If $E$ is compact set in $\mathbb{R}^n$ with $0< H^s(E)<\infty$, where $H^s$ is $s$-dimensional Hausdorff measure and $0<s<n$, then fix $x$ and define $$ m(r):= f(B(x,r)) :=H^s(E\cap B(x,r))$$
If $m(r)\leq r^s$, then $$ \int_{B(x,1)} \frac{df(y)}{|x-y|^t} <\infty$$ for $0<t< s$.
Proof of special case: If $m(r)=r^s$, then it is trivial.
How can we prove this in general?
Use the Layer cake representation:
$$|x-y|^{-t} = \int_0^\infty \chi_{\{y : |x-y|^{-t} > \lambda\}} \,d\lambda = \int_0^\infty \chi_{B(x, \lambda^{-1/t})} \,d\lambda $$ Integrate both sides with respect to restricted Hausdorff measure $\mu=H^s_{|E}$, changing the order of integration on the right (Tonelli's theorem) $$\int |x-y|^{-t} \,d\mu(y) = \int_0^\infty \mu(B(x, \lambda^{-1/t})) \,d\lambda $$ By assumption, $$\mu(B(x, \lambda^{-1/t})) \le \min(\lambda^{-s/t}, H^s(E))$$ and the integral $$ \int_0^\infty \min(\lambda^{-s/t}, H^s(E)) \,d\lambda $$ converges because $s/t > 1$.