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 $\lambda$,


$$\lim\_{r\to0}\frac{\mu(B(x,r))}{\lambda(B(x,r))}<\infty \quad \mbox{ as } \quad r\to0, \quad \lambda\mbox{-a.e.}$$



If a measure $\mu$ and Lebesgue measure $\lambda$ are singular, is the derivative of $\mu$ with respect to $\lambda$ $\infty$, $\mu$-a.e.?
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.


