Generated Sigma Algebras

Question

I am not a mathematician, rather I pick up on topics on the go, when I need something for the topic I am studying in the given time. So I am sorry if this is trivial to most of you and apologies for any conceptual mistakes I might make in the description - I'll try to be as precise as possible.

At the moment, I am studying Probability Theory, from this course: https://www.youtube.com/playlist?list=PL5B3KLQNAC5jT6yjV1199ji1zUy1YUp6P , [for the purposes of understanding Stochastic Calculus for Finance (Vol. II - S. Shreve)], and I stumbled upon sigma algebras.

While I do understand the concept; if we have a set which is a collection of subsets of Omega (i.e. if we have a collection of events) denoted by F, then F is a sigma-algebra if it satisfies the following three conditions;

1. Omega belongs in F,
2. F is closed under complements,
3. F is closed under countable Unions

So far so good and I also understand the properties that derive from the definition as well as how they are derived. In addition I know that we have the trivial sigma algebra, the smallest sigma algebra on Omega and the Discrete Sigma Algebra, which is the power set of Omega, being the largest sigma algebra on Omega.

My problem is with generated sigma algebras. I do understand the definition; Let A be an arbitrary collection of subsets of Omega, then sigma(A) is the generated sigma algebra, generated from A and is the smallest sigma algebra containing A. Further, we can find the smallest sigma algebra by intersecting all sigma algebras containing A, as the intersection of sigma algebras is also a sigma algebra.

The last part is the one I don't understand and confuses me. I get that we have the power set of Omega that definitely contains the collection A - But what exactly do we mean by intersecting all the sigma algebras containing A to find the smallest one containing A? Does it mean that if we have a sigma algebra containing the collection A and another collection of subsets, B (which is a sigma algebra containing A, but i get that it is not the smallest) and intersect it with the power set of Omega, we generate sigma(A), which is indeed the smallest and more refined to answer the questions that we need in our problem? But, where exactly does the bigger sigma algebra (on the collections A and B) come from?

If anyone could provide a more intuitive explanation or even better give an example (finite, like a die roll), I would be very grateful.

Many thanks for your time in reading this! :)

Answer 1

The point is not to take a $\sigma$-algebra $B$ containing $A$ and intersect it with the powerset of $\Omega$. (Note that if $B$ is any collection of subsets of $\Omega$, and $P(\Omega)$ is the powerset of $\Omega$, then $B\cap P(\Omega)=B$. So intersecting with the powerset of $\Omega$ doesn't do much.) The point is that if we have two $\sigma$-algebras $B_1$ and $B_2$ that contain $A$, then $B_1\cap B_2$ is also a $\sigma$-algebra containing $A$ (exercise). Moreover, $B_1\cap B_2$ is going to be smaller than $B_1$ and $B_2$ (unless one of $B_1$ or $B_2$ contains the other).

So $\sigma(A)$ takes this idea to its extreme: we intersect all $\sigma$-algebras containing $A$. In symbols, let $\mathscr{B}$ be the set of $\sigma$-algebras on $\Omega$ that contain $A$. Then $\sigma(A)=\bigcap_{B\in\mathscr{B}}B$. Then $\sigma(A)$ is a $\sigma$-algebra containing $A$ (exercise), and if $B$ is a $\sigma$-algebra containing $A$ then $\sigma(A)\subseteq B$ by definition. So it makes sense to call $\sigma(A)$ the smallest $\sigma$-algebra containing $A$, or the $\sigma$-algebra generated by $A$.

Now you ask where these bigger $\sigma$-algebras come from, and that very much depends on the particular example. In general, the collection $\mathscr{B}$ above could be quite complicated. The most we can say in general is that there is always at least one $\sigma$-algebra in $\mathscr{B}$, namely, the powerset of $\Omega$.

The construction of $\sigma(A)$ described above is good for a definition, but rather difficult to put into practice because it might be difficult or at least very time consuming to calculate $\mathscr{B}$. Given a particular $A$, if one wants to get a more explicit description of $\sigma(A)$ then this usually involves calculating families of sets that must be in any $\sigma$-algebra containing $A$ until you come up with a family that is itself a $\sigma$-algebra. Indeed, if you can come up with a collection $B$ which is a $\sigma$-algebra containing $A$ and must be contained in any $\sigma$-algebra that contains $A$, then it follows that $\sigma(A)=B$.

When $\Omega$ is finite the brute force idea is a little more reasonable because you can just start closing $A$ under intersections and complements until you get an algebra.

Answer 2

Consider, for example, the real line and the collection of all one-point sets $\{\{x\}: x \in \mathbb R\}$. One sigma algebra containing these sets is the power set. Another one is the family of all countable sets and their complements. There are many others too. Now take the sets that are common to all such sigma algebras. That gives the sigma algebra generated by our family. In this case this turns out to be exactly countable sets and their complements. Reason: Any sigma algebra that contains singletons must contain all countable sets (since they are countable unions of singletons). It must also contain their complements. Hence the smallest sigma algebra containig all singeltons is exactly the family of all countable sets and their complements.