# Is there a counterpart of a base of a topology for a $\sigma$-algebra?

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. – Ilya Oct 31 '11 at 16:36
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
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
This question and its answers might be relevant. – Arturo Magidin Oct 31 '11 at 17:10
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

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? – 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
@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