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
          1






          active

          oldest

          votes


















          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











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058670%2fopen-balls-in-mathbbrd-are-jordan-measurable%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058670%2fopen-balls-in-mathbbrd-are-jordan-measurable%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Ellipse (mathématiques)

                    Image
                    Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                    Read more

                    Quarter-circle Tiles

                    Image
                    up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                    Read more

                    Mont Emei

                    Image
                    Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                    Read more