Open balls in $mathbb{R}^d$ are Jordan Measurable












2












$begingroup$


I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.




Show that an open Euclidean
ball $B(x, r) := {y in mathbb{R}^d
: |y − x| < r}$ in $mathbb{R}^d$ is Jordan measurable, with Jordan
measure $c_d r^d$
for some constant $c_d > 0$ depending only on
$d$.




Is there an elementary way to approach this problem?






real-analysis







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.




    Show that an open Euclidean
    ball $B(x, r) := {y in mathbb{R}^d
    : |y − x| < r}$     in $mathbb{R}^d$ is Jordan measurable, with Jordan
    measure $c_d r^d$
    for some constant $c_d > 0$ depending only on
    $d$.




    Is there an elementary way to approach this problem?






    real-analysis







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.




      Show that an open Euclidean
      ball $B(x, r) := {y in mathbb{R}^d
      : |y − x| < r}$       in $mathbb{R}^d$ is Jordan measurable, with Jordan
      measure $c_d r^d$
      for some constant $c_d > 0$ depending only on
      $d$.




      Is there an elementary way to approach this problem?






      real-analysis







      share|cite|improve this question









      $endgroup$




      I'm trying to solve the following question from Terrence Tao's An Introduction to Measure Theory.




      Show that an open Euclidean
      ball $B(x, r) := {y in mathbb{R}^d
      : |y − x| < r}$       in $mathbb{R}^d$ is Jordan measurable, with Jordan
      measure $c_d r^d$
      for some constant $c_d > 0$ depending only on
      $d$.




      Is there an elementary way to approach this problem?






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 1 at 17:18









      user82261user82261

      23517




      23517






















          1 Answer
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          oldest

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          1












          $begingroup$

          I think it's a fairly involved calculation to prove this from scratch (using covers).



          On the other hand, we have



          $1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),



          $2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),



          $3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$



          so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and



          $c(B)=int 1_B=C_npi{(n/2)}r^n.$



          Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.



          Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.



          So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:



          Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.



          A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.



          The result follows.






          share|cite|improve this answer











          $endgroup$













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            1 Answer
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            active

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            1












            $begingroup$

            I think it's a fairly involved calculation to prove this from scratch (using covers).



            On the other hand, we have



            $1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),



            $2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),



            $3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$



            so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and



            $c(B)=int 1_B=C_npi{(n/2)}r^n.$



            Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.



            Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.



            So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:



            Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.



            A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.



            The result follows.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              I think it's a fairly involved calculation to prove this from scratch (using covers).



              On the other hand, we have



              $1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),



              $2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),



              $3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$



              so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and



              $c(B)=int 1_B=C_npi{(n/2)}r^n.$



              Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.



              Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.



              So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:



              Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.



              A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.



              The result follows.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                I think it's a fairly involved calculation to prove this from scratch (using covers).



                On the other hand, we have



                $1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),



                $2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),



                $3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$



                so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and



                $c(B)=int 1_B=C_npi{(n/2)}r^n.$



                Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.



                Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.



                So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:



                Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.



                A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.



                The result follows.






                share|cite|improve this answer











                $endgroup$



                I think it's a fairly involved calculation to prove this from scratch (using covers).



                On the other hand, we have



                $1). $ the content of a Jordan measurable set $S$ is $c(S)=int 1_S$ (this is easy to prove),



                $2). $ if $c(partial S)=0$ then $S$ Jordan measurable, (this requires some effort, but is straightforward),



                $3). $ the volume of an $n-$ dimensional sphere of radius $r$ has the form $C_npi{(n/2)}r^n,$



                so it suffices to prove that $c(partial B)=0$ because then we have that $B$ is Jordan measurable and



                $c(B)=int 1_B=C_npi{(n/2)}r^n.$



                Since $partial B=partial overline B$, we may work with the closed ball. Furthermore, without loss of generality, we may assume that $x=0, r=1$.



                Now, the graph of the continuous function $f$, from the $n-1$-ball: $ xmapsto sqrt{1-|x|^2},$ is the boundary of the upper hemisphere of the unit $n$-ball.



                So, to conclude the proof, we only need show that the graph of $f, $ Gr$(f)$, has Jordan content zero:



                Let $epsilon>0.$ Since the closed ball is compact and $f$ is continuous, there is a $delta>0$ such that $|x-y|<deltaRightarrow |f(x)-f(y)|<epsilon.$ Partition $[0,1]^{n-1}$ into cubes $Q_k:1le kle M$, choosing $M$ large enough so that $x,yin Q_kRightarrow |f(x)-f(y)|<epsilon.$ Choose $x_kin Q_k$ for each $1le kle M.$ Finally, define $R_k={(x,y):xin Q_k; |y-f(x_k)|<epsilon}.$ Then, by construction, Gr$(f)$ is contained in $bigcup_k R_k$ and $sum^M_{k=1}|R_k|<M|Q_k|(2epsilon)=2epsilon.$ Thus, $c^*($Gr$f)=0$.



                A symmetry argument or the above analysis applied to the map $ xmapsto -sqrt{1-|x|^2},$ shows that the boundary of the lower hemisphere also has Jordan content zero.



                The result follows.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 2 at 5:04

























                answered Jan 1 at 23:29









                MatematletaMatematleta

                11.4k2920




                11.4k2920






























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