I think the following is true but I can not prove it:
$$\int_0^1\int_0^1\frac{1}{|x-y|^a}\,dx\,dy<\infty$$
where $dx$ and $dy$ are Lebesgue measures, and $0<a<1$.
If the above integral makes sense as a Riemann integral, it clearly makes sense as a Lebesgue integral too. That said, I am stuck.
Since we are dealing with a non-negative integrand function, we may simply exploit Fubini's theorem and symmetry (no matter if we are dealing with Lebesgue-integrability or improper Riemann-integrability): $$\begin{eqnarray*} \iint_{(0,1)^2}\frac{dx\,dy}{|x-y|^a} = 2 \iint_{0<y<x<1}\frac{dx\,dy}{(x-y)^a}&=&2\int_{0}^{1}\int_{0}^{x}\frac{1}{(x-y)^a}\,dy\,dx\\&\stackrel{y\mapsto x(1-t)}{=}&2\iint_{(0,1)^2}\frac{x}{x^at^a}\,dt\,dx\\&=&\frac{2}{(1-a)(2-a)} \end{eqnarray*}$$ which is finite for any $a\in(0,1)$.