This problem is from a Ph.D Qualifying exam for real analysis.
Suppose that $F$ is a closed set in $\mathbb{R}$ such that $m(G)<\infty$, where $G=\mathbb{R}\backslash F$ and $m$ is the Lebesgue measure on $\mathbb{R}$. Let $\delta(x)=\inf\{|x-y|:y\in F\}$. Define a function $J(x)$ by $$J(x)=\int_{\mathbb{R}}\frac{\delta(y)}{|x-y|^2}dy,\:x\in \mathbb{R}$$
Prove that $J(x)<\infty$ a.e. in $F$.
The hint of the problem suggested to show that $\int_F J(x)dx$ is finite, and the result will directly follow.
My attempt: $\int_F J(x)dx= \int_F \int_{\mathbb{R}}\frac{\delta(y)}{|x-y|^2}dy dx =\int_{\mathbb{R}} \int_{F}\frac{\delta(y)}{|x-y|^2}dx dy$
Here the integrand is nonnegative, we can interchange the integral by Fubini-Tornelli theorem.
However, I got stuck in calculating the inner integral $\int_{F}\frac{\delta(y)}{|x-y|^2}dx$, or simply $\int_{F}\frac{dx}{|x-y|^2}$. How should I calculate this stuff? I attempted to split the integral by $\int_{\mathbb{R}}\frac{dx}{|x-y|^2}-\int_{G}\frac{dx}{|x-y|^2}$, but $\int_{\mathbb{R}}\frac{dx}{|x-y|^2}$ does not converge, so I gave up.
What approach should I do?
Any hints or advices will help a lot!
If $y\in F$, then $\delta(y)=0$. Using Fubini (or Tonneli in fact), $$\int_F J(x)dx=\int_F\int_{\mathbb R\backslash F}\frac{\delta(y)}{|x-y|^2}dydx=\int_{\mathbb R\backslash F}\delta(y)\int_F\frac{1}{|x-y|^2}dxdy.$$
If $y\notin F$, then $d=\delta(y)>0$ and $(y-d,y+d)\cap F=\emptyset$, i.e. $F\subset (-\infty ,y-d]\cup[y+d,\infty )$ and $$\int_{F}\frac{1}{|x-y|^2}dx\leq \int_{-\infty }^{y-d}\frac{1}{(x-y)^2}dx+\int_{y+d}^\infty \frac{1}{(x-y)^2}dx=\frac{2}{d}=\frac{2}{\delta(y)}.$$ Therefore, $$\int_F J(x)\leq \int_{\mathbb R\backslash F}\delta(y)\frac{2}{\delta(y)}dy=2m(\mathbb R\backslash F)<\infty .$$