Sigma algebra generated by a sigma algebra and another set I'm really stuck. 
If $\mathcal{A}$ is a sigma-algebra in $X$ and $C\subset X$ does not belong to $\mathcal{A}$, how do I need to show that the sigma-algebra $\sigma(\mathcal{A}\cup \{C\})$ (so generated by $\mathcal{A}$ and $C$) is equal to the collection of sets of the form $K=(A \cap  C) \cup (B \cap C^c) $, with $A,B \in \mathcal{A}$.
Hope somebody can help me how to solve this problem.
 A: $$[(A\cap C)\cup(B\cap C^{\complement})]^{\complement}=(A^{\complement}\cap C)\cup(B^{\complement}\cap C^{\complement})$$ so the collection of sets of that form is closed under complements.
$$\bigcup_{n=1}^{\infty}[(A_n\cap C)\cup(B_n\cap C^{\complement})]=((\bigcup_{n=1}^{\infty}A_n)\cap C)\cup ((\bigcup_{n=1}^{\infty}B_n)\cap C^{\complement})$$
so the collection of sets of that form is closed under countable unions.
Then it must be a $\sigma$-algebra that evidently contains $\mathcal A$ as a subcollection and has $C$ as one of its elements.
Further it is evident that any $\sigma$-algebra with these properties will contain every element of the form $(A\cap C)\cup(B\cap C^{\complement})$ where $A,B\in\mathcal A$.

edit:
$\begin{aligned}\left[\left(A\cap C\right)\cup\left(B\cap C^{\complement}\right)\right]^{\complement} & =\left(A\cap C\right)^{\complement}\cap\left(B\cap C^{\complement}\right)^{\complement}\\
 & =\left(A^{\complement}\cup C^{\complement}\right)\cap\left(B^{\complement}\cup C\right)\\
 & =\left(A^{\complement}\cap B^{\complement}\right)\cup\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right)\\
 & =\left(A^{\complement}\cap B^{\complement}\cap C\right)\cup\left(A^{\complement}\cap C\right)\cup\left(A^{\complement}\cap B^{\complement}\cap C^{\complement}\right)\cup\left(B^{\complement}\cap C^{\complement}\right)\\
 & =\left(A^{\complement}\cap C\right)\cup\left(B^{\complement}\cap C^{\complement}\right)
\end{aligned}
$
