Question: Given an $\alpha\in(0,n)$, does there exist a measurable set $E\subset\mathbb R^n$ with $\mu(E)\in(0,\infty)$ such that $$\int_{E^c}\int_{E}\frac{dy\, dx}{|x-y|^{n+\alpha}}<+\infty$$ ? (Here $E^c=\mathbb R^n\setminus E$.)
Background:
Currently, I am reading Hitchhiker’s guide to the fractional Sobolev spaces and in Remark 6.6, it stated that an inequality derived earlier $$\int_{E^c}\int_{E}\frac{dy \,dx}{|x-y|^{n+sp}}\ge C(n,s,p)\mu(E)^{\frac{n-sp}{n}}\qquad (\star)$$
can be seen as a special example of the following fractional Sobolev inequality $$\iint_{\mathbb R^n\times\mathbb R^n} \frac{|f(x)-f(y)|^p}{|x-y|^{n+sp}}dx\,dy\ge C'(n,s,p)||f||_{L^{p^\star}(\mathbb R^n)}^{p}$$ with $f=\chi_E$, the indicator function on $E$ where $E$ is a measurable subset of $\mathbb R^n$ with finite measure. Here, $s\in(0,1)$, $p\in[1,\infty)$, $n>sp$ and $p^\star=\frac{np}{n-sp}$. $C(n,s,p)$ denotes a constant depending only on $n,s,p$ and likewise for $C'$.
I am totally fine with the derivations. However, looking at $(\star)$, I wonder if there exists non-trivial examples of $E$ such that the L.H.S. converges; otherwise $(\star)$ is just trivial.
One obvious set to try is the unit ball. In @Ninad Munshi's answer to this question, it gives an explicit form of $$\int_{B(0,1)}\frac{dy}{|x-y|^{n+\alpha}}=\frac{\operatorname{vol}(S^{n-2})}{2\alpha(1-\alpha^2)}\cdot\frac{1}{|x|}\cdot\left(\frac{|x|-\alpha}{(|x|-1)^\alpha}-\frac{|x|+\alpha}{(|x|+1)^\alpha}\right)$$
for $\alpha\in(0,1)$. The R.H.S. is obviously not integrable over $B(0,1)^c$.
