So, let us have $\chi_{B(x,r)}$ , which are characteristic functions of $B(x,r)$. As limit $r \to 0$ of sets $B(x,r)$ is a point $x$, I think that a limit of $\chi_{B(x,r)}$ as $r \to 0$ should be:
$f(y) = 1$, if $y = x$ and $f(y) = 0$ otherwise, or some, well weird kind of Dirac delta, with the crucial thing
$\int_{R^n} f(y)g(y)dy=g(x)$ for well enough good $g$. Am I right or crazy?
The pointwise limit of $\chi_{B(x,r)}$ as $r\to 0$ is indeed a function, say $f$, which is $1$ at $x$ and zero elsewhere, as you suspected. That's because (i) $\chi_{B(x,r)}(x)=1$ for all $r>0$, but (ii) $\chi_{B(x,r)}(y)=0$ for $|x-y|>r$, and (iii) as $r$ becomes smaller and smaller, $|x-y|>r$ eventually holds for any $y\ne x$.
The function $f$ in $L^1$ is the same as the zero function because, in $L^1$, all pairs of functions which differ only on a set of measure zero are seen as being both the same function. The relevant set of measure zero here is $\{x\}$; $f$ and the zero function differ only in $\{x\}$. But you did not mention $L^1$.
As for $\lim_{r\to 0} \int_{\Bbb{R}^n}\chi_{B(x,r)}g(y)\,dy=\lim_{r\to 0} \int_{B(x,r)}g(y)\,dy$, for continuous $g$ that will be zero, since (i) $|g(y)|$ is bounded for $y$ in a neighborhood of $x$, (ii) the integral is bounded by the measure of $B(x,r)$ times the maximum of $|g(y)|$, and (iii) the latter product tends to zero as $r\to0$ because the measure of $B(x,r)$ does.
To get a Dirac delta, you probably want $\chi_{B(x,r)}/m(B(x,r))$ or something of that sort, $m(B(x,r))$ being the measure (volume) of $B(x,r)$.
By the dominated convergence theorem, you have $$\lim _{r\to 0} \chi_{B(x,r)} = 0 $$in the sence of $L^1(\Bbb R^n)$. As a corollary, in the sence of $D'(\Bbb R)$, too.
edit
Let for simplicity $r\le 1$. Then, $\forall r \in [0,1]$ we have $$0\le \chi_{B(x,r)} \le \chi_{B(x,1)}$$ and $$\lim_{r\to0}\chi_{B(x,r)}(y) = \begin{cases}0,&y\ne x,\\1,&y=x.\end{cases}$$ For obvious reasons, $\forall r\ge 0$ we have $\chi_{B(x,r)} \in L^1(\Bbb R)$.
Therefore, by the dominated convergence theorem, the sequence $\chi_{B(x,r)}$ converges in $L^1$ and the limit is equal to the the pointwise limit. As this pointwise limit is zero almost everywhere, we can conclude that $\lim_{r\to0}\chi_{B(x,r)}=0$ in the sense of $L^1$.