If I show that $$\frac 12\sup\limits_{R>0}\int\limits_{R\le|x|\le 2R}\lvert K(x)\rvert dx\le \sup\limits_{R>0}\frac 1R\int\limits_{|x|\le R}\lvert K(x)\rvert |x| dx\le 2\sup\limits_{R>0}\int\limits_{R\le|x|\le 2R}\lvert K(x)\rvert dx$$, where $K$ is defined as $K(x)=\frac{\Omega\left(\frac{x}{|x|}\right)}{\lvert x \rvert ^n}$ and $\Omega\in L^1(\mathbb S^{n-1})$, then I show the equivalence between $(5.3.4)$ and $(5.3.6)$ (this is actually an exercise in the book of Loukas Grafakos, Classical Fourier Analysis $3$rd ed.)
$\Omega$ is not explicitly defined, it's just radial function, If I write it in spherical coordinates, I still cannot conclude anything, I think there's another property of $K$ needed.
$\int_{B_{2R}\setminus B_R} K(|x|) dx = \int_R^{2R} K(r) r^{n-1} dr \int_{\mathbb{S}^{n-1}} d\Omega.$
Another Question: Is the LHS equivalent to
$\sup\limits_{R/2>0}\int\limits_{R/2\le|x|\le R}\lvert K(x)\rvert dx$
or to
$\sup\limits_{\epsilon>0}\int\limits_{\epsilon\le|x|\le R}\lvert K(x)\rvert dx$
(Is the ratio or the difference relevant here ?)
Note that $$\sup \limits_{R > 0} \int_{R \le |x| \le 2R} |K(x)| \, dx = \sup \limits_{S > 0} \int_{S/2 \le |x| \le S} |K(x)| \, dx$$
holds, as the change of variable $R = S/2$ shows.
Let $M_n = \{x \mid 2^{-(n + 1)} R \le |x| \le 2^{-n} R\}$. Then by monotone convergence we have:
$$\begin{align*} \frac{1}{R} \int\limits_{|x| \le R} |K(x)| |x| \, dx &= \frac{1}{R} \sum \limits_{n = 0}^\infty \int_{M_n} |K(x)| |x| \, dx \le \frac{1}{R} \sum \limits_{n = 0}^\infty \int_{M_n} |K(x)| 2^{-n} R\, dx \\ &\le \frac{1}{R} \sum \limits_{n = 0}^\infty A_1 2^{-n} R = 2A_1 \end{align*}$$
The other inequality is easier to see: $$\frac{1}{2} \int \limits_{R/2 \le |x| \le R} |K(x)| \, dx \le \frac{1}{2} \int\limits_{R/2 \le |x| \le R} |K(x)| \frac{|x|}{R/2} \le \frac{1}{R} \int \limits_{|x| \le R} |K(x)| |x|\, dx$$