If two functions are equal almost everywhere, the first is continuous a.e., is the second?












3












$begingroup$



If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?




I think this is true.



My “proof”:




Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.



Define $D_3 = { x in E: g(x) text{ discontinuous}}$.



If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.



Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.



So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.




Does this proof work?



Thanks!



Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.






real-analysis analysis measure-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    How do you justify $D_3 subseteq D_1 cup D_2$ ?
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:46










  • $begingroup$
    @YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
    $endgroup$
    – Jane Doe
    Dec 12 '18 at 20:47












  • $begingroup$
    Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:51








  • 4




    $begingroup$
    Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
    $endgroup$
    – Andrés E. Caicedo
    Dec 12 '18 at 20:52










  • $begingroup$
    @Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
    $endgroup$
    – James Yang
    Dec 12 '18 at 20:54


















3












$begingroup$



If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?




I think this is true.



My “proof”:




Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.



Define $D_3 = { x in E: g(x) text{ discontinuous}}$.



If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.



Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.



So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.




Does this proof work?



Thanks!



Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.






real-analysis analysis measure-theory







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    How do you justify $D_3 subseteq D_1 cup D_2$ ?
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:46










  • $begingroup$
    @YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
    $endgroup$
    – Jane Doe
    Dec 12 '18 at 20:47












  • $begingroup$
    Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:51








  • 4




    $begingroup$
    Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
    $endgroup$
    – Andrés E. Caicedo
    Dec 12 '18 at 20:52










  • $begingroup$
    @Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
    $endgroup$
    – James Yang
    Dec 12 '18 at 20:54
















3












3








3





$begingroup$



If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?




I think this is true.



My “proof”:




Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.



Define $D_3 = { x in E: g(x) text{ discontinuous}}$.



If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.



Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.



So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.




Does this proof work?



Thanks!



Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.






real-analysis analysis measure-theory







share|cite|improve this question









$endgroup$





If $f = g$ a.e. in $E in mathfrak{M}$ (the Lebesgue measurable sets)
and $f$ is continuous a.e. in $E$, is $g$ continuous a.e. in $E$?




I think this is true.



My “proof”:




Let us denote $D_1 = { x in E: f(x) text{ discontinuous}}$,
$m(D_1) = 0$ and $D_2 = { x in E: f(x) neq g(x)}$, $m(D_2) = 0$.



Define $D_3 = { x in E: g(x) text{ discontinuous}}$.



If $f$ is identically $g$, then it is clear that the result follows as
$D_3 = D_1$.



Otherwise, we have $D_3 subseteq D_1 cup D_2$, and so $m^*(D_3) leq m^*(D_1 cup D_2) leq m^*(D_1) + m^*(D_2) = 0$.



So, $m(D_3) = 0$ and hence $g$ is continuous almost everywhere.




Does this proof work?



Thanks!



Edit: for clarity, $m$ denotes the Lebesgue measure and $m^*$ the Lebesgue outer measure.






real-analysis analysis measure-theory




real-analysis analysis measure-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 20:43









Jane DoeJane Doe

330113




330113








  • 2




    $begingroup$
    How do you justify $D_3 subseteq D_1 cup D_2$ ?
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:46










  • $begingroup$
    @YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
    $endgroup$
    – Jane Doe
    Dec 12 '18 at 20:47












  • $begingroup$
    Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:51








  • 4




    $begingroup$
    Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
    $endgroup$
    – Andrés E. Caicedo
    Dec 12 '18 at 20:52










  • $begingroup$
    @Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
    $endgroup$
    – James Yang
    Dec 12 '18 at 20:54
















  • 2




    $begingroup$
    How do you justify $D_3 subseteq D_1 cup D_2$ ?
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:46










  • $begingroup$
    @YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
    $endgroup$
    – Jane Doe
    Dec 12 '18 at 20:47












  • $begingroup$
    Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 20:51








  • 4




    $begingroup$
    Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
    $endgroup$
    – Andrés E. Caicedo
    Dec 12 '18 at 20:52










  • $begingroup$
    @Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
    $endgroup$
    – James Yang
    Dec 12 '18 at 20:54










2




2




$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46




$begingroup$
How do you justify $D_3 subseteq D_1 cup D_2$ ?
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:46












$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47






$begingroup$
@YvesDaoust My reasoning was that if $f(x) = g(x)$ on $Esetminus D_1$ then since $f(x)$ is continuous there, $g(x)$ ought to be as well?
$endgroup$
– Jane Doe
Dec 12 '18 at 20:47














$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51






$begingroup$
Continuity of $f$ in $Esetminus D_1$ does not imply continuity of $g$, because of $D_2$.
$endgroup$
– Yves Daoust
Dec 12 '18 at 20:51






4




4




$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52




$begingroup$
Consider the case where $f$ is identically 0, and $g$ is the characteristic function of the rationals.
$endgroup$
– Andrés E. Caicedo
Dec 12 '18 at 20:52












$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54






$begingroup$
@Andrés E. Caicedo, you're absolutely right I was just about to fix my mistake with that example. That should be the answer.
$endgroup$
– James Yang
Dec 12 '18 at 20:54












3 Answers
3






active

oldest

votes


















6












$begingroup$

The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.



(Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
      $endgroup$
      – Andrés E. Caicedo
      Dec 12 '18 at 20:57






    • 1




      $begingroup$
      @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
      $endgroup$
      – Yves Daoust
      Dec 12 '18 at 20:59












    • $begingroup$
      The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
      $endgroup$
      – Andrés E. Caicedo
      Dec 12 '18 at 21:02





















    0












    $begingroup$

    No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:



    $D_2cap (E setminus D_1) subseteq (E setminus D_3)$



    which means:



    $D_3 subseteq (E setminus D_2) cup D_1$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.



      (Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.



        (Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.



          (Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)






          share|cite|improve this answer









          $endgroup$



          The claim is not true. An easy counterexample is obtained by letting $f$ be identically 0 and $g$ be the characteristic function of the rationals. We have $f=g$ a.e., $f$ is everywhere continuous and $g$ is nowhere continuous.



          (Clearly, the restriction of $g$ to the irrationals is continuous (being constant), but this is not enough to ensure that $g$ is continuous on the irrationals.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 20:59









          Andrés E. CaicedoAndrés E. Caicedo

          65.4k8158247




          65.4k8158247























              0












              $begingroup$

              $f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 20:57






              • 1




                $begingroup$
                @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
                $endgroup$
                – Yves Daoust
                Dec 12 '18 at 20:59












              • $begingroup$
                The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 21:02


















              0












              $begingroup$

              $f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 20:57






              • 1




                $begingroup$
                @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
                $endgroup$
                – Yves Daoust
                Dec 12 '18 at 20:59












              • $begingroup$
                The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 21:02
















              0












              0








              0





              $begingroup$

              $f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.






              share|cite|improve this answer









              $endgroup$



              $f$ continuous on $Esetminus D_1$ implies $f$ continuous on $Esetminus(D_1cup D_2)$. And of course, $g$ continuous on $Esetminus(D_1cup D_2)$. Then the measure of $D_3$ is at most that of $D_1cup D_2$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 12 '18 at 20:55









              Yves DaoustYves Daoust

              127k673226




              127k673226








              • 1




                $begingroup$
                Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 20:57






              • 1




                $begingroup$
                @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
                $endgroup$
                – Yves Daoust
                Dec 12 '18 at 20:59












              • $begingroup$
                The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 21:02
















              • 1




                $begingroup$
                Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 20:57






              • 1




                $begingroup$
                @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
                $endgroup$
                – Yves Daoust
                Dec 12 '18 at 20:59












              • $begingroup$
                The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
                $endgroup$
                – Andrés E. Caicedo
                Dec 12 '18 at 21:02










              1




              1




              $begingroup$
              Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
              $endgroup$
              – Andrés E. Caicedo
              Dec 12 '18 at 20:57




              $begingroup$
              Not quite. The restriction of $g$ to a set $F$ may be continuous without $g$ itself being continuous on $F$.
              $endgroup$
              – Andrés E. Caicedo
              Dec 12 '18 at 20:57




              1




              1




              $begingroup$
              @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
              $endgroup$
              – Yves Daoust
              Dec 12 '18 at 20:59






              $begingroup$
              @AndrésE.Caicedo: what difference do you make between "restricted to $F$" and "on $F$" ??
              $endgroup$
              – Yves Daoust
              Dec 12 '18 at 20:59














              $begingroup$
              The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
              $endgroup$
              – Andrés E. Caicedo
              Dec 12 '18 at 21:02






              $begingroup$
              The domain of $g$ restricted to $F$ is $F$. The domain of $g$ may be larger, and points not in $F$ may affect whether $g$ is continuous at some point in $F$. See my answer for an example. There, with $F$ the irrationals, $g|F$ is continuous because it is constant, but $g$, defined everywhere, is not continuous at any point.
              $endgroup$
              – Andrés E. Caicedo
              Dec 12 '18 at 21:02













              0












              $begingroup$

              No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:



              $D_2cap (E setminus D_1) subseteq (E setminus D_3)$



              which means:



              $D_3 subseteq (E setminus D_2) cup D_1$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:



                $D_2cap (E setminus D_1) subseteq (E setminus D_3)$



                which means:



                $D_3 subseteq (E setminus D_2) cup D_1$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:



                  $D_2cap (E setminus D_1) subseteq (E setminus D_3)$



                  which means:



                  $D_3 subseteq (E setminus D_2) cup D_1$






                  share|cite|improve this answer









                  $endgroup$



                  No your proof does not work because the fact that $f(x)=g(x)$ on $Esetminus D_1$ implies $g(x)$ is continuous from the continuity of $f(x)$ tells you that:



                  $D_2cap (E setminus D_1) subseteq (E setminus D_3)$



                  which means:



                  $D_3 subseteq (E setminus D_2) cup D_1$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 21:00









                  John11John11

                  1,0321821




                  1,0321821






























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