I would like to prove that the number of simple jump discontinuities of any function is countable.

Can someone point me some material where the proof is or explain the proof here?


  • $\begingroup$ The point is that any jump discontinuity has a neighborhood with no other jump discontinuity, and that the real line is Lindelof. @xavierm02: there are only countably many integers. $\endgroup$ – ronno Dec 22 '12 at 8:55
  • $\begingroup$ Have to seen this paper jstor.org/stable/2689945?seq=1#page_scan_tab_contents ? $\endgroup$ – onurcanbektas May 20 '18 at 11:53

Let $f:(a,b)\to \mathbb{R}$ and $$A=\left\{x\in (a,b):f\text{ has a jump discontinuity at $x$}\right\}$$ Now $$A=A^{+}\cup A^{-}$$ where $$A^{+}=\left\{x\in (a,b):\lim_{y\to x^+}f(y)>\lim_{y\to x^-}f(y)\right\}$$ and $$A^{-}=\left\{x\in (a,b):\lim_{y\to x^+}f(y)<\lim_{y\to x^-}f(y)\right\}$$ I will show $A^{+}$ is countable and leave the rest to you. Fix $x\in A^{+}$ and then $\exists q\in \mathbb{Q}$ so that $$\lim_{y\to x^+}f(y)>q>\lim_{y\to x^-}f(y)$$ (why???). This means that $\exists \delta>0$ so that $$x-\delta<y<x<z<x+\delta\implies f(z)>q>f(y)$$ and so (why?) $\exists n\in \mathbb{N}$ so that $$x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)$$ If we let $$A_{q,n}=\left\{x\in (a,b):x-\frac1n<y<x<z<x+\frac1n\implies f(z)>q>f(y)\right\}$$ ($q\in \mathbb{Q}$,$n\in \mathbb{N}$) then by our previous discussion $$A^{+}\subseteq\bigcup_{q\in \mathbb{Q}}\bigcup_{n\in \mathbb{N}}A_{q,n}$$ Therefore the problem moves to proving that $A_{q,n}$ is countable. This follows from the fact $A_{q,n}$ is isolated (show this!).

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    $\begingroup$ Wow! This is an amazing solution. I really like it. Very elegant! $\endgroup$ – elaRosca Dec 22 '12 at 9:20
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    $\begingroup$ The proof for the other case follows in similar manner, just with the sign changed in the RHS of the implication, where we define Aq,n. q exists by density of Q in R. $\endgroup$ – elaRosca Dec 22 '12 at 9:29
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    $\begingroup$ @elaRosca Indeed. How would you answer the first "why"? $\endgroup$ – Nameless Dec 22 '12 at 9:30
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    $\begingroup$ Q is dense in R. $\endgroup$ – elaRosca Dec 22 '12 at 9:31
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    $\begingroup$ I wrote the code in latex but it does not seem to recognize it here. $\endgroup$ – elaRosca Dec 22 '12 at 9:56

The argument below is essentially the one outlined in Robert Israel's post here, but I tweak it a bit to show that there are only countably many removable discontinuities as well.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function. The key idea is that we can control the amount of fluctuation in $f$ (and hence the size of jumps) on the left (resp., right) side of a point $x$ where the left limit (resp., right limit) exists by taking points sufficiently close to $x$. We cannot guarantee that there are no jumps in a neighborhood of a jump discontinuity; for example, the function $g:[-1,1]\rightarrow\mathbb{R}$ given by

$$g(x) = \begin{cases} \phantom{-}1 & \text{if}\ x\leq0 \\ 1/n & \text{if}\ n \text{ is a positive integer and } 1/(n+1)<x\leq 1/n \end{cases}$$

has a jump discontinuity at and in every neighborhood of $0$ (a more pathological example is given in iballa's comment on Koushik's post; see also Brian Scott's post here for details). However, it is true that we can make jumps around a jump discontinuity as small as desired by taking a sufficiently small neighborhood (but we actually only use a slightly weaker result -- see below). To that end, we note that the definition of the left limit and the triangle inequality give the

Lemma. If $f(x-)=\lim_{t\rightarrow x^-} f(t)$ exists then for any $\varepsilon > 0$ we have some $\delta>0$ such that $$\mathrm{diam} f(x-\delta,x) < \varepsilon. \Box$$

Now for any $x\in\mathbb{R}$ where $f(x-), f(x+)$ exist, put


and for any $\varepsilon>0$, let

$$\mathcal{J}(\varepsilon)=\{ x\in\mathbb{R} : f(x-),f(x+) \text{ exist and } M(x)>\varepsilon \}.$$

