# Smallest of sigma-algebra

Can anyone show me how to solve this question extracted from Michael Taylor's book:

If $\{\mathcal{F}_{\alpha} : \alpha \in A\}$ is the collection of all $\sigma$-algebras of subsets of $X$ that contain $\mathcal{C}$, show that $$\bigcap_{\alpha \in A} \mathcal{F}_{\alpha} = \mathcal{F}$$ is a $\sigma$-algebra of subsets of $X$, containing $\mathcal{C}$, and is in fact the smallest such $\sigma$-algebra. One says $\mathcal{F}$ is the $\sigma$-algebra generated by $\mathcal{C}$ and writes $\mathcal{F} = \sigma(\mathcal{C})$

• You just have to verify the defining properties of a $\sigma$ algebra for the intersection and that $\cal{C}$ is contained in that intersection. Chase through the definitions. Since each member of the intersection has the corresponding properties this is actually straightforward. – user20266 May 29 '12 at 17:55
• Note that $\mathcal{F}\subset\mathcal{F}_\alpha$ for all $\alpha\in A$. This proves minimality. – Norbert May 29 '12 at 17:55
• As I spell out below, it is not only for sigma-algebras that you use the technique that gets you the answer to this question; it's for many other things as well. – Michael Hardy May 29 '12 at 19:05

Recall the definition of a $\sigma$-algebra: it is a collection of subsets of some particular $X$, such that:

1. It is closed under countable unions;
2. It is closed under taking complement (relative to $X$).

From this, countable intersections follow by DeMorgan's Laws.

Now suppose that $X_i\in\mathcal F$ for $i\in\mathbb N$. So for all $\alpha\in A$ we have $X_i\in\mathcal F_\alpha$. Since $\mathcal F_\alpha$ is a $\sigma$-algebra we have that $\bigcap X_i\in\mathcal F_\alpha$ for all $\alpha\in A$ and therefore $\bigcap X_i\in\bigcap\mathcal F_\alpha=\mathcal F$.

For complements, the principle is the same.

Furthermore, the same idea works to show that $\mathcal F$ contains $\mathcal C$. Lastly we need to show that this is indeed the smallest:

Suppose that $\mathcal S$ is a $\sigma$-algebra which contains $\mathcal C$, then for some $\alpha\in A$ we have $\mathcal S=\mathcal F_\alpha$, so it took part in the intersection which generated $\mathcal F$, therefore $\mathcal F\subseteq\mathcal S$.

If an XXXXX is defined as a set closed under certain operations, then any intersection of XXXXXs is another XXXXX. The proof is as follows. Suppose $\{A_\alpha : \alpha\in I\}$ is a set of XXXXXs and $u,v\in \bigcap\limits_{\alpha\in I}A_\alpha$. Then for every $\alpha\in I$, $u,v\in A_\alpha$. Performing the operations on $u,v$ under which every XXXXX is closed, we conclude that what we get is in $A_\alpha$. Since that's true of every value of $\alpha$, what we get is in $\bigcap\limits_{\alpha\in I}A_\alpha$. Therefore $\bigcap\limits_{\alpha\in I}A_\alpha$ is an XXXXX.

DO NOT take "$u,v$" to mean just two things; construe it if necessary as a countably infinite or even bigger class of things, if the operations under which every XXXXX is closed require that.

Now instead of "an XXXXX", just say "an XXXXX of which $\mathcal{C}$ is a subset", and apply that argument.

That's the answer to your first question. But if it's homework, it should be spelled out in more detail than what appears above.

Now what about "smallest"? Now we need the assumption that $\{A_\alpha : \alpha\in I\}$ is not just a collection of XXXXs containing $\mathcal{C}$ as a subset, but is the collection of all XXXXXs containing $\mathcal{C}$ as a subset.

It's the smallest simply because for every $\alpha_0$, $\displaystyle A_{\alpha_0}\supseteq \bigcap_{\alpha\in I}A_\alpha$. "Smallest" means simply that it's a subset of all others.

• What???${}{}{}$ – Norbert May 29 '12 at 17:56
• @Norbert: This way he gets the first answer. This is the equivalent of "FIRST!!" comments in other sites. – Asaf Karagila May 29 '12 at 17:58
• I do not like the XXXXX notation. I get lost when I try to count the Xs. :-) – user20266 May 29 '12 at 17:58
• @MichaelHardy You can delete your question, edit it and then undelete. – Norbert May 29 '12 at 18:05
• I don't think I would get anything for being "first". My contribution that is not in the other answer is mainly that it is not only in the case if sigma-algebras that this argument works; it's lots and lots of things. – Michael Hardy May 29 '12 at 18:11