Let $F$ be a closed subset of $[0,1]$ of positive Lebesgue measure. Let $\delta(x)$ be defined as $\delta(x) = \operatorname{dist}(x, F)$. Consider $$M(x) = \int_{0}^{1}\frac{\delta(y)}{|x -y|^2} \, dy$$ Prove that for almost every point $x \in F$, $M(x) < \infty$.
My thoughts so far are the following: We wish to show that $M \in L^{1}(F)$, which is more than sufficient to complete the proof. Thus, consider $$\int_{F} \int_{0}^{1}\frac{\delta(y)}{|x -y|^2} \, dy \,dx$$ From here, I would like to proceed by using Fubini Theorem to switch the order of integration to $$\int_{0}^{1}\int_{F} \frac{\delta(y)}{|x -y|^2} \, dx \,dy$$ From here, I am not really sure what to do.
Note that $$ M(x) = \int_{F^c\cap I}\frac{\delta(y)}{|x-y|^2}dy $$ where $I=[0,1]$. Hence by Tonelli's theorem, we have $$\begin{eqnarray} \int_F M(x)dx &=& \int_{F^c\cap I}\delta(y)\left(\int_F\frac{1}{|x-y|^2}dx\right)dy\\ &\le&\int_{F^c\cap I}\delta(y)\left(\int_{|z|\ge \delta(y)}\frac{1}{|z|^2}dz\right)dy\\ &= &\int_{F^c\cap I}\delta(y)\frac{2}{\delta(y)}dy= 2|F^c \cap I| \le 2. \end{eqnarray}$$