In the book 'Geometry of sets and measures in Euclidean space' by P. Mattila, theorem 3.4 states that
Let $\mu$ and $\nu$ be uniformly distributed Borel regular measures on a seperable metric space $X$. Then there is a constant $c$ such that $\mu = c\nu$.
The beginning of the proof given goes as follows:
Define $g(r) := \mu(B(x,r))$, $h(r) = \nu(B(x,r))$ for $x \in X$ and $0 < r < \infty$. Notice that the limit
$$\lim_{r \downarrow 0}\left(\nu(U\cap B(x,r))/h(r)\right)$$
exists and equals 1 for $x \in X$. By Fatou's lemma and Fubini's theorem, it follows that \begin{align}\mu(U) &= \int_U \lim_{r \downarrow 0}h(r)^{-1}\nu(U \cap B(x,r)) d\mu x \\ &\leq \liminf_{r\downarrow 0}h(r)^{-1} \int\nu(U\cap B(x,r)) d\mu x \\ &= \liminf_{r\downarrow 0} h(r)^{-1}\int_U \mu(B(y,r))d\nu y \\ &= \left(\liminf_{r\downarrow 0}g(r)/h(r)\right)\nu(U) \end{align}
The proof continuous, but the problem I'm having is in the above argumentation. The use of Fatou's lemma is self-evident, and the change of integration domain from $U$ to the entire space can be motivated. However, how is Fubini's theorem applied to yield row 3? The reapperance of $U$ and the integrand $\mu(B(y,r))$ confuses me a bit.
Consider the double integral $$ \int_{X\times U} \chi_{\{|x-y|\le r\}}\,d(\mu_x \otimes \nu_y) $$ The function being nonnegative, we can convert it to an iterated integral in either order (Tonelli's theorem). One way is
$$ \int_{X }\int_U \chi_{\{|x-y|\le r\}}\,d\nu_y \,d \mu_x = \int_{X }\nu(U\cap B(x,r))\,d \mu_x $$ the other way is $$ \int_{U }\int_X \chi_{\{|x-y|\le r\}} \,d \mu_x\,d\nu_y = \int_{U }\mu( B(y,r))\,d \nu_y $$