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Feb 23, 2023 at 7:38 comment added IljaKlebanov Thank you, Nate!
Feb 23, 2023 at 3:34 comment added Nate Eldredge @iljusch: Let $E$ be the support. If it is not separable then it contains an uncountable set $S$ with no limit points. So for each $x \in S$ we can find an open ball $B(x, r_x)$ containing no other points of $S$. Then the balls $B(x, r_x/2)$ are pairwise disjoint. But since $x$ is in the support, each of these uncountably many disjoint balls must have positive measure. This is impossible for a finite or $\sigma$-finite measure.
Feb 22, 2023 at 23:59 comment added IljaKlebanov Thanks for your reply! I am not familiar with this result, could you please point me to a reference? So your argument would be: Consider the support of the (regular finite) measure, which is separable. Then take the closure of its linear span, which is a (possibly smaller) separable Banach space and work therein. Is this correct?
Feb 22, 2023 at 14:22 comment added Dirk Werner Wouldn't the support of a (regular finite) measure be separable?
Feb 22, 2023 at 8:51 comment added IljaKlebanov @Dirk Werner : Preiss and Tiser claim to show this for general Banach spaces in the title and the abstract, but the Theorem itself is formulated and proved for separable Banach spaces only. Do you know whether the general statement holds? Thanks in advance!
Mar 7, 2020 at 17:33 comment added Dirk Werner On the positive side, the answer to whether $\mu$ is zero is yes for Banach spaces; as proved by D. Preiss and J. Tiser, Measures in Banach spaces are determined by their values on balls. Mathematika 38, No. 2, 391-397 (1991). Zbl 0755.28006
Mar 4, 2020 at 18:34 history edited Nate Eldredge CC BY-SA 4.0
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Mar 4, 2020 at 18:27 history answered Nate Eldredge CC BY-SA 4.0