Prove that the number of jump discontinuities is countable for any function 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?
Thanks.
 A: 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!).
A: 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
$$M(x)=\max\{|f(x)-f(x-)|,|f(x)-f(x+)|\},$$
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
$$f(t_0 -)=\lim_{n\rightarrow\infty} f(t_0-1/n) \in [f(t_0)-\varepsilon, f(t_0)+\varepsilon]$$
and
$$f(t_0 +)=\lim_{n\rightarrow\infty} f(t_0+1/n) \in [f(t_0)-\varepsilon, f(t_0)+\varepsilon].$$
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.
A: 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 
A: 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.
