Let $F \subseteq \mathbb R^d$, $ 0 < s < t$ and let $\mu$ be a measure on $\mathbb R^d$ that satisfies $\mu(F) = 1$ and $\mathrm{supp}(\mu) \subseteq F$. Let's further assume that
$$ \int_{\mathbb R^d} \int_{\mathbb R^d} \frac{d \mu(x) d \mu(y)}{|x - y|^s} = \infty. $$
I now want to show that
$$ \int_{\mathbb R^d} \int_{\mathbb R^d} \frac{d \mu(x) d \mu(y)}{|x - y|^t} = \infty. $$
My initial thought was to maybe rewrite the integrand in the second integral as something like $\frac 1{|x - y|^t} = \frac 1{|x - y|^s} \frac 1{|x - y|^{t - s}}$ and to then somehow compare that to the original integral, but I'm not really sure how to go from there. I thought about integration by parts but I don't think that holds for general measures, and I'm not sure how else to proceed.
If $|x-y| \leq 1$ then $|x-y|^{t} \leq |x-y|^{s}$ and $\frac 1 {|x-y|^{t}} \geq \frac 1 {|x-y|^{t}}$. Note that the integral over $\{(x,y): |x-y| >1\}$ is finite for both the integrands $\frac 1 {|x-y|^{t}}$ and $\frac 1 {|x-y|^{s}}$ (because $\mu (F)=1$ and $\mu$ is supported by $F$). Hence finiteness of the integral depends only on integral over $|x-y| \leq 1$.