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.

Solution 1:

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!).

Solution 2:

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.