$\iint |x-y|^{-t} \,d\mu\, d\mu < \infty$ iff $t<1$

Question

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.

Answer 1

$|x|^{-t}$ 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, $$ \begin{align} \int_0^1\int_0^1|x-y|^{-t}\,\mathrm{d}x\,\mathrm{d}y &=\int_0^1\int_{-y}^{1-y}|x|^{-t}\,\mathrm{d}x\,\mathrm{d}y\\ &\le\int_0^1\frac2{1-t}\,\mathrm{d}y\\ &=\frac2{1-t} \end{align} $$

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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 $t\ge1$, $$ \begin{align} \int_0^1\int_0^1|x-y|^{-t}\,\mathrm{d}x\,\mathrm{d}y &=\int_0^1\int_{-y}^{1-y}|x|^{-t}\,\mathrm{d}x\,\mathrm{d}y\\ &\ge\int_0^1\int_0^{1-y}|x|^{-t}\,\mathrm{d}x\,\mathrm{d}y\\[9pt] &=\infty \end{align} $$