I would like to prove the following:

Let $g$ be a monotone increasing function on $[0,1]$. Then the set of points where $g$ is not continuous is at most countable.

My attempt:

Let $g(x^-)~,g(x^+)$ denote the left and right hand limits of $g$ respectively. Let $A$ be the set of points where $g$ is not continuous. Then for any $x\in A$, there is a rational, say, $f(x)$ such that $g(x^-)\lt f(x)\lt g(x^+)$. For $x_1\lt x_2$, we have that $g(x_1^+)\leq g(x_2^-)$. Thus $f(x_1)\neq f(x_2)$ if $x_1\neq x_2$. This shows an injection between $A$ and a subset of the rationals. Since the rationals are countable, $A$ is countable, being a subset of a countable set.

Is my work okay? Are there better/cleaner ways of approaching it?

  • Related posts: 56831 and 14458. – Srivatsan Nov 23 '11 at 7:15
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    This is the standard proof. It looks fine to me, except that I'd call $f$ an injection "of $A$ into the rationals", rather than "between $A$ and a subset of the rationals". – Robert Israel Nov 23 '11 at 7:19
  • @RobertIsrael: Thanks. Point well noted. – AKM Nov 23 '11 at 13:50
  • @Srivatsan: Thanks for the links – AKM Nov 23 '11 at 13:50
  • you have writen g(x+)<f(x)<g(x-) is it not posible dat g(x+)<=f(x)<g(x-) since a function is also discntnous when left and right hand limit are not equal but function value is equal to one of th limit – user28374 Apr 5 '12 at 4:42
up vote 22 down vote accepted

This looks beautiful to me: or, more truthfully, it looks like exactly what I would write.

If anything else can be asked of this argument, maybe it is a justification that monotone functions have discontinuities as you have described. I happen to have recently written this up in lecture notes for a "Spivak calculus" course: see $\S 3$ here. Although the fact is quite well known, many texts do not treat it explicitly. I think this may be a mistake: in the the same section of my notes, I explain how this can be used to give a quick proof of the Continuous Inverse Function Theorem.

  • Where did we use that $f$ is defined on $[0,1]$? Would the argument work if $f$ is defined on $\mathbb{R}$? Thank you. – Leo Feb 23 '14 at 7:56
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    The argument would work for functions defined on any interval in $\mathbb{R}$. – Pete L. Clark Feb 23 '14 at 17:04
  • the link is doesn't exist anymore. – Bhaskar Vashishth Aug 12 '15 at 20:22
  • I don't understand how can we define $f(x)\in \Bbb{Q}$, to define the injection $\phi$ we said that for any $x\in A\mapsto q$ where $q$ is a rationnel between right limit and left, but how can we define $q$ 'formally'? – JeSuis Sep 7 '16 at 18:46
  • @JeSuis: Given a bijection $\mathbb{Z}^+ \rightarrow \mathbb{Q}$, you could choose the $x_n$ lying in the interval so as to minimize $n$. But, honestly, I think it would be better if you were happy not making such a definitive choice: it does not matter for the argument, so why worry about it? – Pete L. Clark Sep 7 '16 at 18:59

Just for slight variation, another proof.

Assume $f:[a,b]\to \mathbb R$ is monotonically increasing and let $D$ be its set of discontinuities. For every $x\in D$ let $c_x=\lim_{t\to x+}f(t)-\lim_{t\to x-}f(t)$ (since $f$ is monotone the one-sided limits exist (and are finite)). As $x$ was a point of discontinuity it follows that $c_x>0$. Now, let $S$ be the sum of all $c_x$. More formally, consider the set $T$ of all finite sums of elements of the form $c_x$, and let $S$ be its supremum.

Now, for every finite sum $s=c_{x_1}+ \cdots +c_{x_n}$ (we may assume the points $x_i$ appear in their natural order in $[a,b]$) using monotonicity of $f$ it follows that the total variation of $f$, that is $f(b)-f(a)$, is not less than the sum $s$ (intuitively, because that sum is the sum of total variations at points along the way from $a$ to $b$). In symbols: $s\le f(b)-f(a)$. It follows that the supremum also satisfies $S\le f(b)-f(a)$. But a sum of infinitely many positive elements can be bounded only if there are countably many elements. Thus, $D$ is countable.

  • "But a sum of infinitely many positive elements can be bounded only if there are countably many elements"- why? Can you give a rigorous proof? – user 170039 Sep 6 '15 at 12:51
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    @user170039 for every $n>1$ there can be only finitely many elements $s$ satisfying $s>1/n$. – Ittay Weiss Sep 6 '15 at 22:41
  • But what if for each $n$ there are $n+1$ elements in $[\frac{1}{n+1}, \frac{1}{n}]$. And what even is a sum of uncountably many elements? Can you define it? – Blrp Jun 7 '17 at 19:18

Your proof is "almost fine" why?.

What you show is that $f$ is 1-1,$f: A \to \mathbb{Q}$, then $|A| \leq |\mathbb{Q}|$. In order to define such function as you did, you should show that your function is well defined. That is, for any $x \in A$ $f(x) = q_1$ and $f(x) = q_2$ then $q_1 = q_2$. Why? because of the definition of a function and as you did, clearly we can have for the same $x$ two different rationals in the interval $(g^{-}(x),g^{+}(x))$ since the rationals are dense. How can you solve this issue? just define $f$ as follows, for each $x \in A$, and any enumeration $\{q_n\}_{n=1}^{\infty}$ of $\mathbb{Q}$

$$f(x) = q_x = \min_{n\in\mathbb{N}}\{\{q_n\}\cap(g^{-}(x),g^{+}(x))\}$$

Then we can not have two different outputs for the same $x$.

  • By the way, you can also use the maximum or just the min or max distance from the mid-point of the interval and many other choices that will guarantee that your function is well defined. – Richard Clare Sep 18 '17 at 22:30

I would suggest you use the axiom of choice to find a unique $f(x)$ for any $x\in A$. Indeed, you can define $Q(x)=\{F_x\in \mathbb{Q}: g(x^-)<F_x<g(x^+)\}$ for all $x\in A$. Then notice that: 1) $Q(x)\neq \emptyset~\forall x\in A$, 2) $Q(x)\cap Q(y)=\emptyset~\forall x\neq y$ in $A$. 1) implies, by the axiom of choice that there is a function $f:A\to \cup_{x\in A}Q(x)\subset \mathbb{Q}$ such that $f(x)\in Q(x)$ for all $x\in A$ and 2) implies that this function is an injection.

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