Timeline for Regular Borel measures and the measure of a singleton
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2018 at 2:00 | comment | added | André Porto | Ok. I've got it now. Thanks a lot. About the correction in your first comment, it has no effect. By the assumption on $\mu$, the conditions $C_1$: "$\mu(A)<1/3$" and $C_2$: "$\mu(A)<2/3$" are equivalent, and any value at the interval $[1/3,2/3]$ is forbidden and would fit to the contradiction. Finally, the negation of $C_1$ (which is precisely the negation of $C_2$) is $\mu(A)\geq 1/3$ which, by the assumption on $\mu$ is equivalent to $\mu(A)> 2/3$. | |
| Oct 21, 2018 at 0:20 | comment | added | erz | @AndréPorto Assume that the supremum of the masses of atoms is not attained. Then there is a sequence $B_n$ of atoms such that $\mu(B_n)$ grows to $\alpha>0$. Consider $C_n=B_n\backslash (B_1\cup...\cup B_{n-1})$. Since $B_n$ is an atom $\mu(B_n\cap B_k)$ is either $\mu(B_n)$ (which is impossible since $\mu(B_n\cap B_k)\le \mu(B_k)<\mu(B_n)$, or $0$. Hence, $C_n$ is a sequence of disjoint sets, with $\mu(C_n)=\mu(B_n)$ converging to $\alpha>0$. Contradiction. | |
| Oct 20, 2018 at 21:49 | comment | added | André Porto | @erz, is there such a thing as an atom with largest mass? I ask that because unions of atoms are not atoms. | |
| Oct 20, 2018 at 21:46 | comment | added | André Porto | Actually, it lacks to prove that $\mu(L)=\mu(A)$ in the argument of the last comment. | |
| Oct 20, 2018 at 21:42 | comment | added | André Porto | If we assume that $\mu$ is inner regular and that $S$ a $T_2$ (no need of compactness) we conclude that any atom has a singleton in which $\mu$ is concentrated. Fix an atom $A$ and the family $F$ of compact subsets of $A$ with measure $\mu(A)$. Since $\mu$ is inner regular, $F$ is non-empty. Since $A$ is an atom, the family $F$ has the F.I.P.. Let $L=\cap F\neq\emptyset$. If we prove that $\mu(L)=\mu(A)$, then $L$ is the least element of $F$ and, as such, using inner regularity again, we prove that $L$ has no proper non-empty measurable subset. Since $S$ is $T_2$, $L$ must be singleton. | |
| Oct 20, 2018 at 21:25 | comment | added | erz | I think you meant to say: if $\mu(A)<\frac{2}{3}$, then ... and so $\mu$ takes a forbidden value $\frac{1}{3}$. @AndréPorto Let $A_1$ be an atom with the largest mass, let $A_2$ be an atom in $S\backslash A_1$ with the largest mass, and so on. Since $\mu(S)=1$ we have $\mu(A_n)\to 0$, from where $\mu$ is atomless on $S\backslash A$. | |
| Oct 20, 2018 at 20:53 | comment | added | André Porto | Just one question: how do you know that the family $A_1, A_2, ...$ is countable? | |
| Oct 20, 2018 at 15:50 | history | answered | Dirk Werner | CC BY-SA 4.0 |