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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.

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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
    
See also math.stackexchange.com/questions/147612/… –  Nate Eldredge May 21 '12 at 12:08
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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?

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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? –  Thomas E. May 18 '12 at 4:29
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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? –  Thomas E. 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
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