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.....


2 Answers 2


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 $$f(x) = \sum_{r\in\mathbb{Q},r\le x}\frac1{g(r)^2+g(r)}$$ 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 $\epsilon\ge\frac1n$. 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 $y\in (x,x+\delta)$, $$f(y)-f(x)=\sum_{r\in\mathbb{Q},x<r\le y}\frac1{g(r)^2+g(r)} \le \sum_{j=n+1}^{\infty}\frac1{j^2+j}=\frac1{n+1}<\epsilon$$ From that, $\lim_{y\to x^+}f(y)=f(x)$ for all $x$. We have limits from the right.

For limits from the left, consider the variant function $$f^*(x)=\sum_{r\in\mathbb{Q},r< x}\frac1{g(r)^2+g(r)}$$ 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 $\epsilon\ge \frac1n$. Find $\delta$ such that $(x-\delta,x)$ contains none of the $n$ points $r_i$ with $g(r_i)\le n$. Then, for $y\in (x-\delta,x)$, $$f^*(x)-f(y) = \sum_{r\in\mathbb{Q},y\le r< x}\frac1{g(r)^2+g(r)} \le \sum_{j=n+1}^{\infty}\frac1{j^2+j}=\frac1{n+1}<\epsilon$$ From that, $\lim_{y\to x^-}f(y) = f^*(x)$ for all $x$, and we have limits from the left.

Of course, these limits $\lim_{y\to x^+}f(y)=f(x)$ and $\lim_{y\to x^-}f(y) = f^*(x)$ 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.


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

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


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