Uniformly distributed measures are unique up to multiplicative constant. I am reading the book Geometry of sets and measures in euclidean spaces of Mattila and Im having trouble understanding the last part of the proof of theorem 3.4
Definition: Let $X$ be a metric space and $\mu$ a Borel measure over $X$, if for all $0<r<\infty$ and $x,y \in X$ we have $$0 < \mu(B_r(x)) = \mu(B_r(y)) < \infty $$ we say that it is uniformly distributed
Then he proves
Theorem 3.4:
Let $\mu,\nu$ be uniformly distributed Borel regular measures on a separable metric space $X$, then there is a constant $c$ such that $\mu=c\nu$
He proves that for all open sets $U$ we have $\mu(U) = c\nu(U)$ and then says that by theorem 1.10 (2) and borel regularity of $\mu,\nu$ we have $\mu = c\mu$.
Theorem 1.10 (2) says: Let $\mu$ be a Borel regular measure on $X$, $A$ a $\mu$ measurable set and $\epsilon > 0$, If there are open sets $V_1,V_2,\dots$ such that $A \subset \bigcup_{i=1}^\infty V_i$ and $\mu(V_i) < \infty$ for all i, then there is an open set $V$ such that $A \subset V$ and $\mu(V\setminus A) < \epsilon$
And Borel regularity means that for all $A \subset X$ there is a Borel set $B$ such that $A \subset B$ and $\mu(A) = \mu(B)$
I understand the rest of proof except that part, could anyone please clarify it?
 A: The separability of $X$ implies that we can find a countable family of open balls $\{U_i\}_{i\geq 1}$ that cover $X$. Also, by the assumption on $\nu$, we have $\nu(U_i) < \infty$ for each $i$. This shows that Theorem 1.10(2) can be applied to any Borel-measurable subset of $X$.
Now let $A \in \mathcal{B}(X)$ be any Borel subset of $X$. Then for each choice $\varepsilon = \varepsilon_n = \frac{1}{n}$ (or with any choice of $(\varepsilon_n)_{n\geq 1}$ of positive numbers that vanish as $n\to\infty$), we can invoke Theorem 1.10(2) to find an open set $V_n$ such that
$$ A \subseteq V_n, \qquad\text{and}\qquad \nu(V_n \setminus A) < \varepsilon_n. $$
From this, we obtain
\begin{align*}
\mu(A)
\leq \mu(V_n)
= c\nu(V_n)
\leq c\nu(A) + c\varepsilon_n,
\end{align*}
and letting $n \to \infty$ shows that $\mu(A) \leq c \nu(A)$. Arguing similarly, with the role of $\mu$ and $\nu$ interchanged, we also obtain the inequality $c\nu(A) \leq \mu(A)$. Therefore we have the equality $\mu(A) = c\nu(A)$ for any Borel set $A$.
I am not sure why the author also mentioned Borel regularity, since I see no reason to deal with non-Borel-measurable sets in this argument.
