Rudin RCA 7.22. If both $f$ and the maximal function $Mf$ are integrable in $R^k$ then $f=0$ a.e. This is Exercise 7.22 (a) from Rudin's RCA. 
Assume that both $f$ and its maximal function $Mf$ are in $L^1(R^k)$. Prove that then $f(x)=0$ a.e.  
Hint: To every other $f \in L^1(R^k)$ corresponds a constant $c = c(f)>0$ such that 
$$(Mf)(x) \ge c|x|^{-k}$$ whenever $|x|$ is sufficiently large. 
I am stuck with this problem. Indeed how can I prove the hint to begin with? I would greatly appreciate any help.
 A: For the hint, first notice that by dominated convergence:
$$
\|f\|_{L^1} = \int_\mathbb{R^k} |f(y)|\mathrm{d}y = \lim_{r \to \infty} \int_{B_r} |f(y)| \mathrm{d}y
$$
for any $f \in L^1 (\mathbb{R}^k)$. Note also that $\|f\|_{L^!} \neq 0$ thanks to the assumption that $f$ is not a.s. 0. Now, the maximal function is given by:
$$
(Mf)(x) = \sup_{B_r \ni x} \frac{1}{C(k)|r|^k}\int_{B_r(z)}|f(y)|\mathrm{d}y
$$
where $B_r(z)$ is a ball centered around $z$ with radius $r$ that contains our point $x$. By the above, we may fix $\epsilon>0$ such that for $r$ sufficiently large, (i.e for $r \geq R_0$), we have:
$$
 \|f\|_{L^{1}} - \int_{B_r(z)}|f(y)|\mathrm{d}y <\varepsilon
$$
Fix $x_0$ such that $|x| \geq R_0$. Then, by definition of $(Mf)(x)$, we have:
$$
(Mf)(x) = \sup_{B_r \ni x} \frac{1}{C(k)|r|^k}\int_{B_r(z)}|f(y)|\mathrm{d}y
\geq \frac{1}{C(k)|x_0|^k }\int_{B_{|x_0|}(z)}|f(y)|\mathrm{d}y
$$
For $x \geq x_0$, we may apply the above $\varepsilon$ inequality, and conclude that:
$$
 \frac{1}{C(k)|x_0|^k }\int_{B_|x_0|(z)}|f(y)|\mathrm{d}y \geq \frac{1}{C(k)|x_0|^k } (\|f\|_{L^1} -\epsilon) \geq \frac{1}{C(k)|x|^k}(\|f\|_{L^1} - \epsilon) 
$$
Multiplying all constants through on the RHS gives us a constant depending only on $f$ and dimension.
A: This immediate from two facts:
(i) He might have been more explicit about the definition; in fact if $f\in L^1$ the maximal function is $$Mf(x)=\sup_{r>0}\frac 1{m(B(x,r))}\int_{B(x,r)}|f|.$$
(ii) If $A>0$ is fixed and $r=|x|+A$ then $$B(0,A)\subset B(x,r);$$also $r\sim |x|$ as $|x|\to\infty$.
