Proving equality of sigma-algebras Let $C_1$ and $C_2$ are two collections of subsets of the set $\Omega$. We want to show that if $C_2$ $\subset$ $\sigma$[$C_1$] and $C_1$ $\subset$ $\sigma$[$C_2$], then $\sigma$[$C_1$]=$\sigma$[$C_2$] where $\sigma$[$C$] is defined to be the sigma-algebra generated by a collection $C$ of subsets of $\Omega$ (or the smallest sigma-algebra).
I'm having trouble characterizing the smallest sigma-algebra of sets in $C_1$ and $C_2$. Would the smallest sigma-algebra of $C_1$ be {$\emptyset$, $\Omega$, $C_1$, $\Omega$ $\not$ $C_1$} or would we have to characterize them in terms of the sets in $C_1$ which is the countable subcollection of sets in $\Omega$?
 A: First of all, $\{\varnothing,\Omega,C_1,\lnot C_1\}$ is not a $\sigma$-algebra on $\Omega$. Recall that $C_1$ is a collection of subsets of $\Omega$ and a $\sigma$-algebra on $\Omega$ is a collection of subsets of $\Omega$. So $C_1$ should be a subset of $\sigma(C_1)$ and not an element of it. 
You can define $\sigma(C)$ as the intersection of all $\sigma$-algebras which contain $C$. But you can also start closing $C$ under complements and countable unions. The problem is that this process might take you $\omega_1$ steps to complete (and without the axiom of choice, possibly longer).
So while it is very very useful to understand the internal construction of "the smallest $\sigma$-algebra such that ...", it is unnecessary in this case. The second definition is easier to use.
HINT: Note that if $C_2\subseteq\sigma(C_1)$ then $\sigma(C_2)\subseteq\sigma(C_1)$ (by definition of "smallest"). 
A: Use twice the fact that, if $A$ and $B$ are two subsets of $2^\Omega$ such that $A\subseteq\sigma(B)$, then $\sigma(A)\subseteq\sigma(B)$, once for $(A,B)=(C_1,C_2)$, and once for $(B,A)=(C_1,C_2)$.
The fact itself is direct using the identity
$$\sigma(B)=\bigcap_{F\in\mathfrak F} F,
$$
where $\mathfrak F$ is the collection of sigma-algebras $F$ such that $B\subseteq F$.
