Newton's Theorem For my Measure & Integration course, I've been asked to prove the following:
Let $g$ be a function taking values on $\Bbb{R_+}$, $f: \Bbb{R}^3 \to \Bbb{R}$ such that $f(x) = g(|x|)$. Suppose that $f \in L^1(\Bbb{R}^3, dx)$, and let
$$ \Phi(x) = \int_{\Bbb{R}^3} \frac{f(y)}{|x-y|}dy. $$
For all $r=|x| > 0$,

a) Show that $$\Phi(x) = \frac{4π}{r}\int_0^r g(s)s^2\ \mathrm ds+ 4\pi\int_r^{\infty}g(s)s\ \mathrm ds.$$


b) Deduce that $$\Phi(x) = \int_{\Bbb{R}^3} \frac{f(y)}{max\{|x|,|y|\}}\mathrm dy.$$


c) Show that $$\mu\{x: \Phi(x) > t\} < \infty\qquad\text{for all }t>0,$$
where $\mu$ denotes the Lebesgue measure


d) Conclude that $\Phi(x) = \Phi^*(x)$, where $\Phi^*(x)$ is the symmetric decreasing rearrangement of the function $\Phi(x)$.

After a lot of torture I managed to prove part a), but I'm completely stuck on the following parts. Any advice would be very very welcome!
 A: In part a, you showed that $\Phi$ itself is radial, it depends only on $r = |x|$.
Thus, $\Phi(x)$ is equal to its average value of the sphere of radius $|x|$.
I.e.
$$
\Phi(x) = \frac{1}{4\pi |x|^2}\int_{|u|=|x|} \Phi(u) du
$$
which by Fubini is
$$
= \frac{1}{4\pi |x|^2} \int_{\mathbb{R}^3}f(y) \int_{|u|=|x|} \frac{1}{|u-y|}du\, dy.
$$
This inner integral is a standard computation of a surface integral. You can see it's calculation e.g. here Surface integral of $f(x) = \frac{1}{ \Vert x -x_0 \Vert } $ over sphere.
Thus the last line is
$$
\frac{1}{4\pi |x|^2} \int_{\mathbb{R}^3}f(y) \frac{{4\pi |x|^2}}{\max(|x|, |y|)} dy
= \int_{\mathbb{R}^3}f(y) \frac{1}{\max(|x|, |y|)} dy
$$
as desired for part b.
By part b, we know that as $x \to \infty$ the integrand tends to $0$. Since the integrands with $|x|>1$ are dominated by the integrand with for a fixed $x_0$ on the unit sphere, we can apply dominated convergence to see that $\Phi \to 0$ as $|x| \to \infty$ (important that $f \in L^1$ here). This obviously implies part c as the measure in question is dominated by a that of a big ball.
For part d, just note that $\Phi$ is symmetric by b, and clearly decreasing in $|x|$ by c, so it is its own decreasing symmetric rearrangement.
