Prove that a set of norm p is of measure zero












2












$begingroup$


The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.



($||x||_p$ is the $p$ -norm).



I know i need to prove that the boundary of $A$ is of measure $0$.
And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.



I started by declaring the function $f(x) = (||x||_p)^p -1$.
So $B=$ {$x in R^n || f(x)=0$}.



My direction was to prove that the graph of the function is of measure zero






real-analysis measure-theory multivariable-calculus







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.



    ($||x||_p$ is the $p$ -norm).



    I know i need to prove that the boundary of $A$ is of measure $0$.
    And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.



    I started by declaring the function $f(x) = (||x||_p)^p -1$.
    So $B=$ {$x in R^n || f(x)=0$}.



    My direction was to prove that the graph of the function is of measure zero






    real-analysis measure-theory multivariable-calculus







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.



      ($||x||_p$ is the $p$ -norm).



      I know i need to prove that the boundary of $A$ is of measure $0$.
      And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.



      I started by declaring the function $f(x) = (||x||_p)^p -1$.
      So $B=$ {$x in R^n || f(x)=0$}.



      My direction was to prove that the graph of the function is of measure zero






      real-analysis measure-theory multivariable-calculus







      share|cite|improve this question









      $endgroup$




      The question is to show that the set $A = $ {$x in R^n | (||x||_p)^p le 1$} is a Jordan set for $0 le p le 1$.



      ($||x||_p$ is the $p$ -norm).



      I know i need to prove that the boundary of $A$ is of measure $0$.
      And also the boundary is the set $B = $ {$x in R^n | (||x||_p)^p = 1$}.



      I started by declaring the function $f(x) = (||x||_p)^p -1$.
      So $B=$ {$x in R^n || f(x)=0$}.



      My direction was to prove that the graph of the function is of measure zero






      real-analysis measure-theory multivariable-calculus




      real-analysis measure-theory multivariable-calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 20 '18 at 11:28









      Gabi GGabi G

      463110




      463110






















          2 Answers
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          active

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          1












          $begingroup$

          The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice answer, but I am looking for something that will use the graph of the function I wrote
            $endgroup$
            – Gabi G
            Dec 20 '18 at 14:52



















          0












          $begingroup$

          To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$



          Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$



          Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice answer, but I am looking for something that will use the graph of the function I wrote
              $endgroup$
              – Gabi G
              Dec 20 '18 at 14:52
















            1












            $begingroup$

            The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Nice answer, but I am looking for something that will use the graph of the function I wrote
              $endgroup$
              – Gabi G
              Dec 20 '18 at 14:52














            1












            1








            1





            $begingroup$

            The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$






            share|cite|improve this answer









            $endgroup$



            The set $A$ is closed and bounded hence compact and therefore is measurable. Suposse that $mu (A) =k>0,$ then since $Asubset [-2,2]^n $ and $$bigcup_{vin [0,1] } (vA)subset [-2,2]^n $$ and $(vA)cap (uA) =emptyset $ for $uneq v$ and $mu (tA) =t^nmu (A),$ then we get a contradiction with the fact that $mu ([-2,2]^n ) =4^n <infty.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 20 '18 at 12:45









            MotylaNogaTomkaMazuraMotylaNogaTomkaMazura

            6,607917




            6,607917












            • $begingroup$
              Nice answer, but I am looking for something that will use the graph of the function I wrote
              $endgroup$
              – Gabi G
              Dec 20 '18 at 14:52


















            • $begingroup$
              Nice answer, but I am looking for something that will use the graph of the function I wrote
              $endgroup$
              – Gabi G
              Dec 20 '18 at 14:52
















            $begingroup$
            Nice answer, but I am looking for something that will use the graph of the function I wrote
            $endgroup$
            – Gabi G
            Dec 20 '18 at 14:52




            $begingroup$
            Nice answer, but I am looking for something that will use the graph of the function I wrote
            $endgroup$
            – Gabi G
            Dec 20 '18 at 14:52











            0












            $begingroup$

            To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$



            Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$



            Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$



              Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$



              Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$



                Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$



                Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.






                share|cite|improve this answer











                $endgroup$



                To prove that the graph of your function has measure zero, argue as follows: let $Q$ be the $n-$ dimensional cube, centered at the origin, with edge length $k:kin mathbb N.$ Let $f$ be your function. And let $epsilon>0.$



                Since $f$ is uniformly continuous on $Q$, we can partition the cube into sub-cubes $Q_i$ such that $s, tin Q_iRightarrow |f(s)-f(t)|<epsilon.$ Then $G(f)subseteq bigcup Q_itimes [m_i,M_i]$ where $m_i=min_{Q_i} f(x)$ and $M_i=max_{Q_i} f(x).$



                Now, the measure of the graph of $f$ on $Q$ satisfies $mu(G_f)le epsiloncdot text{vol} Q =k^nepsilon.$ Thus, $mu(G_f)=0$ on $Q$. To conclude, note that $mathbb R^n$ is a countable union of such cubes $Q$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 20 '18 at 15:34

























                answered Dec 20 '18 at 15:19









                MatematletaMatematleta

                11.6k2920




                11.6k2920






























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