function only contains jump discontinuity but is not piecewise continuous












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Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
$lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....










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    1












    $begingroup$


    Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



    Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
    $lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



    piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



    My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....










    share|cite|improve this question









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      1












      1








      1


      1



      $begingroup$


      Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



      Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
      $lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



      piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



      My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....










      share|cite|improve this question









      $endgroup$




      Does there exists a function defined on $[0,1] rightarrow mathbb{R}$ such that it contains only jump discontinuity but it is not piecewise continuous?



      Jump discontinuity at a point $a$ means $lim_{x rightarrow a^{-}}f(x)$ and
      $lim_{x rightarrow a^{+}}f(x)$ both exists but not equal



      piecewise continuous means every finite subinterval only contains a finite number of discontinuous points and they are all jump discontinuity



      My first thought is Dirichlet function and but it appears that it is not the function that I am looking for.....







      real-analysis calculus






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      asked 2 hours ago









      JoeJoe

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          $begingroup$

          OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



          The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
          $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
          Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

          Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
          $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
          From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



          For limits from the left, consider the variant function
          $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
          This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
          $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
          From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



          Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



          With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



          I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






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            $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



            $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






            share|cite|improve this answer











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              $begingroup$

              OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



              The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
              $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
              Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

              Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
              $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
              From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



              For limits from the left, consider the variant function
              $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
              This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
              $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
              From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



              Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



              With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



              I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



                The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
                $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
                Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

                Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
                $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



                For limits from the left, consider the variant function
                $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
                This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
                $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



                Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



                With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



                I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



                  The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
                  $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
                  Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

                  Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
                  $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



                  For limits from the left, consider the variant function
                  $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
                  This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
                  $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



                  Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



                  With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



                  I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.






                  share|cite|improve this answer









                  $endgroup$



                  OK, next thought - the function $f(frac pq)=frac1q$ and zero elsewhere. That's closer; it has limits of zero everywhere. But then each rational is a removable discontinuity, not a jump discontinuity. Closer, but still not it.



                  The next idea after that: let's build an increasing function with jumps at every rational. Let $g$ be an enumeration of the rationals; for each rational $r$, $g(r)$ is a different positive integer $n$. Then, define
                  $$f(x) = sum_{rinmathbb{Q},rle x}frac1{g(r)^2+g(r)}$$
                  Since $sum_n frac1{n^2+n}$ converges (to $1$), that sum is finite for every $x$.

                  Choose some arbitrary $x$ and $epsilon>0$. Let $n$ be such that $epsilongefrac1n$. There are only finitely many values $r_1,r_2,dots,r_n$ with $g(r_i)le n$. If we choose $delta$ such that $(x,x+delta)$ contains none of these $r_i$, then for $yin (x,x+delta)$,
                  $$f(y)-f(x)=sum_{rinmathbb{Q},x<rle y}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^+}f(y)=f(x)$ for all $x$. We have limits from the right.



                  For limits from the left, consider the variant function
                  $$f^*(x)=sum_{rinmathbb{Q},r< x}frac1{g(r)^2+g(r)}$$
                  This $f^*$ is equal to $f$ except at the rationals, where $f(r)-f^*(r)=frac1{g(r)^2+g(r)}$. Again, choose arbitrary $x$ and $epsilon>0$, and let $n$ be such that $epsilonge frac1n$. Find $delta$ such that $(x-delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)le n$. Then, for $yin (x-delta,x)$,
                  $$f^*(x)-f(y) = sum_{rinmathbb{Q},yle r< x}frac1{g(r)^2+g(r)} le sum_{j=n+1}^{infty}frac1{j^2+j}=frac1{n+1}<epsilon$$
                  From that, $lim_{yto x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.



                  Of course, these limits $lim_{yto x^+}f(y)=f(x)$ and $lim_{yto x^-}f(y) = f^*(x)$ differ for every rational $x$, so there's a jump discontinuity at every rational.



                  With $f$ discontinuous at a dense set of points, it fails to be continuous on any interval, and can't be a piecewise continuous function. Done. We have our example.



                  I defined this as a function from $mathbb{R}$ to $mathbb{R}$, but it's easy to get a function on a smaller interval. Restricting $f$ works, as does using an enumeration of the rationals in that smaller interval.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 54 mins ago









                  jmerryjmerry

                  12.3k1628




                  12.3k1628























                      2












                      $begingroup$

                      $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                      $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                        $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                          $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$






                          share|cite|improve this answer











                          $endgroup$



                          $$f(x)=sum_{k=1}^infty 2^{-k} ( 2^kx-lfloor 2^kxrfloor)$$ has a jump discontinuity at every $frac{n}{2^k}$ and it is continuous everywhere else



                          $$g(x)=sum_{k=1}^infty 2^{-k} ( lfloor 2^kxrfloor-2lfloor 2^{k+1}xrfloor)$$ is easier to see : if $x in [frac{N}{2^k},frac{N+1}{2^k}]$ then $g(x) = frac{N}{2^k}+ O(2^{-k})$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 21 mins ago

























                          answered 26 mins ago









                          reunsreuns

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