The Banach–Ulam problem asks: are there any nontrivial measures on a set $X$ equipped with the σ-algebra $2^X$ of all subsets of $X$?
All measures are by definition countably additive.
Here a measure on $(X,2^X)$ is trivial if it can be obtained by the following construction. Start with a set $X$, a map $f\colon X\to\mathbf{R}$, and a σ-ideal? $I$ on $X$. Set $\mu(A)=\sum_{x\in A}f(x)$ if $A\in I$ and $\mu(A)=\infty$ otherwise. Then $\mu$ is a measure on $(X,2^X)$.
Once trivial measure on $(X,2^X)$ are excluded, we can further reduce to the case of probability measures on $(X,2^X)$ that vanish on all singleton subsets of $X$.
Recall that the additivity of a poset $P$ is the smallest cardinal $\kappa$ such that $P$ has a subset of cardinality $\kappa$ without an upper bound.
Recall that the additivity of a measure $\mu$ is the smallest cardinal $\kappa$ such that there is a disjoint family of cardinality $\kappa$ of measurable subsets such that the additivity property of $\mu$ fails for this family.
Recall that any measure $\mu$ on $(X,\Sigma)$ induces a σ-ideal? $N_\mu$ such that $A\in N_\mu$ if $A\subset X$ and there is $B\in\Sigma$ such that $A\subset B$ and $\mu(B)=0$.
A cardinal $\kappa$ is real-valued-measurable if there is a $\kappa$-additive probability measure $\mu\colon 2^\kappa\to\mathbf{R}$ that vanishes on all singleton subsets of $\kappa$.
If the probability measure only takes values 0 and 1, then $\kappa$ is known as a measurable cardinal.
A cardinal $\kappa$ is atomlessly-measurable if there is an atomless $\kappa$-additive probability measure $\mu\colon 2^\kappa\to\mathbf{R}$.
Here an atom of a measure $\mu$ on $(X,\Sigma)$ is $A\in\Sigma$ such that $\mu(A)\ne0$ and if $B\in\Sigma$ satisfies $B\subset A$, then either $\mu(B)=0$ or $\mu(A\setminus B)=0$.
(Ulam?.) Suppse $(X,2^X,\mu)$ is a nontrivial probability space.
The additivity of $\mu$ equals the additivity of the σ-ideal? $N_\mu$ and is a real-valued measurable cardinal.
Any real-valued-measurable cardinal is either atomlessly measurable or measurable.
Any atomlessly measurable cardinal is weakly inaccessible? and not greater than $\mathfrak{c}$.
Any measurable cardinal is strongly inaccessible?.
The Lebesgue measure on $\mathbf{R}$ extends to $2^{\mathbf{R}}$ if and only if there is an atomlessly measurable cardinal.
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Last revised on May 15, 2020 at 21:44:59. See the history of this page for a list of all contributions to it.