Following is a question and it's solution from folland's real analysis chapter 3 exercise 22.
Q22. If $f \in L^1\left(\mathbb{R}^n\right), f \neq 0$, there exist $C, R>0$ such that $H f(x) \geqq C||^{-n}$ for $|x|>R$. Hence $m(\{x: H f(x)>\alpha\}) C / \alpha$ when $\alpha$ is small, so the estimate in the maximal theorem is essentially sharp.
A22. Assuming $\|f\|_1>0$, there exists $R \in(0, \infty)$ such that $\int_{B_R(0)}|f| d m>0$ (otherwise the monotone convergence theorem implies that $\left.\int|f| d m=\lim _{N \rightarrow \infty} \int_{B_N(0)}|f| d m=0\right)$. If $x \in \mathbb{R}^n \backslash B_R[0]$, then $B_R(0) \subseteq B_{2|x|}(x)$ so $$ H f(x) \geq A_{2|x|}|f|(x)=\frac{1}{m\left(B_{2|x|}(x)\right)} \int_{B_{2|x|}(x)}|f| d m \geq \frac{1}{m\left(B_{2|x|}(0)\right)} \int_{B_R(0)}|f| d m=\frac{1}{|x|^n m\left(B_2(0)\right)} \int_{B_R(0)}|f| d m . $$
Note that $C:=\frac{1}{m\left(B_2(0)\right)} \int_{B_R(0)}|f| d m$ is positive and independent of $x$. Now, if $\alpha \in\left(0, \frac{C}{2 R^n}\right)$ and $x \in \mathbb{R}^n$ such that $R<|x|<\left(\frac{C}{\alpha}\right)^{1 / n}$ then $H f(x) \geq C|x|^{-n}>C \frac{\alpha}{C}=\alpha$ and hence $B_{\left(\frac{C}{\alpha}\right)^{1 / n}}(0) \backslash B_R[0] \subseteq\left\{x \in \mathbb{R}^n \mid H f(x)>\alpha\right\}$. Thus $$ m\left(\left\{x \in \mathbb{R}^n \mid H f(x)>\alpha\right\}\right) \geq m\left(B_{\left(\frac{C}{\alpha}\right)^{1 / n}}(0)\right)-m\left(B_R[0]\right)=\left(\frac{C}{\alpha}-R^n\right) m\left(B_1(0)\right)>\frac{C m\left(B_1(0)\right)}{2 \alpha} $$
My question is how do you obtain the following inequality $$=\frac{1}{m\left(B_{2|x|}(x)\right)} \int_{B_{2|x|}(x)}|f| d m \geq \frac{1}{m\left(B_{2|x|}(0)\right)} \int_{B_R(0)}|f| d m$$.
I understand $$m(B_{2|x|}(x))=m(B_{2|x|}(0))$$ but how is $$ \int_{B_{2|x|}(x)}|f| d m \geq \int_{B_R(0)}|f| d m$$
By definition of $x$, $B_R(0)\subset B_{2|x|}(x)$, so by monotonicity of integrals, \begin{align} \int_{B_R(0)}|f|\,dm\leq \int_{B_{2|x|}(x)}|f|\,dm. \end{align}