Let $\mu$ be a positive Borel measure on $\mathbb R^d$ with $0 < \mu(\mathbb R^d) < \infty$. Let $I_s(\mu) < \infty$, where $I_s(\mu)$ is defined by $$I_s(\mu) := \int_{\mathbb R^d} \int_{\mathbb R^d} \frac{d \mu(y)}{|x - y|^s} d \mu(x)$$
I now want to show that $\mu(\{x\}) = 0$ for all $x \in \mathbb R^d$.
I started by assuming for the sake of contradication that there is an $x_0 \in \mathbb R^d$ with $\mu(\{x_0\}) =: c > 0$. Then we have $\infty \geq \mu(A) \geq c$ for all Borel sets $A \subseteq \mathbb R^d$ with $x \in A$. I presume I can use this to evaluate first the inner and then the outer integral and arrive at the conclusion that $I_s(\mu) = \infty$, but I'm a little stuck on the details. I don't really know any explicite values of $\mu$, so how do I calculate/solve the integrals here?
Hint: $\int_{\mathbb R^{d}} f(x,y)\,d\mu (x)<\infty$ for almost every $y$ by Fubini's Theorem. [$f$ is the integrand in the double integral]. Suppose $\mu (E)=0$ and $\int_{\mathbb R^{d}} f(x,y)\,d\mu (x)<\infty$ for $y \notin E$. Celarly $\mu \{y\}=0$ if $y \in E$. If $y \in E^{c}$ I leave it to you to show (by considering the behavior of $f$ near $y$) that we must have $\mu \{y\}=0$. [Basically I have reduced the double integral to a single integral. I hope this makes it easy for you].