Prove that $\sigma(U \cup\{C\})=\{(A \cap C)\cup(B \cap C^\complement)| A,B \in U\}$

Question

Let $U$ be a $\sigma$-Algebra ($\sigma-algebra$) and $C \subset \mathcal{P}$(power-set) gets added to $U$. How can I prove:

$$\sigma(U \cup\{C\})=\{(A \cap C)\cup(B \cap C^\complement)| A,B \in U\}$$

Answer 1

Observe that:

$$\bigcup_{n=1}^{\infty} \left[(A_n\cap C)\cup(B_n\cap C)\right]=\left[(\bigcup_{n=1}^{\infty} A_n)\cap C\right]\cup\left[(\bigcup_{n=1}^{\infty} B_n)\cap C\right]$$

$$\left[\left(A\cap C\right)\cup\left(B\cap C^{\complement}\right)\right]^{\complement}=\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right)$$

This tells us that the collection on RHS is closed under the formation of countable unions and complements, so is a $\sigma$-algebra.

It contains $C$ as element and $\mathcal U$ as subcollection, so will contain $\sigma(\mathcal U\cup\{C\})$ as subcollection.

Conversely any $\sigma$-algebra that contains $C$ as element and $\mathcal U$ as subcollection will evidently also contain the collection on RHS.