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

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!

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 $$C_{n}=\left\{c\in C\mid \frac{1}{n} Each $$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 $$a\leq c_1<\cdots that $$\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.
• 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 2, 2022 at 16:38
• Okay, edited to answer the real question. @Medo Sep 2, 2022 at 16:58
• Nice answer. Especially the remark that such sets would be uncountable and not just infinite. +1 Sep 3, 2022 at 8:08