Prove that $M$ is $\sigma$-algebra Let $X$ be a set and let $S$ be a $\sigma$-algebra on $X$ with $E\subseteq X$.
Some subset, $M := \{ (A \cap E) \cup (B\setminus E) | A,B \in S \} $.
Prove that $M$ is $\sigma$-algebra on $X$ and moreover prove that $M$ is the smallest $\sigma$-algebra on $X$ that contains all members of the family $S\cup\{E\}$.
Please help with this problem or at least offer some hints on how to show that is the smallest.
 A: $M$ is a sigma-algebra.
Proof. Note that if $A \in S$, then $A \in M$ (by letting $B=A$), so $S \subset M$. 
We now show that $M$ is closed under countable unions. Let $[(A_n \cap E) \cup (B_n \cap E^c)]_n \subset M$. Let $A = \cup_n A_n$ and $B = \cup_n B_n$. Then, $A, B \in S$ and
\begin{align}
\bigcup_n [(A_n \cap E) \cup (B_n \cap E^c)] &= \bigcup_n (A_n \cap E) \cup \bigcup_n (B_n \cap E^c)\\
&= [A \cap E] \cup [B \cap E^c] \in M.
\end{align}
Finally, we show that $M$ is closed under complementation. Let $(A \cap E) \cup (B \cap E^c) \in M$. Then, $(A^c \cap E) \cup (B^c \cap E^c) \in M$ because $S$ is a sigma-algebra, $A^c \cap B^c \in M$ because $S \subset M$, and
\begin{align}
[(A \cap E) \cup (B \cap E^c)]^c &= [((A \cap E) \cup B) \cap ((A \cap E) \cup E^c)]^c\\
&=[(A \cup B) \cap (B \cup E) \cap (A \cup E^c)]^c\\
&= (A \cup B)^c \cup (B \cup E)^c \cup (A \cup E^c)^c\\
&= (A^c \cap B^c) \cup (B^c \cap E^c) \cup (A^c \cap E) \in M,
\end{align}
where, in the final step, we have used the fact that $M$ is closed under unions. This completes the proof.
M is the smallest sigma-algebra containing $S \cup \{E\}$.
Proof. Let $N$ be the smallest sigma-algebra containing $S \cup \{E\}$. From the above result, we know that $M$ is a sigma-algebra, and it's clear that $M$ contains $S \cup \{E\}$. Hence, $N \subset M$. It suffices, then, to show that $M \subset N$. And in order to show that it suffices to show that $N$ contains all sets of the form 
$$(A \cap E) \cup (B \cap E^c), \ \ A,B \in S.$$
But this is clearly the case because, by assumption, $N$ is a sigma-algebra containing $S \cup \{E\}$.
