TITLE: $\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $
QUESTION [5 upvotes]: $\mathcal{C}_1$, $\mathcal{C}_2$ are collections of subsets of $X$,then Im having hard time seeing why the  following is true. Can someone explain them to me?
$\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $
$\sigma(\sigma( \mathcal{C} )) = \sigma( \mathcal{C} ) $
where $\sigma( \mathcal{F} ) $ is the sigma field generated by the collection $\mathcal{F}$ of subsets of $X$
thanks

REPLY [4 votes]: Do you remember the proof of the existence of the $\sigma$-field generated by a set? You define $\mathcal{A} = \{\sigma-\text{fields containing}\ $C$\}$ (you note that this family is not empty) and you take the intersection of the elements of $\mathcal{A}$.
Once you have this both the properties follow very easily

$\mathcal{A} = \{\sigma-\text{fields containing}\ C_2\} \subset \{\sigma-\text{fields containing}\ C_1\} = \mathcal{B}$. Then $$\sigma(C_1) = \bigcap_{B \in \mathcal{B}}B \subset \bigcap_{A \in \mathcal{A}}A = \sigma(C_2).$$
Clearly $\sigma(C) \subset \sigma(\sigma(C))$. The other inclusion is given by $\sigma(C) \in \{\sigma-\text{fields containing}\ \sigma(C)\}.$