It is well-known that:
Let $\mu$ is a Radon measure on $\mathbb{R}^n$. If $A \subseteq \mathbb{R}^n$ is $\mu$-measurable, then for $\mu$-a.e. $x \in \mathbb{R}^n \setminus A$, $$ \lim\limits_{r \to 0} \frac{\mu (A \cap B(x, r))}{\mu(B(x, r))} = 0$$
I am asked to show the converse is true, but I have no idea.
Any hint would be good.