Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Update post on Jan 9, 2012:

Given a sigma algebra $\mathcal{F}$ on a set $X$, and a partition $\mathcal{C}$ of $X$. If I am correct, then:

$\mathcal{C}$ is a generator of $\mathcal{F}$, if and only if any measurable subset is a union of some members of $\mathcal{C}$.

Such class of subsets (partition plus the part after "if and only if" characterizes it) to the sigma algebra is like a base to a topology. Allow me to call it the "base" of the sigma algebra.

I wonder if any sigma algebra always has a "base"? If a sigma algebra has finitely many measurable subsets, then there exists a "base". If there is a "base", must the sigma algebra has finitely many measurable subsets?

Thanks and regards!


Original post:

A base of a topology is defined as a collection of open sets such that every open set is a union of some of them.

I was wondering if there is a similar concept for a $\sigma$-algebra? My question arose from a notice that a class of subsets that form a partition of the universe seems like a "base" for the $\sigma$-algebra it generates.

Actually I am curious if there is a general concept for a class of subsets closed under some set operation(s).

Thanks and regards!

share|improve this question
    
Does the construction $\sigma(\mathcal C)$ which is the intersection of all $\sigma$-algebras containing $\mathcal C$ answer your question? Then you need no conditions on $\mathcal C$ to generate $\sigma$-algebra, while you do have such conditions for the base. –  Ilya Oct 31 '11 at 16:36
2  
The first two "the"s in this question should be "a"s. A topology in general has many bases. –  Chris Eagle Oct 31 '11 at 16:40
1  
Sometimes in the construction @Gortaur refers to $\mathcal{C}$ is called a base. More common seems to be a countable generating set for a standard probability space (look for "base" on that page) –  t.b. Oct 31 '11 at 16:46
2  
This question and its answers might be relevant. –  Arturo Magidin Oct 31 '11 at 17:10
2  
I have spent two weeks looking for the original paper by Rokhlin. It is actually not that hard to find- if you look for the name it was published under, Rohlin: ma.huji.ac.il/~matang02/rohlin.pdf –  Michael Greinecker Dec 31 '11 at 19:05

1 Answer 1

up vote 4 down vote accepted

Here are some simple examples which are enough to answer your Jan 9, 2012 questions. Denote by $\mathcal S(X)=\{\{x\};x\in X\}$ the set of singletons of a set $X$.

The power set $2^\mathbb Z=\{A;A\subseteq\mathbb Z\}$ of $\mathbb Z$ is a sigma-algebra on $\mathbb Z$ with $\mathcal S(\mathbb Z)$ as "base". But $2^\mathbb Z$ is neither finite nor countable. In fact, there is no such thing as an infinite countable sigma-algebra.

The Borel sigma-algebra $\mathcal B(\mathbb R)$ has no "base" since any of its bases should contain every singleton, hence the base could only be $\mathcal S(\mathbb R)$, but the subset $\mathbb R_+$ is in $\mathcal B(\mathbb R)$ and is neither countable nor co-countable.

share|improve this answer
    
Why does the basis have to be made of singletons? –  Asaf Karagila Jan 9 '12 at 7:08
    
Thanks! If such a "base" exists, must it be countable? –  Tim Jan 9 '12 at 7:10
    
@AsafKaragila: Didier didn't say it had to. Your question is a good one though. –  Tim Jan 9 '12 at 7:20
2  
@Asaf: If C is a base and every singleton is measurable, every singleton belongs to C. If C is also a partition of X then C=S(X). So, either one asks that a base is also a partition, and then, B(R) has no base (this is the option in my post), or one does not ask that, and then, every sigma-algebra has itself as a base (and the question is empty). –  Did Jan 9 '12 at 7:44
    
@Didier: Of course one can ask for some "minimality" in the base. –  Asaf Karagila Jan 9 '12 at 8:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.