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Fix a finitely additive measure $\mu$ on $\mathbb R^2$ that is consistent with the Lebesgue measure. Does Lebesgue's density theorem hold for such a $\mu$, i.e., is it true that for every $A$ we have $\lim_{\epsilon} \frac{\mu(B_\epsilon\cap A)}{\mu(B_\epsilon)}=0$ or $1$ $\mu$-almost everywhere?

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The answer is no. To see this first observe that if Lebesgue's density theorem holds for $\mu$, then for every set $A$ having $\mu(A)>0$, there is a $\delta>0$ such that for all $t\in \mathbb R^2$ with $|t|<\delta$, we have $A\cap (A+t)\neq \varnothing$. Thus, to provide a counterexample, it suffices to construct a finitely additive measure $\mu$ extending Lebesgue measure, a set $A$ with $\mu(A)>0$, and sequence of elements $t_n\in \mathbb R^2$ such that $|t_n|\to 0$ while $A\cap(A+t_n)=\varnothing$ for each $n$.

Here we construct such $\mu$, $A$, and $t_n$:

Let $\mu$ be a finitely additive, translation invariant measure on $\mathbb R^2$ extending Lebesgue measure.

Let $\xi:\mathbb R^2 \to \mathbb R/\mathbb Z$ be a non-Lebesgue measurable homomorphism. The important property of $\xi$ that we will use is the following:

(*) For all $\varepsilon>0$, the image of $B_\varepsilon(0)$ is dense in $\mathbb R/\mathbb Z$.

Property (*) can be proved in a similar manner to the density of the image of a non-measurable homomorphism of $\mathbb R$ to itself: see the Wikipedia article on Cauchy's functional equation, for example.

The finite additivity of $\mu$ implies that at least one of the sets $A_i=\xi^{-1}([\frac{i}{3},\frac{i+1}{3})), i =0,1,2$ has positive $\mu$-measure, so let $A$ be such an $A_n$. Now use (*) to choose a sequence of elements $t_n\in \mathbb R^2$ such that $\xi(t_n)\in (\frac{1}{3},\frac{2}{3})$ for all $n$ and $|t_n|\to 0$. Then $\xi(A)\cap \xi(A+t_n)=\varnothing$, as $\xi(A)$ and $\xi(A+t_n)$ are disjoint intervals of length $1/3$. Thus $A\cap (A+t_n)=\varnothing$, as well.

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    $\begingroup$ If I understand well, you assume that I want the limit to be 1 for $\mu$-almost all points of $A$. While this is indeed part of Lebesgue's density theorem, I didn't require it. Also, do I see well that in your construction it is essential that $A$ is unbounded? Do you think that you could also construct a bounded counterexample? $\endgroup$
    – domotorp
    Aug 8, 2019 at 0:03
  • $\begingroup$ I did assume the stronger version of the density theorem; I don’t immediately see a short proof of the stronger version from your hypothesis. You can get a bounded set from the example in this answer by taking the intersection of a bounded set of positive measure with the preimage of [i/3,(i+1)/3). $\endgroup$ Aug 8, 2019 at 0:34
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    $\begingroup$ Why would such a bounded set have positive measure? Or do you mean that instead of $\mathbb R^2$ you map from a bounded set? Then why would that be translation invariant? $\endgroup$
    – domotorp
    Aug 8, 2019 at 0:58
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    $\begingroup$ The union of the $A_i$ is all of $\mathbb R^2$, so for a given $S\subset \mathbb R^2$ we have $S=\bigcup_{i\leq 3} A_i\cap S$. If $\mu(S)>0$ then $\mu(A_i\cap S)>0$ for some $i$, as well. If $S$ is bounded and $\mu(A_i\cap S)>0$, then $A_i\cap S$ will also be a counterexample to the generalized density theorem. $\endgroup$ Aug 8, 2019 at 16:56

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