This is a Theorem from Mattila's Book Geometry of sets and measures in Euclidean spaces:
Let $\mu$ and $\nu$ be uniformly distributed Borel regular measures on a separable metric space $X$.
There is a small detail I don't get in the proof. First one define, for $0<r<\infty$, $x\in X$, $$g(r)=\mu(B(x,r)), \ \ \nu(B(x,r)),$$ where $B(x,r)$ is a ball at $x$ with radius $r$. Then, for open $U\in X$, one deduce the inequalities $$\mu(U)\leq\left(\liminf_{r\downarrow 0}\frac{g(r)}{h(r)}\right)\nu(U),\ \text{ and }\ \nu(U)\leq\left(\liminf_{r\downarrow 0}\frac{h(r)}{g(r)}\right)\mu(U).$$
Here is the question: It says that "It follows that the limit $c=\lim_{r \downarrow 0}(g(r)/h(r))$ exists and $\mu(U)=c \nu(U)$", why does the limit exists? It should be easy but I dont see why right now.
Sincerely Ingvar
From this post one realizes that $$(\liminf_{n\to \infty} x_n)^{-1} = \limsup_{n\to \infty} x_n^{-1}$$ Using the following relation $$\liminf_{r \to 0}f(r) = \liminf_{n \to \infty} f(1/n)$$ and the given conditions in your question we have that $$\left(\liminf_{r \to 0} \frac{h(r)}{g(r)}\right)^{-1}\nu(U)\leq \mu(U) \leq \left(\liminf_{r \to 0} \frac{g(r)}{h(r)}\right)\nu(U)$$ and thus $$\limsup_{r\to 0}\frac{g(r)}{h(r)} \leq \liminf_{r\to 0}\frac{g(r)}{h(r)}$$ Since the reverse inequlity always is true, we have equality and the limit exists.