$\mathcal M(\Bbb R^n)$ denotes the set of all Borel measures $\mu$ on $\Bbb R^n$ with $0 < \mu(\Bbb R^n) < \infty$ and compact support $\operatorname{spt}\mu \subset \Bbb R^n$.
Suppose $\mu\in \mathcal M(\Bbb R^n)$ and $I_s(\mu) = \gamma(n,s)\int |\hat\mu(x)|^2 |x|^{s-n}\, dx < \infty$. Then, $$|\hat\mu(x)|\le |x|^{-s/2}$$ for 'most' $x$ with large norm. Here, 'most' simply means what is needed in order that the above integral would be finite. For example, we must have $$\lim_{R\to\infty} R^{-n} \mathcal L^n(\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}) = 0$$
I am trying to see why $\lim_{R\to\infty} R^{-n} \mathcal L^n(\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}) = 0$ in this case. By definition, $$\lim_{R\to\infty} R^{-n} \mathcal L^n(\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}) = 0$$ means that for all $\epsilon > 0$, there exists $R_\epsilon > 0$ such that $$R\ge R_\epsilon \implies |R^{-n} \mathcal L^n(\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}) | < \epsilon$$ Suppose the negation is true, i.e., there exists $\epsilon > 0$ such that for every $R > 0$, there exists $r \ge R$ such that $$\mathcal L^n(\{x\in B(0,r): |\hat\mu(x)| > |x|^{-s/2}\}) \ge \epsilon r^n$$ which implies $$\mathcal L^n(\{x\in B(0,r): |\hat\mu(x)| > |x|^{-s/2}\}) \ge \epsilon R^n$$
How should I proceed? Thank you!
Note: $\mathcal L^n$ denotes the $n$-dimensional Lebesgue measure.
Reference: Fourier Analysis and Hausdorff Dimension by Pertti Mattila.
Let $$A_R:=\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}$$ and let $$D_R:=A_R \setminus B(0,\sqrt{R})\,.$$ Then for all $x \in D_R$, we have $R^{-n} \le |x|^{-n} \le |\hat\mu(x)|^2 |x|^{s-n}\,,$ so $$R^{-n} \mathcal L^n(D_R) \le \int_{D_R} |\hat\mu(x)|^2 |x|^{s-n}\, dx \le \int_{ \Bbb R^n} |\hat\mu(x)|^2 |x|^{s-n} {\bf 1}_{\{|x|\ge \sqrt{R}\}} \,dx \to 0$$ as $R \to \infty$ by Lebesgue's dominated convergence theorem.
Finally, $$R^{-n} \mathcal L^n(A_R) \le R^{-n} \mathcal L^n(D_R) +R^{-n} \mathcal L^n(B(0,\sqrt{R})) \to 0 \quad \text{as} \;\: R \to \infty \,.$$