Let $\mu$ and $\lambda$ be Radon measures on $\mathbb{R^n}$. Show that $\mu$ and $\lambda$ are mutually singular iff $D(\mu,\lambda,x)=\infty$ for $\mu$ almost all $x \in \mathbb{R^n}$.
I have looked at this for a while but to no avail and I cannot see how to approach this. Good hints not fully answers please!
Define $\displaystyle \underline{D}(\mu,\lambda,x) = \liminf_{r \to 0^+} \frac{\mu(B(x,r))}{\lambda(B(x,r))}$. This is defined up to a set with $\lambda$ measure zero.
Suppose that $\mu$ and $\lambda$ are mutually singular. There exists a set $E$ with the property that $\mu(E) = \lambda(\mathbb R^n \setminus E) = 0$. Define $A_k = \{x \in \mathbb R^n \setminus E : \underline{D}(\mu,\lambda,x) \le k\}$. It is a standard result that if $\underline{D}(\mu,\lambda,x) \le t$ for all $x \in A$, then $\mu(A) \le t \lambda(A)$. In particular, $\mu(A_k) \le k \lambda(A_k) = 0$ so that $\mu(A_k) = 0$. Consequently, $$ \mu(\{x : \underline D(\mu,\lambda,x) < \infty\}) = \mu ( \cup A_k) = 0. $$
On the other hand, $D(\mu,\lambda,x)$ is known to exist and be finite for $\lambda$-almost all $x \in \mathbb R^n$. Let $E = \{x : {D}(\mu,\lambda,x) < \infty\}$. Then $\lambda(\mathbb R^n \setminus E) = 0$, and if $D(\mu,\lambda,x) = \infty$ for $\mu$-almost all $x$ you have $\mu(E) = 0$ so that $\lambda$ and $\mu$ are singular.