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Nov 4, 2022 at 14:55 comment added Arno @AaronHill Ah, of course. Thanks for clearing up my confusion.
Nov 3, 2022 at 21:46 comment added Aaron Hill I don't think finite unions is that easy. When you are partitioning the set $A \cup B$ into pieces you must be using half-open intervals. That means that elements of your partition may have a combination of points from $A$ and points from $B$. You can't get all of the points in $A$ to go in $[0, \frac{\epsilon}{2})$ and all of the points of $B$ to go in $[\frac{\epsilon}{2}, \epsilon)$. Maybe the way you can get the points of $A$ to fit in a small interval is incompatible (somehow) with the way you can get the points of $B$ to fit in a small interval.
Nov 3, 2022 at 19:29 comment added Arno For the union, it should just work to fit $A$ into $[0,\varepsilon/2)$ and $B$ into $[\varepsilon/2,\varepsilon)$, right? More generally, the natural candidate would seem to be to set $\mu(A)$ to be the $\inf$ of all $\varepsilon$ where it works. Problem just is that this should be finitely sub-additive, but I don't see why it would finite additive.
Nov 3, 2022 at 19:05 comment added Aaron Hill I don't know whether the union of two small sets must be small, but suspect that it must. (That there is no Banach-Tarski decomposition in one dimension, using only finitely many pieces, seems like it might be relevant here.) Since countably many translates of a Vitali set covers the line, this notion of small is not closed under countable unions.
Nov 3, 2022 at 18:33 comment added Pierre PC If such a measure exists, then in particular a union of two small sets has to be small. Do you know if this is indeed the case? (If so, then the measure giving 0 to small sets and infinite mass to the others fits the bill, but it is not particularly enlightening.)
Nov 3, 2022 at 17:31 history asked Aaron Hill CC BY-SA 4.0