Suppose $F$ is a closed set in $\mathbb{R}$, whose complement has finite measure, and let $\delta(x)$ denote the distance from $x$ to $F$, that is $$\delta(x)=d(x,F)=\inf\{ | x -y | : y \in F \}\,\,\,$$
Consider $$I(x) = \int_{\mathbb{R}} \frac{\delta(y)}{ |x-y|^2} dy$$
Show that $I(x) < \infty$.
$\textbf{My Attempt:}$ I actually found the solution to this problem but I do not fully understand it.
$$ \int_{F} I(x) dx = \int_{F} \int_{\mathbb{R}} \frac{\delta(y)}{|x-y|^2} dy dx = \int_{F} \int_{\mathbb{F^c}} \frac{\delta(y)}{|x-y|^2} dy dx $$
Let us assume everything is measurable so we can use Tonelli's theorem to sway the limits of integration.
Then $$ \int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx dy$$
$\textbf{This is the part where I am a bit lost...}$
Observe that $$ F \subset \{ x \in \mathbb{R} : |x-y| \ge \delta(y)$$
What was the motivation for seeing this??
Then, $$ \int_{F} \frac{1}{|x-y|^2} dx \le 2 \int_{\delta(y)}^{\infty} \frac{1}{x^2}= \frac{2}{\delta(y)}$$
The line above is particularly confusing to me. Can anyone clarify the reasoning behind using $\delta(y)$ in the limit of the integral ?
Inclusion $F\subset\{x:|x-y|\geq \delta(y)\}$ (note that $y\in F^c$) follows from $$(\forall x\in F)\qquad |x-y|\geq \inf_{z\in F}|z-y|=\delta(y),$$ hence $$\int_F \frac1{|x-y|^2} dx \leq \int_{\{x:|x-y|\geq \delta(y)\}} \frac1{|x-y|^2}dx. $$ Let $z:=|x-y|$. Note that for any $s>0$, $x>y$ such that $z=x-y=s$, there exists also a unique $x'<y$ such that $|x'-y|=s$. Let us consider $x>y$, hence $dz=d(x-y)=dx$, and $$\int_{\{x:|x-y|\geq \delta(y)\}} \frac1{|x-y|^2}dx \leq 2 \int_{\delta(y)}^{+\infty} \frac{1}{z^2} dz .$$ The factor 2 arises exactly because we found for any $s>0$, $x>y$ such that $z=x-y=s$, there exists also a unique $x'<y$ such that $|x'-y|=s$.