Let $\mathcal{B}$ be a collection of bounded, measurable subsets of $\mathbb{R}^{n}$, such that for every $x$ there exists a sequence $\left\{R_{k}(x)\right\}\subset\mathcal{B}$ containing $x$ and with diameter tending to zero. Given a measurable set $E$, we say that $\mathcal{V}\subset\mathcal{B}$ is a Vitali covering of $E$ if for every $x$ there exists a sequence $\left\{R_{k}(x)\right\}\subset\mathcal{V}$ containing $x$ and with diameter tending to zero.
Suppose that for every measurable set $A$ with $\left|A\right|<\infty$, every Vitali covering $\mathcal{V}$ of $A$, and every $\varepsilon>0$, there exists a countable subcollection $\left\{R_{k}\right\}\subset\mathcal{V}$ satisfying the following properties:
- $\left|A\setminus\bigcup R_{k}\right|=0$
- $\left|\bigcup R_{k}\setminus A\right|\leq\varepsilon$
- $\left\|\sum_{k}\chi_{R_{k}}\right\|_{L^{1}}\leq C\left|A\right|$
where $C$ is a constant independent of $A$, $\mathcal{V}$, and $\varepsilon$. Does it follow that $\mathcal{B}$ satisfies the density property $$\lim_{k\rightarrow\infty}\dfrac{\left|A\cap R_{k}\right|}{\left|R_{k}\right|}=\chi_{A}(x) \ \text{ a.e. (*)}$$ where $x\in R_{k}\in\mathcal{B}$ for all $k$ and the diameter of $R_{k}$ tends to zero?
My motivation for asking this question is that when $q>1$ and $1/q+1/p=1$, a variant of this theorem holds when we replace condition 3 with $$\left\|\sum_{k}\chi_{R_{k}}\right\|_{L^{q}}\leq C\left|A\right|^{1/q}$$ and (*) by $$\lim_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f=f(x) \ \text{ a.e.}$$ for every sequence $\left\{R_{k}\right\}$ as above and $f\in L_{loc}^{p}$.