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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!

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    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.2011-10-31
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    @Gortaur: Can you be more specific what $\mathcal{C}$ is, and how the "base" is defined?2011-10-31
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    @Gortaur: That's more analogous to a sub-base.2011-10-31
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    The first two "the"s in this question should be "a"s. A topology in general has many bases.2011-10-31
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    @ChrisEagle: Thanks!2011-10-31
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    @ChrisEagle: thanks, I'm confused with "a/the" as usual2011-10-31
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    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](http://en.wikipedia.org/wiki/Standard_probability_space) (look for "base" on that page)2011-10-31
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    @Tim: since in the first version of your question you were talking about 'the base', I don't think I've understood your question *and how the "base" is defined*2011-10-31
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    @t.b.: Thanks! The concept there is a little unexpected, because it is defined with respect to a measure, while I didn't expect a measure in my question.2011-10-31
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    [This question](http://math.stackexchange.com/questions/54172/the-sigma-algebra-of-subsets-of-x-generated-by-a-set-mathcala-is-the-s) and its answers might be relevant.2011-10-31
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    Well, you can still define a base in this sense without referring to a measure. The writeup there isn't particularly good, but I couldn't find an online reference, and the original references to Rokhlin are a bit hard to find.2011-10-31
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    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: http://ma.huji.ac.il/~matang02/rohlin.pdf2011-12-31
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    @MichaelGreinecker: Thanks for the link! I am unable to find out how it will help with my questions. By the way, I just provided a more clear version of my questions.2012-01-09
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    You wrote that you define base as "the part after if and only if". This means your definition is: *any measurable subset is a union of some members of C.* As pointed out in Didier's, then $\mathcal F$ is already a base. I guess you wanted to include the part about partition in your definition of base; perhaps you could edit your post in a such way that it is clear, what your definition of base is.2012-01-09
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    @MartinSleziak: Thanks! How about now?2012-01-09
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    Thanks for the edit, Tim - this formulation seem to be unambiguous.2012-01-09

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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.

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    Why does the basis have to be made of singletons?2012-01-09
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    Thanks! If such a "base" exists, must it be countable?2012-01-09
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    @AsafKaragila: Didier didn't say it had to. Your question is a good one though.2012-01-09
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    @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).2012-01-09
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    @Didier: Of course one can ask for some "minimality" in the base.2012-01-09