I'm struggling to understand part of a proof in Falconer's book on Fractal Geometry. It's Theorem 4.13 (a):
Let $F$ be a subset of $\mathbb{R}^n$. If there is a mass distribution $\mu$ on $F$ with $I_s(\mu)=\iint\frac{\mathrm{d}\mu(x)\mathrm{d}\mu(y)}{|x-y|^s}<\infty$, then $\mathcal{H}^s(F)=\infty$ and $\dim_HF\geq s$.
I don't understand the bold bit below.
Suppose that $I_s(\mu)<\infty$ for some mass distribution $\mu$ with support contained in $F$. Define $$F_1=\left\{x\in F:\;\overline{\lim_{r\rightarrow0}}\ \mu(B(x,r))/r^s>0\right\}.$$ If $x\in F_1$ we may find $\varepsilon>0$ and a sequence of numbers $\{r_i\}$ decreasing to $0$ such that $\mu(B(x,r_i))\geq\varepsilon r_i^s$. Since $\mathbf{\mu(\{x\})=0}$ (otherwise $\mathbf{I_s(\mu)=\infty}$) it follows from the continuity of $\mu$ that, by taking $q_i$ ($0<q_i<r_i$) small enough, we get $\mu(A_i)\geq\frac{1}{4}\varepsilon r_i^s$ ($i=1,2,\dots$), where $A_i$ is the annulus $B(x,r_i)\backslash B(x,q_i)$. ...
Why is this $I_s(\mu)$ infinite if $\mu(\{x\})\neq 0$?
Apologies as well, I couldn't upload an image of the proof.
If $f=f(x, y)$ is continuous, $f(x, y)\ge 0$ for all $(x, y)\in D\times D$ and $x_0\in D$, then $$ \int_{D\times D}f(x, y)d\mu(x)d\mu(y)\ge \int_{\overset{|x-x_0|\le \delta}{|y-y_0|\le \delta}}f(x, y)d\mu(x)d\mu(y)\to \mu(\{x_0\})^2 f(x_0, y_0)$$ as $\delta \to 0$. It is not difficult to adapt this to the case of $$ f(x, y)=|x-y|^{-s}, $$ in which case the right-hand side is infinite if $\mu(\{x_0\})\ne 0$ (if $s>0$, of course).