$\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $

$\mathcal{C}_1$, $\mathcal{C}_2$ 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

It involves two steps: $$ \mathcal{C}_1\subseteq\mathcal{C}_2\;\;\Longrightarrow\;\;\mathcal{C_1}\subseteq\sigma(\mathcal{C_2})\;\;\Longrightarrow\;\;\sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2). $$ The first implication is obvious since $\mathcal{C}_2\subseteq\sigma(\mathcal{C}_2)$. The second implication holds by definition of $\sigma(\mathcal{C}_1)$, i.e. it is the smallest sigma-algebra that contains $\mathcal{C}_1$. Since $\sigma(\mathcal{C}_2)$ is sigma-algebra that also contains $\mathcal{C}_1$, then it necessarily has to be larger than $\sigma(\mathcal{C}_1)$ (or equal to).

The second implication explained more mathematically: the definition of $\sigma(\mathcal{C}_1)$ is $$ \sigma(\mathcal{C}_1)=\bigcap_{\mathcal{E}\in\Sigma(\mathcal{C}_1)}\mathcal{E}, $$ where $\Sigma(\mathcal{C}_1)=\{\mathcal{E}\mid \mathcal{E}\text{ is a sigma-algebra and }\; \mathcal{C}_1\subseteq\mathcal{E}\}$. If $\mathcal{C}_1\subseteq\sigma(\mathcal{C}_2)$, then $\sigma(\mathcal{C}_2)\in\Sigma(\mathcal{C}_1)$ and thus $\sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2)$ obviously.

This also explains why $\sigma(\sigma(\mathcal{C}))=\sigma(\mathcal{C})$, since $\sigma(\sigma(\mathcal{C}))$ is the smallest sigma-algebra containing $\sigma(\mathcal{C})$, and this is of course $\sigma(\mathcal{C})$ itself.

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 $\sigma(C) \in \{\sigma-\text{fields containing}\ \sigma(C)\}.$