Absolute continuity of "approximately uniform" measures Let $\mu$ be a Borel probability measure on the torus $\mathbb{T} = \mathbb{R}/\mathbb{Z}$.  Suppose $\mu$ is "approximately uniform" in the sense that there exists a constant $C > 0$ such that
$$
C^{-1} \ \leq \ \frac{\mu(B_r(x))}{\mu(B_r(y))} \ \leq \ C
$$
for all $x,y \in \mathbb{T}$ and all $r>0$ (here $B_r(x)$ is the ball of radius $r$ centered at $x$).  Does this imply that $\mu$ is absolutely continuous with respect to Lebesgue measure on $\mathbb{T}$?
 A: Yes.
Claim: For $n \in \mathbb{N}$, $\mu(B_{1 / 2n}(0)) \lesssim \mathcal{L}^{1}(B_{1 / 2n}(0))$, with the implicit constant independent of $n$.
Before we prove this claim, let's see why it helps. By approximate uniformity, the claim actually gives us that $\mu(B_{1 / 2n}(x)) \lesssim \mathcal{L}^{1}(B_{1 / 2n}(x))$ for all $x \in \mathbb{T}$, where the implicit constant is the same for all $x$ and $n$. Moreover, one can either be more careful in the proof or use the claim and some reasoning to see that $\mu(B_{r}(x)) \lesssim \mathcal{L}^{1}(B_{r}(x))$ for all $r \leq 1$, with the implicit constant independent of $x$ and $r$. Having the constant independent of both $x$ and $r$ gives absolute continuity.
Proof of claim: The balls $B_{1 / 2n}\big(\frac{k}{n}\big)$ are disjoint for $0 \leq k < n$, so
$$1 \geq \sum_{k = 0}^{n - 1} \mu\Big(B_{1 / 2n}\Big(\frac{k}{n}\Big)\Big) \geq (1 + (n - 1) C^{-1}) \mu(B_{1 / 2n}(0)),$$
which means that
$$\mu(B_{1 / 2n}(0)) \leq \frac{1}{1 + (n - 1) C^{-1}}.$$
The right-hand side is comparable to $\mathcal{L}^{1}(B_{1 / 2n}(0))$ with the implicit constant independent of $n$, which finishes the proof.
