Borel measure taking finitely many values I was doing some problems in measure theory and came across this rather interesting problem:

Let $\mu$ be a Borel measure on $\mathbb{R}$ with $\mu(\mathbb{R})=1$ and $\mu(A) \in \mathbb{Q} \cap [0,1]$ for every Borel set $A$. Show that $\mu$ only takes finitely many values.

How does one even approach this problem? Any help would be appreciated.
 A: Hint: Argue by contraposition: assuming $\mu$ attains infinitely many values, show that there are infinitely many disjoint sets of strictly positive measure. Then show that there is a sequence $(A_n)_n$ of disjoint sets such that $0<\mu(A_n)<\mu(A_{n-1})/4$. Given such sequence, show that the set $\{\sum_{n\in X} \mu(A_n)| X\subseteq \mathbf N \}$ is uncountable (by showing that distinct subsets of $\mathbf N$ yield distinct sums).
A: From The Lebesgue  Deocompsition of a measure we have that $\mu=fdm+\nu$
where $m$ is the Lebesgue measure and $\nu$ is singular with respect to the Lebesgue measure.
Also we can decompose $\nu$ to $\nu_1$ and $\nu_2$ where $\nu_1$ is a continuous measure and $\nu_2=\sum_{x \in A}\mu(\{x\})\delta_x$ where $A$ is a countable set and $\mu(\{x\})>0 ,\forall x \in A$
Note the the functions $F(x)=\int_{-\infty}^xf(x)dm$ and  $G(x)=\nu_1((-\infty,x])$  are continuous functions(since the measures are continuous) with values at $\Bbb{Q} \cap [0,1]$ so from Intermediate Value Theorem they must be constant and so they must be zero.
So we proved that $\mu$ is a discrete measure.
If $A$ is infinite take an enumaration of $A$ to be $A=\{x_n:n\in \Bbb{N}\}$
and prove that the set $\{\sum_{n \in I}\mu(\{x_n\}):I \subseteq \Bbb{N}\}$ is uncountable.
