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.