?



      Is this way correct or can it be proved differently?






      measure-theory




      measure-theory






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      asked 20 hours ago









      Tartulop

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      474






















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          $H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.






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            up vote
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            accepted










            $H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              $H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.






              share|cite|improve this answer























                up vote
                1
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                accepted







                up vote
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                accepted






                $H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.






                share|cite|improve this answer












                $H^{s}$ is countably subadditive and every open cover has a countable subcover. If $Esubset cup U_i$ with $H^{s} (Ecap U_i)=0$ for all $i$ then $H^{s}(E) leq sum H^{s} (Ecap U_i)=0$.







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                share|cite|improve this answer










                answered 20 hours ago









                Kavi Rama Murthy

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