What is the maximal function of a finite Borel measure on $\Bbb R^n$?

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Here is this week's POTW:

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Suppose $\mu$ is a finite Borel measure on $\Bbb R^n$. Define the maximal function of $\mu$ by $$\mathcal{M}\mu(x) = \sup_{0 < r < \infty} \frac{\mu(B(x;r))}{m(B(x;r))}\quad (x\in \Bbb R^n)$$ Here, $m$ denotes the Lebesgue measure on $\Bbb R^n$. Show that if $\mu$ is mutually singular with respect to $m$ (i.e., $\mu \perp m$), then $\mathcal{M}\mu = \infty$ a.e. $[\mu]$.-----

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No one answered this week's problem. You can read my solution below.

Spoiler

It suffices to show that the symmetric derivative $D\mu = \infty$ a.e. [$\mu$]. As $\mu\perp m$, there is a Borel set $A \subset \Bbb R^d$ such that $m(A) = 0 = \mu(\Bbb R^d \setminus A)$. Then $D\mu(x) = \infty$ for all $x\in A\setminus \cup A_n$, where $A_n$ consists of all $x\in A$ for which there is a sequence $r_k \to 0$ such that $\mu(B(x;r_k)) < n\, m(B(x;r_k))$ for all $k$.

By regularity of $\mu$, there are open sets $O_k$ containing $A$ such that $m(O_k) < 3^{-k}$ for all $k$. Letting $n$ and $k$ be fixed, we observe that for each $x\in A_n$, there is an open ball $B(x) \subset O_k$ containing $x$ such that $\mu(B(x)) < n\, m(B(x))$. Let $\Sigma_{k,n}$ be the union of the balls $(1/3)B(x)$ as $x$ ranges over $A_n$. If $K$ is a fixed compact subset of $\Sigma_{k,n}$, then $K$ is covered by finitely many of the balls $(1/3)B(x)$. By a covering lemma, there is a finite set $S$ such that the collection $\{B(x) : x\in S\}$ covers $K$ and $\{(1/3)B(x):x\in S\}$ is disjoint. Now

$$\mu(K) \le \sum_{x\in S} \mu(B(x)) < n \sum_{x\in S} m(B(x)) = 3^d n \sum_{x\in S} m((1/3)B(x)) \le 3^d n\, m(O_j) < 3^{d-j}n$$

By regularity of $\mu$, $\mu(\Sigma_{k,n}) \le 3^{d-j}n$. If $\Sigma_n := \cap_k \Sigma_{k,n}$, then $D\mu(x) = \infty$ for every $x\in A \setminus \cup \Sigma_n$ and $\mu(\Sigma_n) = 0$ for all $n$. Hence, $D\mu = \infty$ a.e. [$\mu$].