Timeline for Is there a density theorem for Banach measure?

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Aug 8, 2019 at 16:56	comment	added	John Griesmer		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.
Aug 8, 2019 at 0:58	comment	added	domotorp		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?
Aug 8, 2019 at 0:34	comment	added	John Griesmer		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).
Aug 8, 2019 at 0:03	comment	added	domotorp		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?
Aug 7, 2019 at 22:48	history	answered	John Griesmer	CC BY-SA 4.0