Prove that jump continuous functions have at most countable number of discontinuities.

Question

Suppose $E:=(E,\|\!\cdot\!\|)$ is a Banach space and $I:=[\alpha,\beta]$ is a compact perfect interval. A function $f:I\to E$ is jump continuous if the one-sided limits $f(\alpha+),f(\beta-)$ and $f(x\pm)$ exist for all $x\in I^\circ$.

The following is Exercise VI.1.5 from Analysis II by Amann & Escher:

Prove that every jump continuous function has at most a countable number of discontinuities.

I know that monotone functions ($\mathbb{R\to R}$) have the same property, whose proof begins by choosing a rational number between $f(x-)$ and $f(x+)$. I tried to imitate, but I don't know what to do when $E$ is not totally ordered as $\mathbb{R}$ is.

The proof of this exercise for the case $E=\mathbb{R}$ can be found here, but I don't know how the proof can be modified for the general case. However, I think the proof here works for the general case, but it does not seem "natural" to me. I'm not familiar with the fact that a discrete subset of $\mathbb{R}$ is countable (which does not appear in Amann & Escher). Also, the exercise comes with no hint, so I think it should be easy in some sense. Maybe I'm missing something. Are there other proofs?

Answer 1

Let $\varepsilon > 0$. Let $(t_n)_n$ be a sequence in $[\alpha , \beta]$ with $t_n \rightarrow t \in [\alpha , \beta]$ and $$\max \{||f(t_n+) - f(t_n)||,||f(t_n-) - f(t_n)||\} > \varepsilon$$ Without loss of generality assume $t_1<t_2<\ldots<t_n<\ldots$ Thus, for every $n$ there exists a $s_n> t_n$ or $s_n < t_n$ with $||f(s_n)- f(t_n)|| > \frac \varepsilon 2$, but with $s_n < s_{n+1} < t$.

So we have $\frac \varepsilon 2 < ||f(s_n)- f(t_n)|| \rightarrow ||f(t-)-f(t-)||=0$. Contradiction.

We conclude, that the set $\{t\in [\alpha , \beta ]: \max \{||f(t+) - f(t)||,||f(t-) - f(t)||\} > \varepsilon \}$ is finite. (Otherwise we find a convergent sequence with Bolzano-Weierstraß)

The set of discontinuities is

$$\bigcup\limits_{n=1}^\infty \{t\in [\alpha , \beta ] : \max \{||f(t+) - f(t)||,||f(t-) - f(t)||> \frac 1 n\}$$

hence countable.

Answer 2

For $x\in (\alpha, \beta)$ let $g(x)=\|f(x)-f(x^-)\| + \|f(x)-f(x^+)\|.$

For any $\epsilon >0$ let $y\in (\alpha,x)$ such that $$ (I)...\quad \forall z\in (y,x)\;(\;\|f(z)-f(x^-)\|<\epsilon).$$ For $z\in (y,x)$ take $z'\in (y,z)$ such that $\|f(z')-f(z^-)\|<\epsilon$ and take $z''\in (z,x)$ such that $\|f(z'')-f(z^+)\|<\epsilon.$

Then $$(II)...\quad \|f(z)-f(z^-)\|\leq \|f(z)-f(x^-)\|+\|f(x^-)-f(z')\|+\|f(z')-f(z^-)\|$$ and $$(III)...\quad \|f(z)-f(z^+)\|\leq \|f(z)-f(x^-)\|+\|f(x^-)-f(z'')\|+\|f(z'')-f(z^+)\|.$$

By $(I),$ each of the RHS terms in $(II)$ and $(III)$ is less than $\epsilon,$ so $$\forall z\in (y,x)\;(0\leq g(z)<6\epsilon).$$

Therefore $g(x^-)=0.$

Similarly, we show that $g(x^+)=0. $

From the case $E=\Bbb R$ we know that $g$ has only countably many discontinuities. So for all but countably many $x\in (\alpha,\beta)$ we have $g(x)=g(x^-)=g(x^+)=0.$ Now $g(x)=0$ implies $f(x)=f(x^-)=f(x^+)$ which implies that $f$ is continuous at $x.$

Addendum. Regarding a discrete $S\subset \Bbb R$ being necessarily countable: For $s\in S$ let $r_s>0$ such that $(s-r_s\;,s+r_s)\cap S=\{s\}$ and let $U(s)=(s-\frac {1}{2}r_s\;,s+\frac {1}{2}r_s).$

For unequal $s_1,s_2$ in S, suppose (by contradiction) that $t\in U(s_1)\cap U(s_2).$ We have $$(r_{s_1}\leq |s_1-s_2|)\; \land \; (r_{s_2}\leq |s_2-s_1|).$$ But we also have $$|s_1-s_2|\leq |s_1-t|+|t-s_2|<\frac {1}{2}r_{s_1}+\frac {1}{2}r_{s_2}\leq $$ $$\leq \frac {1}{2}|s_1-s_2|+\frac {1}{2}|s_2-s_1|=|s_1-s_2|,$$ implying $|s_1-s_2|<|s_1-s_2|,$ which is absurd. Therefore $U(s_1)\cap U(s_2)=\emptyset.$

Therefore $F=\{U(s):s\in S\}$ is a pair-wise disjoint family of non-empty open subsets of $\Bbb R.$ Each $U(s)$ contains a rational that does not belong to any other member of $F,$ so $F$ must be countable because $\Bbb Q$ is countable.

This result easily generalizes from $\Bbb R$ to any separable metric space.