function continuous ae but not borel measurable












2












$begingroup$


It is easy to prove that if $f$ is a function continuous almost everywhere, then $f$ is Lebesgue-measurable by using the property that $mathcal L$ (the Lebesgue-measure) is complete.



Though I've been wondering if the statement "every function continuous ae is Borel-measurable" is true. I feel like it's not but I have a hard time finding a counterexample. So does such a function (continuous ae and not Borel-measurable) exist?






measure-theory lebesgue-measure borel-measures







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    It is easy to prove that if $f$ is a function continuous almost everywhere, then $f$ is Lebesgue-measurable by using the property that $mathcal L$ (the Lebesgue-measure) is complete.



    Though I've been wondering if the statement "every function continuous ae is Borel-measurable" is true. I feel like it's not but I have a hard time finding a counterexample. So does such a function (continuous ae and not Borel-measurable) exist?






    measure-theory lebesgue-measure borel-measures







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      It is easy to prove that if $f$ is a function continuous almost everywhere, then $f$ is Lebesgue-measurable by using the property that $mathcal L$ (the Lebesgue-measure) is complete.



      Though I've been wondering if the statement "every function continuous ae is Borel-measurable" is true. I feel like it's not but I have a hard time finding a counterexample. So does such a function (continuous ae and not Borel-measurable) exist?






      measure-theory lebesgue-measure borel-measures







      share|cite|improve this question









      $endgroup$




      It is easy to prove that if $f$ is a function continuous almost everywhere, then $f$ is Lebesgue-measurable by using the property that $mathcal L$ (the Lebesgue-measure) is complete.



      Though I've been wondering if the statement "every function continuous ae is Borel-measurable" is true. I feel like it's not but I have a hard time finding a counterexample. So does such a function (continuous ae and not Borel-measurable) exist?






      measure-theory lebesgue-measure borel-measures




      measure-theory lebesgue-measure borel-measures






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 3 at 15:17









      BermudesBermudes

      18713




      18713






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $mathfrak c$ Borel subsets of $C$, but $2^{mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is continuous a.e. But ${x: f(x)>0} = E$ is not Borel. So $f$ is not Borel measurable.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            $endgroup$
            – Bermudes
            Jan 3 at 15:42






          • 1




            $begingroup$
            @Bermudes $f(x)=0$ on the complement of $C$.
            $endgroup$
            – Yanko
            Jan 3 at 15:43






          • 1




            $begingroup$
            For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
            $endgroup$
            – GEdgar
            Jan 3 at 15:43








          • 1




            $begingroup$
            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            $endgroup$
            – Bermudes
            Jan 3 at 15:44






          • 2




            $begingroup$
            I will add that to the answer.
            $endgroup$
            – GEdgar
            Jan 3 at 15:45











          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%2f3060657%2ffunction-continuous-ae-but-not-borel-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









          4












          $begingroup$

          Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $mathfrak c$ Borel subsets of $C$, but $2^{mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is continuous a.e. But ${x: f(x)>0} = E$ is not Borel. So $f$ is not Borel measurable.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            $endgroup$
            – Bermudes
            Jan 3 at 15:42






          • 1




            $begingroup$
            @Bermudes $f(x)=0$ on the complement of $C$.
            $endgroup$
            – Yanko
            Jan 3 at 15:43






          • 1




            $begingroup$
            For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
            $endgroup$
            – GEdgar
            Jan 3 at 15:43








          • 1




            $begingroup$
            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            $endgroup$
            – Bermudes
            Jan 3 at 15:44






          • 2




            $begingroup$
            I will add that to the answer.
            $endgroup$
            – GEdgar
            Jan 3 at 15:45
















          4












          $begingroup$

          Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $mathfrak c$ Borel subsets of $C$, but $2^{mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is continuous a.e. But ${x: f(x)>0} = E$ is not Borel. So $f$ is not Borel measurable.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            $endgroup$
            – Bermudes
            Jan 3 at 15:42






