Is there a density theorem for Banach measure?

Question

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?

Answer 1

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.