It is from my text book.
We know that the statement:
if $f\in L^1(\mathbb{R}^n)$ then $Mf\in L^1(\mathbb{R}^n)$
is not true.
In fact, if $Mf\in L^1(\mathbb{R}^n)$, then $f=0$. We can show this. If $a>0$ is arbitrary and $|x|>a$, then $$Mf(x)\geq^{step1} \frac{1}{\lambda (B(x,2|x|))}\int_{B(x,2|x|)}{|f(y)|dy} \geq^{step2} \frac{1}{\lambda (B(0,2|x|))}\int_{B(0,a)}{|f(y)|dy} =^{step3}\frac{const}{|x|^n}\int_{B(0,a)}{|f(y)|dy}. $$ Since $|x|^{n}$ is not integrable for $|x|>a$, it follows that $$ \int_{B(0,a)}{|f(y)|dy}=0. $$
Question 1 (about step 2): I understand the first step. But I dont understand why and for what do we shift the center of our ball to $0$. Also I dont understand how we get this inequality.
Question 2: The same misunderstandings with the step 3. I even dont understands how we get $\frac{const}{|x|^n}$ after the third step.
Please help me to understand. If you just show the direction it will be very helpfull.
Step 2 uses two facts: One is $B(0,a) \subseteq B(x,2|X|)$. [$|y|<a$ implies $|y-x|\leq |y|+a<2|x|$ since $|x| >a$]. Next, Lebesgue measure is translation invariant so the measure of any ball $B(u,R)$ is same as the measure of $B(0,R)$.
Step 3 has a typo. Equality should be changed to $\geq$. The constant is just $a^{n}$. They are using the fact that $\frac {a^{n}} {|x|^{n}} <1$.