          • 1




            $begingroup$
            @Bermudes $f(x)=0$ on the complement of $C$.
            $endgroup$
            – Yanko
            Jan 3 at 15:43






          • 1




            $begingroup$
            For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
            $endgroup$
            – GEdgar
            Jan 3 at 15:43








          • 1




            $begingroup$
            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            $endgroup$
            – Bermudes
            Jan 3 at 15:44






          • 2




            $begingroup$
            I will add that to the answer.
            $endgroup$
            – GEdgar
            Jan 3 at 15:45














          4












          4








          4





          $begingroup$

          Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $mathfrak c$ Borel subsets of $C$, but $2^{mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is continuous a.e. But ${x: f(x)>0} = E$ is not Borel. So $f$ is not Borel measurable.






          share|cite|improve this answer











          $endgroup$



          Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $mathfrak c$ Borel subsets of $C$, but $2^{mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is continuous a.e. But ${x: f(x)>0} = E$ is not Borel. So $f$ is not Borel measurable.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 3 at 15:45

























          answered Jan 3 at 15:27









          GEdgarGEdgar

          62.7k267171




          62.7k267171








          • 1




            $begingroup$
            Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            $endgroup$
            – Bermudes
            Jan 3 at 15:42






          • 1




            $begingroup$
            @Bermudes $f(x)=0$ on the complement of $C$.
            $endgroup$
            – Yanko
            Jan 3 at 15:43






          • 1




            $begingroup$
            For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
            $endgroup$
            – GEdgar
            Jan 3 at 15:43








          • 1




            $begingroup$
            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            $endgroup$
            – Bermudes
            Jan 3 at 15:44






          • 2




            $begingroup$
            I will add that to the answer.
            $endgroup$
            – GEdgar
            Jan 3 at 15:45














          • 1




            $begingroup$
            Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            $endgroup$
            – Bermudes
            Jan 3 at 15:42






          • 1




            $begingroup$
            @Bermudes $f(x)=0$ on the complement of $C$.
            $endgroup$
            – Yanko
            Jan 3 at 15:43






          • 1




            $begingroup$
            For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
            $endgroup$
            – GEdgar
            Jan 3 at 15:43








          • 1




            $begingroup$
            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            $endgroup$
            – Bermudes
            Jan 3 at 15:44






          • 2




            $begingroup$
            I will add that to the answer.
            $endgroup$
            – GEdgar
            Jan 3 at 15:45








          1




          1




          $begingroup$
          Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
          $endgroup$
          – Bermudes
          Jan 3 at 15:42




          $begingroup$
          Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x in C^complement$, then for any $delta > 0$: $f(x) = inf_{y in (xpmdelta)}f(y) = 0$, but why is it that $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
          $endgroup$
          – Bermudes
          Jan 3 at 15:42




          1




          1




          $begingroup$
          @Bermudes $f(x)=0$ on the complement of $C$.
          $endgroup$
          – Yanko
          Jan 3 at 15:43




          $begingroup$
          @Bermudes $f(x)=0$ on the complement of $C$.
          $endgroup$
          – Yanko
          Jan 3 at 15:43




          1




          1




          $begingroup$
          For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
          $endgroup$
          – GEdgar
          Jan 3 at 15:43






          $begingroup$
          For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$.
          $endgroup$
          – GEdgar
          Jan 3 at 15:43






          1




          1




          $begingroup$
          Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
          $endgroup$
          – Bermudes
          Jan 3 at 15:44




          $begingroup$
          Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
          $endgroup$
          – Bermudes
          Jan 3 at 15:44




          2




          2




          $begingroup$
          I will add that to the answer.
          $endgroup$
          – GEdgar
          Jan 3 at 15:45




          $begingroup$
          I will add that to the answer.
          $endgroup$
          – GEdgar
          Jan 3 at 15:45


















          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%2f3060657%2ffunction-continuous-ae-but-not-borel-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

          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

          build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

          Image
          0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
          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