Measure spaces proof This theorem comes from the book Real Analysis by Folland
Note: $M$ is a $\sigma$-algebra
Suppose that $(X,M,\mu)$ is a measure space. Let $\mathcal{N} = \{N\in M: \mu(N) = 0\}$ and $\bar{M} = \{E\cup F: E\in M  \ \ \text{and} \ \ F\subset N \ \ \text{for some} \ \ N\in\mathcal{N}\}$. Then $\bar{M}$ is a $\sigma$-algebra, and there is a unique extention $\bar{\mu}$ of $\mu$ to a complete measure on $\bar{M}$
I believe I need to first show that since $M$ and $\mathcal{N}$ are closed under countable unions then so is $\bar{M}$, but I am not exactly sure how to show this. Then, once I have proven that $\bar{M}$ is a $\sigma$-algebra and given how they defined $\mathcal{N}$ then there must be a unique $\bar{\mu}$ that is a complete measure on $\bar{M}$. I am trying to not look at the proof in the book and do this on my own but I just need some help with the finer details, any suggestions would be greatly appreciated. 
 A: Claim 1: $\bar{\mathcal{M}}$ is a $\sigma$-algebra.
Proof: Since $\emptyset \in \mathcal{N}$ and $\mathcal{M}$ is a $\sigma$-algebra, we have that $X \cup \emptyset = X \in \bar{\mathcal{M}}$.
Next, suppose that $E \in \bar{\mathcal{M}}$.  So $E = A \cup F$ where $A \in \mathcal{M}$ and $F \subset N$ for $N \in \mathcal{N}$.  
Then $X \setminus E = X \setminus(A \cup F) = (X \setminus A) \cap (X \setminus N) \cup ((X \setminus A)\cap (N \setminus F))$.
Now $(X \setminus A) \cap (X \setminus N) \in \mathcal{M}$, while $(X \setminus A) \cap (N \setminus F) \subset N$.
This shows that $X \setminus E \in \bar{\mathcal{M}}$.
Finally, let $E_j \in \bar{\mathcal{M}}$ for $j = 1, 2, \dots$.  Then for each $j$, write $E_j = A_j \cup F_j$, where $F_j \subset N_j$ and $\mu(N_j) = 0$.
Then $ \displaystyle \bigcup_j E_j = \bigcup_j (A_j \cup F_j) = 
\\ \bigcup_j A_j \cup \bigcup_j F_j$
Notice that $\bigcup_j F_j \subset \bigcup_j N_j$ and $\mu(\bigcup_j N_j) = 0$.  
So we have have that $\bigcup_j E_j \in \bar{\mathcal{M}}$.
This shows that $\bar{\mathcal{M}}$ is a $\sigma$-algebra.
Next, let's define $\bar{\mu}(E) = \mu(A)$ where $E = A \cup F$ as above.  You need to show that this definition does not depend on the choice of decomposition of $E$.
