Relating energy integral to the existence of a Frostman measure

Question

I am trying to follow the proof in Mattila's Fourier Analysis and Hausdorff Dimension (bottom of page 19) that given a measure $\mu$ compactly supported on a Borel set $A$ with $0<\mu(A)<\infty$ that has $$I_s(\mu):=\iint |x-y|^{-s}d\mu_xd\mu_y < \infty$$ we can construct a measure (also compactly supported on a Borel set $A$, with $0<\mu(A)<\infty$) that satisfies the Frostman condition, that is, a measure $\nu$ with $$\nu(B(x,r))<r^s.$$ The book says that if $I_s(\mu)<\infty$, then $\int |x-y|^{-s} d\mu_x < \infty$ $\mu$-almost everywhere, which I follow. It also says we can find an $M \in \mathbb{R}^+$ such that $$A:=\left\{y: \int |x-y|^{-s} d \mu_x < M \right\}$$ has $\mu(A)>0$, which also makes sense. But then it claims that $\nu(E) := \mu(E \cap A)$ has $$\nu(B(x,r))\leq 2^sMr^s,$$ and I don't understand where that comes from at all. Can someone explain this step to me?

Answer 1

If $B(x,r)$ is disjoint from $A$, $\,$ then $\nu(B(x,r)=0$. Otherwise, there is some point $y \in B(x,r) \cap A \,. \,$ Then $B(x,r) \subset B(y,2r)$, so $$(2r)^{-s} \nu(B(x,r)) \le \int_{B(y,2r)} (2r)^{-s} \,d\mu(z) \le \int_{B(y,2r)} |z-y|^{-s} \,d\mu(z) \le M \,.$$