Characteristic function of a ball with radius $r$ centered at $x$ Suppose that $\{r_j\}_j$ is a sequence of positive real numbers, and $\{x_j\}_j$ is a sequence in $\mathbb{R}^n$. Suppose also that there are $r \geq 0$ and $x \in  \mathbb{R}^n$, such that
$$\lim_{j\rightarrow \infty}(r_j,x_j) = (r,x)$$
then is it true that $$\lim_{j\rightarrow \infty}\chi_{B(r_j,x_j)} = \chi_{B(r,x)} \textrm{ pointwise?} $$ 
where $B(r_j,x_j)$ is the open ball of radius $r_j$ centered at $x_j$ in $\mathbb{R}^n$ and $B(r,x)$ is the open ball of radius $r$ centered at $x$ in $\mathbb{R}^n$.
So we have a sequence $\{B(r_j,x_j)\}_{1}^{\infty}$ of balls in $\mathbb{R}^n$ and we know that the radius and center are both approaching a certain value and point respectively. So the sequence of ball are approaching a particular one with radius $r$ centered at $x$. Then I believe that the characteristic functions for  $B(r_j,x_j)$ have to converge to the characteristic for $B(r,x)$ when $r_j$ and $x_j$ get close to $r$ and $x$.
 A: 
Suppose that $\{r_j\}_j$ is a sequence of positive real numbers, and $\{x_j\}_j$ is a sequence in $\mathbb{R}^n$. Suppose also that there are $r \geq 0$ and $x \in  \mathbb{R}^n$, such that 
  $$\lim_{j\rightarrow \infty}(r_j,x_j) = (r,x)$$
  then is it true that $$\lim_{j\rightarrow \infty}\chi_{B(r_j,x_j)} = \chi_{B(r,x)} \textrm{ pointwise?} $$ 
  where $B(r_j,x_j)$ is the open ball of radius $r_j$ centered at $x_j$ in $\mathbb{R}^n$ and $B(r,x)$ is the open ball of radius $r$ centered at $x$ in $\mathbb{R}^n$.

Answer: No. $\chi_{B(r_j,x_j)}$ does not converge pointwise to $\chi_{B(r,x)} $ 
Counter-example: Take $n=1$,  $r_j=1+\frac{1}{j+1}$  and $x_j=x=0$. Then 
$$\lim_{j\rightarrow \infty}(r_j,x_j) = (1,0)$$
But $\chi_{B(r_j,x_j)}=\chi_{(-1-\frac{1}{j+1}, 1+\frac{1}{j+1})}$ converges pointwise to $\chi_{[-1,1]}$ and  $\chi_{[-1,1]} \neq \chi_{(-1,1)}=\chi_{B(1,0)}$.
HOWEVER, $$\lim_{j\rightarrow \infty}\chi_{B(r_j,x_j)} = \chi_{B(r,x)} \textrm{ pointwise on } \mathbb{R}^n \setminus S(r,x)$$ 
where $S(r,x)=\{y\in \mathbb{R}^n : |y-x|=r\}$.
Proof: Given any  $y \in \mathbb{R}^n \setminus S(r,x)$, we have two cases: 
Case 1: $|y-x|<r$. Then $\chi_{B(r,x)}(y)=1$.
Let $\epsilon = r - |y-x| $. Since $r_j \to r$ and $x_j\to x$, there is $N$ such that for all $j>N$, $|r_j - r|<\epsilon/2$ and $|x_j-x|<\epsilon/2$. Then 
$$|y-x_j| \leq |y-x|+|x -x_j| < r-\epsilon + \epsilon/2 = r- \epsilon/2 <r_j$$
So, for all $j>N$, $y\in B(r_j,x_j)$, which means that, for all $j>N$, $\chi_{B(r_j,x_j)}(y)=1$. 
So we have that $\lim_{j\to\infty}\chi_{B(r_j,x_j)}(y)=1=\chi_{B(r,x)}(y)$.
Case 2: $|y-x|>r$. Then $\chi_{B(r,x)}(y)=0$.
Let $\epsilon = |y-x|-r $. Since $r_j \to r$ and $x_j\to x$, there is $N$ such that for all $j>N$, $|r_j - r|<\epsilon/2$ and $|x_j-x|<\epsilon/2$. Then 
$$|y-x_j|\geq |(|y-x|-|x -x_j|)| \geq |y-x|-|x -x_j|> r+\epsilon - \epsilon/2 = r + \epsilon/2 >r_j$$
So, for all $j>N$, $y\notin B(r_j,x_j)$, which means that, for all $j>N$, $\chi_{B(r_j,x_j)}(y)=0$. 
So we have that $\lim_{j\to\infty}\chi_{B(r_j,x_j)}(y)=0=\chi_{B(r,x)}(y)$.
By cases 1 and 2, we have proved that $\lim_{j\rightarrow \infty}\chi_{B(r_j,x_j)} = \chi_{B(r,x)} \textrm{ pointwise on } \mathbb{R}^n \setminus S(r,x)$.
A: Consider (in $\mathbb{R}^2$ for example/simplicity) $x_n=((-1)^n\frac{1}{n},0)$ and $r=1$ fixed and the point $x^\ast:=(1,0)$. What do you conclude?
EDIT: I show for example, that for a point $x^\ast\notin B_{x,r}\cup S(r,x)$ the limit is pointwise zero. The same idea applies for points inside the ball.
So, by our assumption on $x^{\ast}$, there exists an $\varepsilon>0$ such that
$$|x^\ast-x|>r+\varepsilon,$$
so that in particular $\chi_{B_{x,r}}(x^\ast)=0$.
By our assumption on the sequences $x_n,r_n$, by the very definition of limit, there exists an $N\in\mathbb{N}$ such that
$$\forall n\geq N, |x_n-x|\leq \varepsilon/4 \text{ and } |r_n-r|\leq \varepsilon/4,$$
but this implies, that for all $n\ge N$
$$|x^\ast-x_n|\geq|x^\ast-x|-|x-x_n|\geq r+\frac{3}{4}\varepsilon\geq r_n+ \frac{1}{2}\varepsilon.$$
Thus $x^\ast\notin B_{x_n,r_n}$ for all $n\ge N$ and so $\chi_{B_{x_n,r_n}}(x^\ast)=0$ for all $n\ge N$. Hence
$$\lim_{n\to\infty}\chi_{B_{x_n,r_n}}(x^\ast)=\lim_{n\to\infty,n\ge N}\chi_{B_{x_n,r_n}}(x^\ast)=0=\chi_{B_{x,r}}(x^{\ast}).$$
