TITLE: Exercise on Radon measures, constructing a convergent sequence
QUESTION [8 upvotes]: Let $\mu$ be a Radon measure on $\mathbb{R}^n$ such that $\mu(B(0, s)) > 0$ for all $s > 0$ and suppose that
$$C = \limsup_{s\ \downarrow\ 0}\frac{\mu(B(0, 2s))}{\mu(B(0, s))} < \infty.$$
For all $r > 0$, let $\mu_r$ be the Radon measure on $\mathbb{R}^n$ defined by the rule
$$\mu_r(E) = \frac{\mu(rE)}{\mu(B(0, r))}\ \text{for all}\ E \subset \mathbb{R}^n.$$
Prove that there exists $r_i\ \downarrow\ 0$ and a Radon measure $\mu_0$ on $\mathbb{R}^n$ such that $\mu_{r_i} \rightharpoonup \mu_0$.

I actually have very little idea of how to approach this problem. I was hoping to make use of the constant $C$ by doing some manipulations such as
$$\mu_r(E) = \frac{\mu(rE)}{\mu(B(0, r))} = \frac{\mu(rE)}{\mu(B(0, 2r))}\frac{\mu(B(0, 2r))}{\mu(B(0, r))}$$
but no such luck. Any hints, or recommendations for Radon measure references, would be greatly appreciated.

REPLY [2 votes]: The comment by Nate Eldredge is spot on. We need $\sup_r \mu_r(B(0,R))<\infty$ for every $R<\infty$; then a compactness theorem for Radon measures (see, e.g., page 55 of the Evans-Gariepy book) delivers the result. 
By the definition of $\mu_r$, 
$$\mu_r(B(0,R))=\frac{\mu(B(0,rR))}{\mu(B(0,r))}$$
Pick $k$ such that $2^k>R$ and apply the assumption  $k$ times: 
$$\limsup_{r\to 0} \frac{\mu(B(0,rR))}{\mu(B(0,r))} \le C^k$$