All Questions
6 questions
-1
votes
1
answer
62
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The existence of a maximal “cross-sectional” filter on the Boolean algebra of measurable subsets of [0, 1] modulo almost everywhere equivalence
Let $\mathcal{B}([0, 1])$ be the Boolean algebra of measurable subsets of $[0, 1]$ modulo almost everywhere equivalence, i.e., two measurable sets which differ only by a Lebesgue null set are ...
4
votes
1
answer
203
views
Generalized limits in Boolean algebras
Let $\mathbb{B}$ be an infinite $\sigma$-complete Boolean algebra. By $\mathbb{B}^\omega$ we denote the countable product of $\mathbb{B}$ with the coordinate-wise operations. Let us call a ...
5
votes
2
answers
488
views
Uncountable atomless subalgebras of the Boolean algebra of all Jordan measurable sets in [0,1]
Definition: Suppose $\mathcal A$ is
the Boolean algebra of all Jordan measurable sets in $I=[0,1]$ (i.e $\mathcal A=\{A\subseteq I: \mu(\partial(A))=0\}$, where $\mu$ is the Lebesgue measure and $\...
1
vote
2
answers
503
views
Question on separability of a measure
Following this question here this question come to mind.
Consider a measured σ-algebra $(S,\mu)$ . Assume that μ is normalized to have total weight 1, and that S is complete (contains all subsets of ...
0
votes
1
answer
371
views
additive measure on countable algebras
I was wondering, can the following theorem be true for finitely additive measures defined on algebras not $\sigma$-algebras. (Theorem is in Bogachev's Measure Theory Vol I).
I was not sure about it,...
1
vote
1
answer
447
views
Stone space of measure algebra [closed]
let $\lambda$ be the Lebesgue measure on the unit interval $I=[0,1]$, and $Leb(I)$ be the Boolean algebra of Lebesgue measurable in $I$ and $\mathcal{N}$ the family of Null sets. The measure algebra $\...