2.5

Part a.) By the definition of the infimum, we have $\forall \epsilon > 0 \hspace{0.25cm} \exists z \in F$ such that $|x - z| < \delta(x) + \epsilon$. Thus, $$\delta(y) \leq |y - z| \leq |y - x| + |x - z| \leq |y - x| + \delta(x) + \epsilon$$ Thus, since $\epsilon$ can be arbitrarily small, we have: $$\delta(y) - \delta(x) \leq |y-x|$$ Now, repeat this argument for $y$, and we have: $$-|y-x| \leq \delta(y) - \delta(x) \leq |y-x| \hspace{0.25cm} \Rightarrow |\delta(y) - \delta(x)| \leq |y-x|$$

Part b.) First observe that if $y \in F$, then $\delta(y) = 0$. Then, certainly: $$I(x) = \int_F \frac{\delta(y)}{|x-y|^2} dy + \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy = \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy$$ Next, notice that since $F^c$ is an open set, for any $x \in F^c$ we know $\exists r > 0$ such that $B_{r}(x) \subset F^c$, and for any $y \in B_{r}(x)$, $\exists M > 0$ such that $\delta(y) \geq M$. (We can always do this by simply choosing $r$ small enough to fit the ball in $F^c$, then using the ball of radius $\frac{r}{2}$.) Therefore, we have: $$\int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \geq M \int_{B_{r}(x)} \frac{1}{|x-y|^2} dy$$ Now, just observe that we can just center $x$ at the origin, and our calculation becomes: $$M\int_{B_{r}(x)} \frac{1}{|x-y|^2} dy = \lim_{b \to 0} \int_b^\infty \frac{1}{y^2}dy = \infty$$ ...as desired.

Part c.) As the hint prescribes, if we can show $I(x)$ is $L^1(F)$ then we have that $I$ must be almost everywhere finite. Indeed, first notice (from part b) that: $$\int_F I(x) dx = \int_{F} \int_{\mathbb{R}} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx = \int_{F} \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx$$ Before I can use Tonelli's Theorem to swap the limits of integration, I need to first check that the function: $$h(x,y) = \chi_{(F \times F^c)}\frac{\delta(y)}{|x-y|^2}$$ ...is measurable. Certainly, $\chi_{(F \times F^c)}$ is measurable since $F$ and $F^c$ are Borel. Next notice that the sets: $$\lbrace y \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} \delta(y) > \alpha \rbrace$$ are measurable, since they're just the union of $F$ and countably-many closed intervals. Lastly, the function $\frac{1}{|x-y|^2}$ is obviously measurable since it's continuous almost everywhere. Thus, since the product of measurable functions are measurable, we have $h(x,y)$ is measurable.

Thus, it follows from Tonelli's theorem that: $$\int_{F} \int_{F^c} \frac{\delta(y)}{|x-y|^2} dy \hspace{0.1cm} dx = \int_{F^c} \int_{F} \frac{\delta(y)}{|x-y|^2} dx \hspace{0.1cm} dy = \ldots$$ $$\ldots = \int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx \hspace{0.1cm} dy$$ Now, this part I found to be kind of tricky (at first). Observe that: $$F \subset \lbrace x \in \mathbb{R} \hspace{0.25cm} | \hspace{0.25cm} |x - y| \geq \delta(y) \rbrace$$ Then we have: $$\int_{F} \frac{1}{|x-y|^2} dx \leq 2\int_{\delta(y)}^\infty \frac{1}{x^2} dx = \frac{2}{\delta(y)}$$ Finally, we have: $$\int_{F^c} \delta(y) \int_{F} \frac{1}{|x-y|^2} dx \hspace{0.1cm} dy \leq 2\int_{F^c} \delta(y) \int_{\delta(y)}^\infty \frac{1}{x^2} dx \hspace{0.1cm} dy \leq 2 m(F^c)$$ ...and since $m(F^c) < \infty$ we have $I$ must be $L^1 (F)$.