Let $f : \mathbb{R}^n \rightarrow \mathbb{C}$ be Lebesgue integrable and $f$ is not identically zero (i.e., $m(\{x \mid f(x) \neq 0\}) >0$). Then there exist constant $C, R > 0$ such that $$\sup_{r>0} \frac{1}{m(B(x,r))} \int_{B(x,r)}|f(y)| \, dy = (Mf)(x) \geq \frac{C}{|x|^n}$$ for $|x| > R$. This fact will show that $Mf$ is not integrable.
Can anyone give me a guide ? For the first assertion, I look at the proof of Weak-$(1,1)$ inequality which provide the upper bound for $Mf$. It seems that the proof related to a clever method of choosing a ball $B(x,r)$. I am not sure how to adapt it to prove this assertion. (If you have an easy way to prove, please suggest it !). For the non-integrability of $Mf$, I do not think that $$\frac{C}{|x|^n} \rightarrow \infty$$ if $|x| \rightarrow \infty$ and $x$ is bounded away from $0$. So I do not see why this inequality will help showing that $Mf$ is not integrable.