Question: Let $\lambda$ be a Radon measure on $\mathbb{R}^n$ and $A \subset \mathbb{R}^n$. Show that
$$ \lim_{r \rightarrow 0} \frac{\lambda(A \cap B(x,r))}{\lambda(B(x,r))}=0 \ \ \ \text{for $\lambda$ almost all $x \in$ $\mathbb{R^n}\setminus A$}$$
if and only if $A$ is $\lambda$-measurable.
I have the following corollary which may be useful,
A measure $\mu$ on $\mathbb{R}^n$ is a Radon measure if and only if it is locally finite and Borel regular.
Im assume that the case "if $A$ is $\lambda$ - measurable...." is easily shown using
$$\mu(C)=\mu(C \cap A)+\mu(C \setminus A) $$
Now if we suppose that $\displaystyle \lim_{r \rightarrow 0} \frac{\lambda(A \cap B(x,r))}{\lambda(B(x,r))}=0 \ \ \ \text{for $\lambda$ almost all $x \in$ $\mathbb{R^n}\setminus A$}$. I cannot see how to proceed. Could someone please guide me as to how to do this.