I'm wrapping up Michael E. Taylor's Measure Theory book and currently tackling the exercises in Chapter 11. I'm encountering difficulties with Exercise 6, specifically at a certain point where I'm stuck. I have at my disposal the standard covering lemmas and classical maximal inequalities.
Let $\Phi: (0, \infty) \rightarrow \mathbb{R}^+$ be a monotone decreasing $C^1$ function with the property that $-s^{n} \cdot \Phi'(s) \in L^1(\mathbb{R}^+)$. Then define $$\varphi_{r}(x) = r^{-n}\cdot \Phi(\frac{\vert x \vert}{r}),$$ and finally set $$M_{\varphi}f(x)= \sup_{r>0} \int_{\mathbb{R}^{n}} \left\vert f(y) \right\vert \cdot \phi_{r}(x-y) \ dy.$$ Now prove that there exists $g \in L^{1}(\mathbb{R}^{+})$ such that $$ \left\vert M_{\varphi}f(x) \right\vert \leq \Vert g \Vert_{L^{1}} \cdot Mf(x)$$ for every measurable $f.$
In my attempt, I proceeded as follows: $$\begin{align*} \int_{\mathbb{R}^{n}} \vert f(y) \vert \cdot r^{-n} \cdot \Phi(\frac{\vert x -y \vert}{r}) \ dy = \int_{\mathbb{R}^{n}} \vert f(x-r\ s) \vert \cdot \Phi(\vert s \vert) \ ds \\ = \int_{0}^{+\infty} \ d\rho \ \int_{S^{n-1}} \rho^{n-1} \ \Phi(\rho) \ \vert f(x-r \ \rho \ \omega) \vert \ d\omega \\ = \int_{0}^{+\infty} \rho^{n-1} \ \Phi(\rho) \ \int_{S^{n-1}} \vert f(x-r \ \rho \ \omega) \vert \ d\omega \ d \rho \\ \int_{0}^{+\infty} \Phi(\rho)\ \frac{d}{d\rho}\left( \int_{0}^{\rho}s^{n-1} \ ds \ \int_{S^{n-1}} \vert f(x-r \ s \omega) \vert \ d\omega \right) \ d \rho \\ = \int_{0}^{+\infty} \Phi(\rho)\ \frac{1}{r^{n}} \frac{d}{d\rho} \left( \int_{0}^{r \ \rho}s^{n-1} \ ds \ \int_{S^{n-1}} \vert f(x- \ s \omega) \vert \ d\omega \right) \ d \rho \\ = \int_{0}^{+\infty} \Phi(\rho)\ \frac{1}{r^{n}} \frac{d}{d\rho} \left( \int_{B_{r\ \rho}(x)} \vert f(y) \vert \ d\omega \right) \ d \rho \\ \text{now there is an integration by parts I cannot fully justify}\\ = \int_{0}^{+\infty} -\Phi'(\rho)\ \frac{1}{r^{n}} \left( \int_{B_{r\ \rho}(x)} \vert f(y) \vert \ d\omega \right) \ d \rho. \end{align*}$$ If I pass to the supremum on the last and first term of my chain of inequalities, I obtain $$ \begin{align*} \left\vert M_{\varphi}f(x) \right\vert \leq \sup_{r>0} \int_{\mathbb{R}^{n}} \vert f(y) \vert \cdot r^{-n} \cdot \Phi(\frac{\vert x -y \vert}{r}) \\ \leq \sup_{r>0} \int_{0}^{+\infty} -\Phi'(\rho)\ \rho^{n} \ \frac{1}{(\rho \ r)^{n}} \left( \int_{B_{r\ \rho}(x)} \vert f(y) \vert \ d\omega \right) \ d \rho \leq \int_{0}^{+\infty} -\Phi'(\rho)\ \rho^{n} \ \sup_{r>0} \left( \frac{1}{(\rho \ r)^{n}} \left( \int_{B_{r\ \rho}(x)} \vert f(y) \vert \ d\omega \right) \right) \ d\rho = Mf(x) \cdot \int_{0}^{+\infty} -\Phi'(\rho)\ \rho^{n} \ d \rho. \end{align*}$$ And we conclude by setting $g(\rho) := -\Phi'(\rho) \ \rho^{n}.$
The only passage not fully justified is the integration by parts. Why can I ignore the boundary term coming from the integration by parts? Perhaps, I can prove the inequality for a compactly supported $\Phi$ and arrive at a general $\Phi$ via approximation.