Let $\mu$ be a Lebesgue measure, $\mathcal{M}$ be the Hardy-Littlewood Centered Maximal Function, and the rest as in the title. Then to this end I define $f_{R}(x):=f(x)\chi_{|x|\le R}$ where $\chi$ is the characteristic function and $R>0$. Then we have $$\begin{aligned}&\mathcal{M}f_{R}(x)\ge\frac{\|f_{R}\|_{L^{1}}}{v_{n}}\cdot\frac{1}{(|x|+R)^{n}},\qquad \text{for } |x|\ge R \text{ and } v_{n} \text{ the volume of the unit ball in } \mathbb{R}^{n} \\ \iff& \sup_{R>0}\frac{1}{\mu(B(x,R))}\int_{B(x,R)}|f_{R}|d\mu\ge\frac{1}{\mu(B(x,|x|+R))}\int_{\mathbb{R}^{n}}|f(x)\chi_{|x|\le R}|dx \\ =&\frac{1}{\mu(B(x,|x|+R))}\int_{B(x,R)}|f(x)|dx \end{aligned} $$
I'm not sure that all my computations are correct. And how can I conclude from here that $f_{R}=0$ a.e.?
Suppose $f$ does not vanish almost everywhere. Then there must be some $R>0$ such that $$ \int_{B(0,R)}|f|dm > 0. $$ For all $x \in \mathbb{R}^n$ with $|x|>R$, \begin{align} (Mf)(x) &\geq \frac{1}{m\left(B(x,2|x|)\right)}\int_{B(x,2|x|)}|f|dm \\ &\geq \frac{c}{|x|^n} \int_{B(0,R)}|f|dm \end{align} where $c$ is a constant. But $1/|x|^n \notin L^1(\mathbb{R}^n)$, hence a contradiction.