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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 $\lambda$,

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

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  • $\begingroup$ Doesn’t mutual singularity guarantee you that $\lambda(B(x,r))/\mu(B(x,r))\to 0$, $\mu$-a.e., which is the same thing? $\endgroup$ – Anthony Quas Mar 15 '18 at 6:30
  • $\begingroup$ Yes, you are right. As the answer show, if $\mu\bot\lambda$ and $B=\{x\in A:\ \lim \lambda(B(x,r))/\mu(B(x,r))>c\}$, then $c\mu(B)\le \lambda(B)=0$. Thanks! $\endgroup$ – ohliv Mar 15 '18 at 18:36
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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.

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  • $\begingroup$ Nice answer. It is a bit hard to read, though. You might want to fix a few typos and otherwise brush up your answer. $\endgroup$ – Iosif Pinelis Mar 15 '18 at 16:25
  • $\begingroup$ Josif, I tried to do that. $\endgroup$ – user83457 Mar 15 '18 at 16:43
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$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\la}{\lambda} \newcommand{\ep}{\epsilon} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}}$

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. $$\frac{\mu(B(x,r))}{\lambda(B(x,r))}\to\infty \quad \mbox{ as } \ r\downarrow0.$$

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