If $\alpha$ is an increasing function of bounded variation, why can't it have uncountable number of discontinuities?

Question

In book 《A Course in Abstract Analysis》,there is a proposition stated:

If $\alpha:[a,b] \to \mathbb{R}$ is an increasing function of bounded variation, it can't have uncountable number of discontinuities

the author argues as follow:

If $\alpha$ is discontinous at the points ${c_n}$, then the sum of the jumps $\Sigma_n[\alpha(c_n+)-\alpha(c_n-)]$ is at most $\alpha(b)-\alpha(a)$.If it has uncountable number of discontinuities, then we can find an $\varepsilon >0$ and an infinite sequence of discontinuities ${c_n}$ such that $\alpha(c_n+)-\alpha(c_n-)\geq \varepsilon$ (Why?) . In the light of what we just pointed out, this would imply that $\alpha(b)-\alpha(a)=\infty$ which is nonsense.

I know that a increasing function should only have countable discontinuities for these discontinuity points should bijectively correspond to a subset of $\mathbb{Q}$ which is obvious countable. But I really can't grasp why "If it has uncountable number of discontinuities, then we can find an $\varepsilon >0$ and an infinite sequence of discontinuities ${c_n}$ such that $\alpha(c_n+)-\alpha(c_n-)\geq \varepsilon$ ". Why there should be an $\varepsilon >0$ ? Any help and hints will be appreciated!

Best regards!

Look at the lengths of the jumps and the set of accumulation points of those lengths. If the only accumulation point of the lengths were $0$, then the set of lengths of jumps would be countable. Therefore, some positive number is an accumulation points of the lengths of the jumps. Take $\epsilon>0$ smaller than that positive accumulation point. — plop, Sep 02 '22 at 15:58
Ultimately, this is because $0$ has a countable neighborhood basis. For example, $(-1/n,1/n)$. — plop, Sep 02 '22 at 16:02
Does this answer your question? How to show that a set of discontinuous points of an increasing function is at most countable — Sourav Ghosh, Sep 02 '22 at 17:43
A function of bounded variation can be expressed as a difference of two increasing functions — Sourav Ghosh, Sep 02 '22 at 17:46
@SouravGhosh Thanks! My question actually lies in why there should always be an $\varepsilon > 0$,which has already been answered below! — user1090529, Sep 03 '22 at 03:40

Thomas Andrews · Accepted Answer · 2022-09-02T17:42:32.210

This is an odd statement to make, because all increasing functions are of bounded variation on a closed interval $[a,b].$

But the crux of the step you are having a problem with is:

Theorem: Let $C$ be an uncountable set, and $f:C\to\mathbb R^+$ be any function. Then there exists an $n$ such that for infinitely many $c\in C,$ $f(c)>\frac1n.$

Proof: If it is not true, let $$C_{n}=\left\{c\in C\mid \frac{1}{n}<f(c)\right\}$$ Each $C_n$ must be finite. But then $C=\bigcup_{n=1}^{\infty} C_n$ is the countable union of finite sets, so it must be countable.

You only need an infinite set in this proof, but there actually must be an $n$ such that $\{c\in C\mid f(c)>1/n\}$ is uncountable, since the countable union of countable sets is countable.

There is nothing magic about $\frac1n$ in this proof - we could have taken any sequence $\epsilon_n$ of positive reals such that $\liminf_{n\to\infty} \epsilon_n=0.$ It just happens that $\frac1n$ is simple.

You can generalize the book's proof for any function of bounded variation, but rather than left- and right- limits, you take:

$$f_+(x)=\limsup_{x'\to x} f(x')\\f_-(x)=\liminf_{x'\to x} f(x').$$

When $f$ is increasing, you can show these are the right and left limits, respectively. We can use the bounded variation property to show $f_+,f_-$ always take finite values, and we easily get $f_+(x)\geq f_-(x)$ with equality at exactly the points of continuity of $x.$

Then we can show, if $f_v(x)=f_{+}(x)-f_{-}(x)$ then for any finite subset $a\leq c_1<\cdots<c_k\leq b$ that $\sum f_v(c_i)$ is less than or equal to the total variance. That takes a little work, but it is not hard.

But it is much easier to prove:

Every function of bounded variation is the difference on monotonic increasing functions, and
All increasing functions have at most countably many discontinuities.

A monotonic function can have at most countable many discontinuities. So do we even need bounded variation if $\alpha$ is increasing. The question should be edited. — Medo, Sep 02 '22 at 16:38
Nice answer. Especially the remark that such sets would be uncountable and not just infinite. +1 — Paramanand Singh, Sep 03 '22 at 08:08

If $\alpha$ is an increasing function of bounded variation, why can't it have uncountable number of discontinuities?

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