Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mu$ be a continuous probability measure on $[0,1]$. Then, the function $g:[0,1] \to [0,1]$ defined by $g(x) = \mu([0,x])$ is called the distribution function of $\mu$. I have proved that $g$ is continuous and increasing, with $g(0)=0$ and $g(1)=1$. Moreover, for every $x \in [0,1]$, $g^{-1}(\{x\})$ is an interval which may be just a point.

Define $A := \{x \in [0,1] : g^{-1}(\{x\})$ contains more than one point $\}$. I'm trying to prove that $A$ is countable, but it is giving me a hard time. My approach is to show that if $A$ is uncountable, then $g$ is not increasing. Any other ideas? I'm sure there is an easier way to prove this.

share|cite|improve this question
Did you want to write continuous from the left instead of continuous? – Martin Sleziak May 17 '12 at 19:20
Yes, I thought it was implicit. – ragrigg May 17 '12 at 19:25
@MartinSleziak: Thank you! Much easier approach... – ragrigg May 17 '12 at 19:34
up vote 3 down vote accepted

If $S$ is any systems of disjoint non-degenerate intervals on real line than it is countable. In particular, this is true for the system $S=\{g^{-1}(x); x\in A\}$, which has the same cardinality as $A$.

This follows from the fact that every interval $I\in S$ contains a rational number so you can get an injective map $S\to\mathbb Q$ by mapping $I$ to some element of the (non-empty) set $I\cap\mathbb Q$. The set $\mathbb Q$ is countable.

In the above proof we have obtained an injection by choosing an element from each set $I\cap\mathbb Q$. In case this is interesting for you, I should mention that we can avoid using Axiom of choice. It suffices to notice that we can explicitly write down some well-ordering of $\mathbb Q$ and then simply choose the element of $A\cap\mathbb Q$ which is minimal with respect to this well-ordering.

By explicitly constructing a well-ordering of $\mathbb Q$ I mean that we are able write a formula in language of ZFC which describes a well-ordering of $\mathbb Q$. To see this it is sufficient to find any bijection between $\mathbb N$ and $\mathbb Q$ (without using AC). Well-ordering can be "transferred" using the bijection.

We can get bijection $\mathbb N\to\mathbb N\times\mathbb N$, e.g. Cantor's pairing function. It can be easily modified to a bijection $\mathbb N\to\mathbb Z\times\mathbb N$. If we want bijection which has $\mathbb Q$ as the codomain, we simply "omit" fractions that appear more than once. E.g. if $f:\mathbb N\to\mathbb Z\times\mathbb N$ then we can get $g:\mathbb Q\to\mathbb N$ by putting $g\left(\frac{p}q\right)=|\{f(k); k<n\}|$, where $n\in\mathbb N$ is the preimage of the pair $(p,q)$, i.e. $f(n)=(p,q)$. (We assume that $p\in\mathbb Z$ and $q\in\mathbb N$ are relatively prime.)

To see that we do not need AC to select a rational number from each non-degenerate interval, see also this question: Open Sets of $\mathbb{R}^1$ and axiom of choice

BTW after posting this answer I found the same proof here: Every collection of disjoint non-empty open subsets of $\mathbb{R}$ is countable?

share|cite|improve this answer
What do you mean by avoiding axiom of choice and yet choosing a well-ordering of $\mathbb{Q}$? Isn't well-ordering equivalent with axiom of choice? – T. Eskin May 18 '12 at 4:29
Axiom of Choice is equivalent to the claim: There exists a well-ordering of every set $A$. However, if we work only with one set $A$ and we already know, that this set is well-ordered, we do not need to use AC. E.g. if $A=\mathbb N$, we know that this set is well-ordered; we do not need AC for that. \\ Here we work with the set $\mathbb Q$. You can write down explicitly formula for a bijection $\mathbb Q\to\mathbb N$. Using this bijection you can get well-ordering of $\mathbb Q$ from the usual ordering of $\mathbb N$. – Martin Sleziak May 18 '12 at 4:36
@Thomas See this question for more details: Open Sets of $\mathbb{R}^1$ and axiom of choice – Martin Sleziak May 18 '12 at 4:40
Thanks for clearing it up. I have never seen the explicit formula for the bijection, you know where I could find it? And how do you know that this function well-orders the rationals? – T. Eskin May 18 '12 at 6:26
I meant formula in the sense formula of language of ZFC. Such a formula is informally described as an algorithm in Asaf's answer to the linked question. Or you could start with some pairing function, which gives you bijection to $\mathbb N\times\mathbb N$. This can be modified to a bijection $\mathbb N\to\mathbb Q$. – Martin Sleziak May 18 '12 at 6:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.