Question. Let $\lambda$ denote Lebesgue measure on $\mathbb R^d$. Rudin shows that for any $x\in \mathbb R^d$, any "regularly" decreasing sequence of Borel sets $E_0(x) \supset E_1(x) \supset \dots$ around $x$(*), and any $f\in L^1(\mathbb R^d)$, we have $$f(x) = \lim_{n\rightarrow \infty} \frac{1}{\lambda(E_n(x))}\int_{E_n(x)} f~d\lambda,$$ for any Lebesgue point $x$, and hence $\lambda$-almost everywhere.
This yields an expression of the density of $\mu = f~d\lambda$ with respect to $\lambda$, which is in an (intuitive) sense "measurable" in $\mu$. My question: What can be said if we replace $\lambda$ with an arbitrary positive measure $\nu$ on $\mathbb R^d$ with respect to which $\mu$ is absolutely continuous, for instance the total variation $|\mu|$. Is it still true that, any $f\in L^1(\nu)$ admits the representation $$f(x) = \lim_{n\rightarrow \infty} \frac{1}{\nu(E_n(x))}\int_{E_n(x)} f~d\nu,$$ for $\nu$-almost every $x$?
Ad (*): precisely, a decreasing sequence $(E_n(x))_{n\in\mathbb N} \in \mathscr B(\mathbb R^d)^{\mathbb N}$ is said to be regularly decreasing iff there is $(r_n)_{n\in\mathbb N}\in \mathbb R_+^\mathbb N$ with $0<r_n \searrow 0$ such that $E_n(x) \subset B_{r_n}(x)$ and, there is $\alpha>0$ with $\lambda(E_n(x))\ge \alpha \lambda(B_{r_n}(x))$.
Motivation. I am concerned with a continuous, finite variation stochastic process $A$ on a filtered probability space $(\Omega,\mathscr F,P)$ (satisfying the usual conditions). Then, by Radon-Nikodym, there is a map $h: \mathbb R_+ \times \Omega \rightarrow \mathbb R$ with $|h| = 1$ which is $P$-almost surely $L^1(\mathbb R_+)$ in the first argument such that $dA = h ~dV$ where $dV$ is the total variation associated to $V$. Question: is $h$ predictable?
Guess. My approach would be to try generalising the above result in the way described. Is this a good idea? One problem I see is that the proof in Big Rudin relies on the specific properties of Lebesgue measure. Are there references on that topic?
Never mind; I found an answer to the question, I think. One has to use Besicovitch's covering theorem instead of the more special one due to Vitali, see for instance Di Benedetto, Real analysis, Birkhäuser 2002. Then one gets a suitable maximal inequality for the maximal function of a complex Borel measure $\mu$ with respect to a given positive Radon measure $\nu$. Then, along the lines of the proof proposed for the Lebesgue measure-case in Big Rudin, one gets - supposing $\mu \ll \nu$ - that the Radon-Nikodym derivative of $\mu$ with respect to $\nu$ is a limit of quotients $$ \frac{\mu(E_n(x))}{\nu(E_n(x))} $$ as $n\rightarrow \infty$.
Please do not hesitate to comment.