If a measure $\mu$ and Lebesgue measure $\lambda$ are singular, is the derivative of $\mu$ with respect to $\lambda$ $\infty$, $\mu$-a.e.? If a positive Radon measure $\mu$ and the Lebesgue measure $\lambda$ are singular, can we show that the derivative of $\mu$ with respect to $\lambda$ is $\infty$, $\mu$-a.e.? Namely, can one show that
$$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\to0, \quad \mu\mbox{-a.e.?}$$
Does this result also hold for any pair of singular Radon measures $\mu\,\bot\,\lambda$?
Actually, a result I can find in [Thm 1.29, Evans-Gariepy, Measure Theory and Fine Properties of Functions] is that, for any Radon measures $\mu$ and $\lambda$,
$$\lim_{r\to0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}<\infty \quad \mbox{ as } \quad r\to0, \quad \lambda\mbox{-a.e.}$$
 A: $\newcommand{\N}{\mathbb N}
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The nice answer by user michael still needs some brushing up, which is what I have done here. I think this could help readers appreciate michael's answer, the central point of which is using the Vitali covering lemma: 
Let $A\subseteq\R^d$ be such that  $\mu(\R^d\setminus A) = 0$ and $\lambda(A) = 0$.  For any real $\epsilon>0$, there is an open set $O_\ep$ such that $ A \subset O_\ep$ and $\lambda(O_\ep) < \epsilon$. Take any real $c>0$ and let 
\begin{equation}
 A_c:=\{x\in A\colon \liminf_{r\downarrow0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}< c\}.  
\end{equation}
For each $x\in A_c$ pick $r_x>0$ such  that 
\begin{equation}
 B(x, r_x) \subset O_\ep\quad\text{and}\quad \frac {\mu(B(x,5r_x))}{\lambda(B(x,5r_x))}<c. 
\end{equation}
By the Vitali covering lemma, there is a countable set $J\subseteq A_c$ such that the balls $B(x,r_x)$ are disjoint for distinct $x\in J$ and 
\begin{equation}
 A_c \subseteq \bigcup_{x\in J} B(x,5r_x). 
\end{equation}
But then 
\begin{multline}
 \mu(A_c) \le  \sum_{x\in J}\mu(B(x,5r_x))\le c \sum_{x\in J} \lambda(B(x,5r_x)) \\ 
 \le c\,5^d \sum_{x\in J} \lambda(B(x,r_x))
 \le c\,5^d \la(O_\ep)\le c\,5^d\ep. 
\end{multline}
Letting $\ep\downarrow0$, we see that $\mu(A_c)=0$, for all real $c>0$. This implies that $\mu$-a.e. 
$$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\downarrow0.$$
A: Let A be  such that  $\mu(A) = 1, \lambda(A) = 0$.  There is an open set $O$ with $ A \subset O, \lambda(A) < \epsilon$. Suppose for convenience that $$\liminf\frac{\mu(B(x,r))}{\lambda(B(x,r))}< a $$ on $A$, otherwise replace $A$ with  the set on which it is true. For each x in A  pick $r_x$ so that $$B(x, r_x) \subset O \quad \text{ and } \quad \frac {\mu(B(x,r_{3x}))}{\lambda(B(x,r_{3x}))}< a .$$  By the Vitali covering lemma there is a disjoint subcollection $J$ with $$A \subset \cup_J B(x, r_{3x})$$ But then $$1 = \mu(A) \le  \sum_J\mu(B(x,r_{3x})) \le a \sum_J \lambda(B(x,r_{3x} )) \le a \cdot 3 \cdot \epsilon $$ and since $\epsilon$  is at your disposal, this is a contradiction.  I've written this for 1 dimension but I think it applies mutatis mutandis  to all.
