a hint for problem in mathematical analysis There is a question having several parts. I solved the part (a) but I have problems with other parts. 
Let 
$f\in L^{1}(\mathbb{R}^{n}),$
with
$f\geq 0\;$ a.e. 
in 
$\mathbb{R}^{n}$
and let 
$g:\mathbb{R}^{n}\to [0,+\infty)$
be the function defined by 
$$ g(x):=\int_{B(x,\vert x\vert/2)}f(y)dy,$$
where 
$\vert \cdot\vert$ 
denotes the Euclidean norm and 
$B(x,r)=\{y\in \mathbb{R}^{n}:\vert y-x\vert <r\}$
for every 
$x\in \mathbb{R}^{n}$
and every
$r\geq 0.$
(a) Prove that 
$g$ 
is continuous. 
(b) Prove that  $g(x)\to 0$
as
$\vert x\vert\to \infty. $
(c) Prove that 
$g$
has a maximum point in 
$\mathbb{R}^{n}.$
Please write only hint and not a whole solution, I have to be accustomed to solve the mathematics problems. A small useful hint will be accepted. Also, if you have seen this problem in other books, please let me know.  
 A: a) Let $x_n \to x$. Apply the dominated convergence theorem to the sequence $f_n(y) := \mathbb{1}_{B(x_n, |x_n|/2)}(y)f(y)$.
b) Suppose that the assertion weren't true. Then, for some $\epsilon > 0$, there would be a sequence $x_n$ with $|x_n | \to \infty$ such that $g(x_n) > \epsilon$. You may assume wlog that the $B(x_n, |x_n |/2)$ are disjoint. Edit: Because these sets are disjoint, for all $n \in \mathbb{N}$ we have that
$$\int_{\mathbb{R}^n} f(y)~\mathrm{d}y \geq \sum_{i=1}^{n}\int_{B(x_i,\vert x_i\vert/2)}f(y)~\mathrm{d}y. $$ 
c) If $g(x) = 0$ for all $x \in \mathbb{R}^n$ the assertion is trivially true. Now let $y \in \mathbb{R}^n$ be such that $g(y) \neq 0$. By b), there is a $R > 0 $ such that $g(x) < g(y)$ for all $|x| > R$. Consider the restriction of $g$ to $[-R,R]$ and use part $a)$ to conclude.
A: b.  Let $|x_k|\to \infty.$ Then $\chi_{B(x_k,|x_k|/2)} \to 0$ pointwise everywhere. Hence so does $f\cdot\chi_{B(x_k,|x_k|/2)}.$ Since $f\cdot\chi_{B(x_k,|x_k|/2)}\le f $ everywhere, the dominated convergence theorem shows $g(x_k)\to 0.$
