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.

We've got the following definition

Let $\mathcal C$ be a class of subsets of $\Omega$. We say that $\sigma(\mathcal C)$ is the $\sigma$-algebra generated by $\mathcal C$ if satisfies that: 1. $\mathcal C\subseteq \sigma(\mathcal C)$. 2. If $\mathcal C\subseteq \mathcal A$, with $\mathcal A$ another $\sigma$-algebra, then $\sigma(\mathcal C)\subseteq \mathcal A$.

I was checking some old problems in my probability notes, and to be honest I really don't understand this, like why do we need this? when do we use this? I can see some things derived from the definition, like $\sigma(\mathcal C)$ is the smallest $\sigma$-algebra that contains $\mathcal C$ or the trivial fact that this is really a $\sigma$-algebra. Also, supposedly, the $\sigma(\mathcal C)$ is the intersection of all the $\sigma$-algebras that contain $\mathcal C$, I don't get how this is something easy to see.

(I was hesitant to add the [measure-theory] tag, since I haven't study that yet)

share|improve this question
1  
A hint for 2: show that any intersection of $\sigma$-algebras is again a $\sigma$-algebra. –  Nate Eldredge Nov 20 '13 at 0:47
    
@NateEldredge yes, I've got that, thanks :) –  Ana Galois Nov 20 '13 at 1:04
add comment

1 Answer 1

up vote 1 down vote accepted

It's hard to directly construct the smallest $\sigma$-algebra containing $\mathcal C$. Instead, you define it as a sort of "infimum" over all $\sigma$-algebras that contain $\mathcal C$.

For the question about intersections, clearly the intersection of all $\sigma$-algebras containing $\mathcal C$ must be contained in $\sigma(\mathcal C)$, since this is one of the elements you are intersecting. Now recall that the intersection of $\sigma$-algebras is again a $\sigma$-algebra. So if the intersection was strictly contained in $\sigma(\mathcal C)$, this would contradict condition $2$ of your definition.

share|improve this answer
    
Oh yes, I see... So it's like saying $\mathcal C\subseteq \sigma(\mathcal C)\subseteq \mathcal A$ for any $\mathcal A$, $\sigma$-algebra, right? Now it seems very direct. If this doesn't bother you, I'd like to know why do we need this? I know that the definition of $\sigma$-algebra is needed for probability measure, but in this particular type, does anything different happen? –  Ana Galois Nov 20 '13 at 1:04
    
@AnaGalois You want to use this definition because you often want to construct a $\sigma$-algebra that contains certain sets. For example, to develop the theory of integration, you want a $\sigma$-algebra that contains all the open sets. But the collection of open sets is not itself a $\sigma$ algebra. So instead you use the $\sigma$-algebra the open sets generate. –  Potato Nov 20 '13 at 1:08
    
wow, that's a handy use. Thanks for the answer. –  Ana Galois Nov 20 '13 at 1:11
    
@AnaGalois Since you're studying probability, here is perhaps a better example: Suppose want to study a collection of events on a probability space. It is not always the case that this collection will be a $\sigma$-algebra, so you can take the $\sigma$-algebra generated by the events to get one. –  Potato Nov 20 '13 at 1:16
    
Yes, that's what I thought after your comment before. Thanks again. –  Ana Galois Nov 20 '13 at 1:23
add comment

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.