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

Question

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://i.stack.imgur.com/4u8vT.jpg. So it seems to be true, I just have no clue why it should be true.

Answer 1

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*}

Answer 2

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.