function continuous ae but not borel measurable












2














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







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    2














    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

























      2












      2








      2







      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













      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 2 days ago









      Bermudes

      184




      184






















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          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



















          • 1




            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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            – Bermudes
            2 days ago






          • 1




            @Bermudes $f(x)=0$ on the complement of $C$.
            – Yanko
            2 days ago






          • 1




            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 $C$.
            – GEdgar
            2 days ago








          • 1




            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            – Bermudes
            2 days ago






          • 1




            I will add that to the answer.
            – GEdgar
            2 days ago











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          4














          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



















          • 1




            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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            – Bermudes
            2 days ago






          • 1




            @Bermudes $f(x)=0$ on the complement of $C$.
            – Yanko
            2 days ago






          • 1




            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 $C$.
            – GEdgar
            2 days ago








          • 1




            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            – Bermudes
            2 days ago






          • 1




            I will add that to the answer.
            – GEdgar
            2 days ago
















          4














          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



















          • 1




            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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            – Bermudes
            2 days ago






          • 1




            @Bermudes $f(x)=0$ on the complement of $C$.
            – Yanko
            2 days ago






          • 1




            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 $C$.
            – GEdgar
            2 days ago








          • 1




            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            – Bermudes
            2 days ago






          • 1




            I will add that to the answer.
            – GEdgar
            2 days ago














          4












          4








          4






          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














          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 2 days ago

























          answered 2 days ago









          GEdgar

          61.9k267168




          61.9k267168








          • 1




            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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            – Bermudes
            2 days ago






          • 1




            @Bermudes $f(x)=0$ on the complement of $C$.
            – Yanko
            2 days ago






          • 1




            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 $C$.
            – GEdgar
            2 days ago








          • 1




            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            – Bermudes
            2 days ago






          • 1




            I will add that to the answer.
            – GEdgar
            2 days ago














          • 1




            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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
            – Bermudes
            2 days ago






          • 1




            @Bermudes $f(x)=0$ on the complement of $C$.
            – Yanko
            2 days ago






          • 1




            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 $C$.
            – GEdgar
            2 days ago








          • 1




            Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
            – Bermudes
            2 days ago






          • 1




            I will add that to the answer.
            – GEdgar
            2 days ago








          1




          1




          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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
          – Bermudes
          2 days ago




          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 $lim_{deltato 0}sup_{y in (xpmdelta)}f(y) = 0$?
          – Bermudes
          2 days ago




          1




          1




          @Bermudes $f(x)=0$ on the complement of $C$.
          – Yanko
          2 days ago




          @Bermudes $f(x)=0$ on the complement of $C$.
          – Yanko
          2 days ago




          1




          1




          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 $C$.
          – GEdgar
          2 days ago






          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 $C$.
          – GEdgar
          2 days ago






          1




          1




          Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
          – Bermudes
          2 days ago




          Oh sure, $C$ is closed, didn't think of that. Thanks a lot!
          – Bermudes
          2 days ago




          1




          1




          I will add that to the answer.
          – GEdgar
          2 days ago




          I will add that to the answer.
          – GEdgar
          2 days ago


















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