Suppose $\mathcal{A}$ is a $\sigma$-algebra on $X$. Let $E \notin \mathcal{A}$. Show $\sigma(\mathcal{A} \cup E)$ consists of elements of the form $(E \cap A) \cup (E^c \cap A')$ where $A, A' \in \mathcal{A}$.
I know the definition of a generated sigma algebra. Am I right in thinking I need to check for an element $x$ to be in $\sigma(\mathcal{A} \cup E)$ it must be of that form because otherwise $\sigma(\mathcal{A} \cup E)$ would not be a sigma algebra?
The standard way to prove statements like this is as follows. (I will let you fill in some of the details.)
(1) Define $\mathcal{E}$ to be the collection of all sets of the form $(E \cap A_1) \cup (E^c \cap A_2)$, $A_1,A_2 \in \mathcal{A}$.
(2) Observe that $\mathcal{E} \subset \sigma(\mathcal{A} \cup E)$.
(3) Observe that $\mathcal{E}$ contains $\mathcal{A}$ and $E$.
(4) Show that $\mathcal{E}$ is a sigma-algebra.
Now, from (3) and (4) it follows that
(5) $\mathcal{E} \supset \sigma(\mathcal{A} \cup E)$
because $\sigma(\mathcal{A} \cup E)$ is the smallest sigma-algebra containing $\mathcal{A}$ and $E$.
Finally, the result follows from (2) and (5).