# Tagged Questions

**2**

votes

**1**answer

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

**1**answer

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

**1**answer

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

**3**answers

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

**1**answer

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

**2**answers

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

**0**answers

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

**0**answers

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 ...