$\iint |x-y|^{-t} \,d\mu\, d\mu < \infty$ iff $t<1$ Consider the measure space $([0,1], \mathcal{L}([0,1]), \mu)$, where $\mu$ is the restriction of the Lebesgue measure to the closed interval $[0,1]$. I wish to show $\iint |x-y|^{-t} \,d\mu\, d\mu < \infty$ if and only if $t<1$. I feel as though I will likely need to use the formula $\int f \, d\mu = \int^{\infty}_0 \mu(\{x:f(x)\geq t\}) \, dt$. I really have trouble with these sort of problems (ones that involve the use of the aforementioned formula). So, I need your help. Explaining the idea as if I was, say, $6$ years old would be great.
 A: $|x|^{-t}$ is locally integrable when $t\lt1$. In fact
$$
\int_{-1}^1|x|^{-t}\,\mathrm{d}x=\frac2{1-t}
$$
This function, being bounded, can be integrated again with no problem. We compute the integral on $[-1,1]$ since $-1\le x-y\le1$ for $x,y\in[0,1]$. In other words,
$$
\begin{align}
\int_0^1\int_0^1|x-y|^{-t}\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^1\int_{-y}^{1-y}|x|^{-t}\,\mathrm{d}x\,\mathrm{d}y\\
&\le\int_0^1\frac2{1-t}\,\mathrm{d}y\\
&=\frac2{1-t}
\end{align}
$$

If $t\gt1$, then
$$
\begin{align}
\int_0^a|x|^{-t}\,\mathrm{d}x
&=\lim_{\epsilon\to0^+}\int_\epsilon^a|x|^{-t}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}\frac{a^{1-t}-\epsilon^{1-t}}{1-t}\\[6pt]
&=\infty
\end{align}
$$
If $t=1$, then
$$
\begin{align}
\int_0^a|x|^{-1}\,\mathrm{d}x
&=\lim_{\epsilon\to0^+}\int_\epsilon^a|x|^{-1}\,\mathrm{d}x\\
&=\lim_{\epsilon\to0^+}(\log(a)-\log(\epsilon))\\
&=\infty
\end{align}
$$
Therefore, if $t\ge1$,
$$
\begin{align}
\int_0^1\int_0^1|x-y|^{-t}\,\mathrm{d}x\,\mathrm{d}y
&=\int_0^1\int_{-y}^{1-y}|x|^{-t}\,\mathrm{d}x\,\mathrm{d}y\\
&\ge\int_0^1\int_0^{1-y}|x|^{-t}\,\mathrm{d}x\,\mathrm{d}y\\[9pt]
&=\infty
\end{align}
$$
