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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)

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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
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.

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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

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