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!


1 Answer 1


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:

  1. Every function of bounded variation is the difference on monotonic increasing functions, and
  2. All increasing functions have at most countably many discontinuities.
  • $\begingroup$ 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. $\endgroup$
    – Medo
    Sep 2, 2022 at 16:38
  • 1
    $\begingroup$ Okay, edited to answer the real question. @Medo $\endgroup$ Sep 2, 2022 at 16:58
  • $\begingroup$ Nice answer. Especially the remark that such sets would be uncountable and not just infinite. +1 $\endgroup$
    – Paramanand Singh
    Sep 3, 2022 at 8:08

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