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

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          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











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          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


















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