Since any point $x$ at which a jump or removable discontinuity occurs lies in $\bigcup_n \mathcal{J}(1/n),$ it suffices to show that each $\mathcal{J}(\varepsilon)$ is countable. Fix $x\in\mathcal{J}(\varepsilon)$ and take $\delta>0$ such that $\mathrm{diam} f(x-\delta,x) < \varepsilon.$ If $t_0$ is an element of $(x-\delta, x)$ such that $f(t_0-), f(t_0+)$ exist then the sequences $f(t_0-1/n), f(t_0+1/n)$ eventually lie in

$$f(x-\delta,x) \subset [f(t_0)-\varepsilon, f(t_0)+\varepsilon],$$

so that

$$f(t_0 -)=\lim_{n\rightarrow\infty} f(t_0-1/n) \in [f(t_0)-\varepsilon, f(t_0)+\varepsilon]$$


$$f(t_0 +)=\lim_{n\rightarrow\infty} f(t_0+1/n) \in [f(t_0)-\varepsilon, f(t_0)+\varepsilon].$$

Consequently, we have $M(t_0)\leq\varepsilon$, and we deduce that $(x-\delta, x)$ and $\mathcal{J}(\varepsilon)$ are disjoint. Letting $q_x$ be any rational number in $(x-\delta, x),$ the map $x\mapsto q_x$ yields an injection $\mathcal{J}(\varepsilon)\rightarrow\mathbb{Q},$ completing the proof.


any jump discontinuity has a neighborhood with no other jump discontinuity, Associate to each such neighbourhood a rational number inside that.so there is a bijection between a subset of rationals and jump discontinuity.

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    $\begingroup$ this shows no. of jump discontinuities can be at most countable $\endgroup$ – Koushik Dec 22 '12 at 9:08
  • $\begingroup$ This is very clear and short, gives a very intuitive answer. Thanks $\endgroup$ – elaRosca Dec 22 '12 at 9:13
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    $\begingroup$ Notice that you can choose a rational in an interval with a canonical way; thus, the result doesn't depend on the axiom of choice. $\endgroup$ – Seirios Dec 22 '12 at 9:16
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    $\begingroup$ In general, to choose an element in a set you need the axiom of choice, but to pick out a rational from $(a,b)$ you can define $q_0 = \min \{ n >0 : \frac{1}{n} < b-a \}$ and $p_0= \min \{ n \geq 0 : \frac{p_0}{q_0} \in (a,b) \}$ if $b>0$ or $p_0= \max \{ n \leq 0 : \frac{p_0}{q_0} \in (a,b) \}$ otherwise, and $p_0/q_0$ works. Here, you only need the archimedean property. $\endgroup$ – Seirios Dec 22 '12 at 9:48
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    $\begingroup$ It seems to me that this is wrong. Suppose we have an enumeration of the rationals $\{q_n\}$. Then the function $f(x) = \sum_{n=1}^{\infty}{\frac{1}{2^n} 1_{[q_n, \infty)} }$ has a jump discontinuity at exactly the rational numbers. $\endgroup$ – iballa Apr 23 '14 at 18:06

As R is union of countable open interval to prove result on R is enough to show on (a,b) a arbitrarily open set .
Claim : Set of Jump discontinuities are countable .It is enough to associate each discontinuity with some countable set here we do by countable rational triple.
$f:(a,b) \to R$
By Jump discontinuities we mean that $f(x-)$ and $f(x+)$ exist but not equal to $f(x)$ So we can make 3 cases 1) $f(x-)$ < $f(x+)$ 2) $f(x-)$ > $f(x+)$ 3)$f(x-)$ = $f(x+)$ $\neq f(x)$
It is enough to show that case 1 and 3
Consider Rational triple (p,q,r) Case1) Consider $f(x-)$ < $f(x+)$ So by denseness of Rational number there exist some rational p such that $f(x-)$ < p < $f(x+)$
a < q < t < x such that f(t) < p As f(x-) < p By definition of f(x-) which is $lim_{t \to x}f(t)=f(x-)$
There exist rational q such that above happening is true from defination of limit
Similarly there exist rational r such that x < t < r < b such that f(t) > p

Now to show uniqueness consider $x \neq y$ and $(p,q,r)$ will hold for both point then without loss of generality consider $x<y$ for (x,y) f(t) < p and f(t) > p which is contradiction

Case 3 )Here $f(x-)=f(x+)=z$ we can associate rational pair (q,r) such that $a<q<t<x$ such that $|f(t)-z|<|f(x)-z|$ and $x<t<r<b$ such that $|f(t)-z|<|f(x)-z|$
Similar to above we can show that it is unique


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