Distribution Functions of Measures and Countable Sets 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.
 A: 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?
