This is a follow-up question to this one.
Is there a ZFC example of an atomless measure that is $2^\omega$-additive, meaning, fewer than continuum many null sets have measurable union that is null?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Are you assuming the collection of measurable sets is a $\sigma$-algebra? $\endgroup$– Monroe EskewMay 6, 2020 at 6:02
-
$\begingroup$ @MonroeEskew, standardly, measures are defined on $\sigma$-algebras. $\endgroup$– KantMay 6, 2020 at 6:15
-
$\begingroup$ I think what you are looking for are real valued measurable cardinals and their existence is equiconsistent to a measurable cardinal. $\endgroup$– Johannes SchürzMay 6, 2020 at 8:38
-
1$\begingroup$ @JohannesSchürz, no, because I do not insist on the measure to be defined for all sets. $\endgroup$– KantMay 6, 2020 at 8:42
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
6
The answer is negative. Suppose $(X,\mu)$ is such a measure space. By the argument in the linked question, there is a partition of $X$ into continuum-many pairwise disjoint $\mu$-null sets. This is done by building a binary tree that splits a given node into two nodes of one half the measure. Each branch corresponds to a point in Cantor space. We induce an atomless $2^\omega$-additive measure $\nu$ on Cantor space via this correspondence.
However, it is consistent with ZFC that the continuum is singular. Under this hypothesis, any $2^\omega$-additive ideal is $(2^\omega)^+$-additive. But each singleton is $\nu$-null, and so the Cantor space is the union of $2^\omega$-many $\nu$-null sets, meaning the whole space has measure 0. But this means that $\mu(X)=0$.
answered May 6, 2020 at 5:46
-
$\begingroup$ It looks like even restricting to finitely additive measures will not help? $\endgroup$– KantMay 6, 2020 at 9:20
-
$\begingroup$ @Kant: What do you mean? That the null sets form a continuum-additive ideal but the measure is not countably additive on positive sets? $\endgroup$ May 6, 2020 at 9:23
-
$\begingroup$ @Kant I think you‘re right. The binary tree construction will transfer the additivity of the null ideal to the induced measure on the reals. But of course we have to assume the measure is defined on a sigma algebra to make sure the intersections along branches are measurable. $\endgroup$ May 6, 2020 at 9:53
-
$\begingroup$ @Kant: Yes, I think so. Take some regular cardinal $\kappa \geq 2^\omega$ and consider the nonstationary ideal on $\kappa$. Then build a complete binary tree of stationary sets with $\kappa$ at the top. Then just define the probability function on finite unions of nodes in the tree such that the immediate children of a node each have probability 1/2 of the parent. The algebra of sets is just the algebra generated by these stationary sets plus/minus a nonstationary set. And the measure of each nonstationary set is zero. $\endgroup$ May 6, 2020 at 10:17
-
$\begingroup$ Thanks. So it seems I can been every cardinal by the additivity of the null ideal? $\endgroup$– KantMay 6, 2020 at 10:38