2
votes
1answer
129 views

The Universal Algebra of a sigma-Algebra

I am searching for the 'dual' algebraic structure of a Sigma Algebra. The notion of duallity is like on the case of the Boolean Algebra and Set Algera. If X is a set, the complement and intersection ...
4
votes
1answer
157 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 ...
0
votes
1answer
86 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 ...
7
votes
3answers
275 views

Which Sigma-Ideals in a Sigma-Algebra are Ideals of Null Sets?

My question is motivated, to be somewhat vague, by an attempt to see how much a measure space is defined by the set of null sets. In other words, assume we are not given a concrete measure on a space ...
0
votes
1answer
118 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 ...
3
votes
2answers
298 views

Cohen algebra (generalization)

Let Bor($X$) = class of all borel subsets of $X$. Cohen algebra is defined as Bor(X) modulo the ideal of meager sets. The Cohen algebra has a combinatorial : it is the unique atomless complete ...
1
vote
0answers
104 views

Can random elements be defined in terms of a measure algebra?

Let $(\Omega,\Sigma,\mu)$ be a probability space, $(X,\mathcal{X})$ be a measurable space and $R(\Omega,X)$ be the set of equivalence classes of measurable functions from $\Omega$ to $X$ under almost ...
13
votes
0answers
475 views

Which complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras?

Projections in an arbitrary commutative von Neumann algebra form a complete Boolean algebra. Moreover, a morphism of commutative von Neumann algebras induces a continuous morphism of the corresponding ...