Size of the maximal function $f^*$ How do we get the last inequality, $c_1\frac{|E|}{|x|^n} \le \chi^*_E(x) \le c_2\frac{|E|}{|x|^n} \text{ for large } |x|$? Why do we need $|x|$ to be large? 

Let $x\in \mathbb R^n$. For any measurable $E$, $$\chi^*_E(x) = \sup\left\{\frac{|E\cap Q|}{|Q|}: Q \text{ has center } x\right\}.$$
If $E$ is bounded and $Q^x$ denotes the smallest cube with center $x$ containing $E$, then $$\frac{|E\cap Q^x|}{|Q^x|} = \frac{|E|}{|Q^x|}.$$
It follows that there are positive constants $c_1$ and $c_2$ such that  $$c_1\frac{|E|}{|x|^n} \le \chi^*_E(x) \le c_2\frac{|E|}{|x|^n} \text{ for large } |x|.$$

 A: I will use the notation $Q_{r}$ to denote the cube centered at $x$ and
side-length $r$. If $r_{x}$ is the side-length of the smallest cube centered
at $x$ and containing $E$, then for every $r>r_{x}$ we have that$$
\frac{|Q_{r}\cap E|}{|Q_{r}|}=\frac{|E|}{|Q_{r}|}=\frac{|E|}{r^{n}}\leq
\frac{|E|}{r_{x}^{n}}=\frac{|E|}{|Q_{r_{x}}|}=\frac{|Q_{r_{x}}\cap
E|}{|Q_{r_{x}}|}.
$$
This shows that all cubes centered at $x$ and side-lenght $r>r_{x}$ are not
good competitors for the supremum, that is,$$
\chi_{E}^{\ast}(x)=\sup\left\{  \frac{|Q_{r}\cap E|}{|Q_{r}|}:\,Q_{r}\text{
has center at }x\text{ and side-length }r\leq r_{x}\right\}  .
$$
Now let $B(0,R)$ be the smallest ball centered at the origin and containing
$E$. Given $|x|>2R$, find $k\geq2$ such that $kR\leq|x|<(k+1)R$. If
$$r<\frac{|x|}{4\sqrt{n}}\leq\frac{(k+1)R}{4\sqrt{n}}\leq\frac{2kR}{4\sqrt{n}
}=\frac{kR}{2\sqrt{n}},$$ then for every $y\in Q_{r}$ we have that
$|y-x|\leq\sqrt{n}r$ and so $|y|\geq|x|-|y-x|\geq kR-\sqrt{n}r>kR-\frac{kR}%
{2}=\frac{kR}{2}\geq R$, which means that $Q_{r}$ does not intersect $B(0,R)$
and, in turn, it does not intersect $E$. Thus, when $kR\leq|x|<(k+1)R$, we can
restrict our attention to cubes of side-length $r\geq\frac{|x|}{4\sqrt{n}}$.
In turn, $\frac{1}{r^{n}}\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}$ and so$$
\frac{|Q_{r}\cap E|}{|Q_{r}|}\leq\frac{|E|}{|Q_{r}|}=\frac{|E|}{r^{n}}%
\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}},
$$
which shows that $\chi_{E}^{\ast}(x)\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}$. 
On the other hand, if $r\geq4|x|$, then for every $y\in E$, $|y_{i}-x_{i}%
|\leq|y_{i}|+|x_{i}|\leq R+|x|<\frac{1}{2}|x|+|x|\leq2|x|\leq\frac{r}{2}$,
which shows that $E$ is contained in $Q_{r}$ and so$$
\frac{|Q_{r}\cap E|}{|Q_{r}|}=\frac{|E|}{|Q_{r}|}=\frac{|E|}{r^{n}}%
$$
and $r_{x}\leq4|x|$. Hence,
$$
\chi_{E}^{\ast}(x)=\sup\left\{  \frac{|Q_{r}\cap E|}{|Q_{r}|}:\,Q_{r}\text{
has center at }x\text{ and side-length }\frac{|x|}{4\sqrt{n}}\leq
r\leq4|x|\right\}  .
$$
Now for any such $\frac{|x|}{4\sqrt{n}}\leq r\leq4|x|$, that is $\frac
{1}{(4|x|)^{n}}\leq\frac{1}{r^{n}}\leq\frac{(4\sqrt{n})^{n}}{|x|^{n}}$, and so$$
\frac{|Q_{r}\cap E|}{|Q_{r}|}=\frac{|Q_{r}\cap E|}{r^{n}}\geq\frac{|Q_{r}\cap
E|}{(4|x|)^{n}}.
$$
In turn,
\begin{align*}
\chi_{E}^{\ast}(x)  & \geq\frac{1}{(4|x|)^{n}}\sup\bigg\{  |Q_{r}\cap
E|:\,Q_{r}\text{ has center at }x\\&\text{ and side-length }\frac{|x|}{4\sqrt{n}
}\leq r\leq4|x|\bigg\}  
=\frac{1}{(4|x|)^{n}}|Q_{r_{x}}\cap E|=\frac{1}{(4|x|)^{n}}|E|.
\end{align*}
As for why do we care,  Zach Boyd already gave a perfectly good answer. In particular the inequality from below shows that the maximal function is not integrable even if $\chi_E$ is.
