$\frac{1}{||x||^n}$ is integrable on complement of $B(0,a) \subset \mathbb{R}^n$

I am trying to prove that for a measurable function $$f$$, the Hardy-Littlewood maximal function is not in $$L^1$$ unless $$f$$ is $$0$$ a.e.

I was able to udnerstand the proof of the inequality here with $$1$$ replaced by an arbitrary constant $$a>0$$ : A question about the Hardy-Littlewood maximal function.

However I want to conclude by saying that Mf is not integrable since $$1/|x|^n$$ is not integrable on $$B(0,a)^c$$. Is this true? Am I thinking in the right direction?

EDIT: I found this in Frank Jones' measure theory book: https://imgur.com/a/tct1vI2. So it seems to be true, I just have no clue why it should be true.

If you know that $$Mf\geq c\|x\|^{-n}$$ for $$\|x\|\geq1$$, then (with $$b$$ the volume of $$B_1(0)$$) \begin{align*} \int_{\mathbb R^n} Mf&\geq\int_{\|x\|\geq 1}\frac{c}{\|x\|^n}\,dx =c\,\int_{\|x\|\geq 1}\int_{\|x\|}^\infty \frac{n}{s^{n+1}}\,ds\,dx\\ \ \\ &=cn\,\int_1^\infty\int_{1\leq\|x\|\leq s}\frac1{s^{n+1}}\,dx \,ds =cn\,\int_1^\infty\frac{m(B_s(0)\setminus B_1(0))}{s^{n+1}} \,ds\\ \ \\ &=-bc+cbn\,\int_1^\infty \frac1s\,ds=\infty. \end{align*}
• I'm wondering if the term after the second equality should be replaced with $cn\int_{1}^\infty \int_{1\leq \| x\| \leq s}\frac{1}{s^{n+1}} dx ds$. After all, $\|x\|\geq 1$
It is true and it follows immediately if you use polar coordinates in $$\mathbb R^{n}$$. Look for 'polar coordinates' in the index of Rudin's RCA.