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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:

  1. $\left|A\setminus\bigcup R_{k}\right|=0$
  2. $\left|\bigcup R_{k}\setminus A\right|\leq\varepsilon$
  3. $\left\|\sum_{k}\chi_{R_{k}}\right\|_{L^{1}}\leq C\left|A\right|$

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 $f\in L_{loc}^{p}$.

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  • $\begingroup$ Do you have a reference for the variant? $\endgroup$
    – epimorphic
    Feb 26, 2015 at 5:55
  • $\begingroup$ @epimorphic: The implication is stated but not proved in A. Cordoba and R. Fefferman, "On Differentiation of Integrals". I have written up (what I believe to be) a proof here. $\endgroup$ Feb 26, 2015 at 21:51
  • $\begingroup$ Did you mean $$\lim\limits_{k\to\infty}\dfrac{\left|A\cap R_{k}\right|}{\left|R_{k}\right|}=\chi_{A}(x) \text{ a.e.}?$$ $\endgroup$
    – Jorkug
    Mar 3, 2015 at 13:28
  • $\begingroup$ @Jorkug: Yes, thank you for pointing that out. I have edited the question. $\endgroup$ Mar 3, 2015 at 13:52

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