Q:

Completion of a $\sigma$-algebra is a $\sigma$-algebra

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let
  \begin{align*}
 \mathcal{N}&= \left\{N\subseteq\Omega:\exists F\in\mathcal{F}, N\subseteq F,\mathbb{P}(F)=0\right\}  \\ \mathcal{G}&=\left\{A\cup N:A\in\mathcal{F},N\in\mathcal{N}\right\} \end{align*}
  Prove that $\mathcal{G}$ is a $\sigma$-algebra.

I've shown that $\Omega\in\mathcal{G}$ and that if $(A_n)_n\subseteq\mathcal{G}$ then $\cup_n A_n\in\mathcal{G}$. How can I show that if $A\in\mathcal{G}$ then $A^c\in\mathcal{G}$?
I also noticed $\mathcal{F}\subseteq\mathcal{G}$ if it helps somehow.

A:

Let $G=A \cup N$ for $A \in \mathcal{F}$ and $N \in \mathcal{N}$. By definition, there exists $F \in \mathcal{F}$ such that $N \subseteq F$ and $\mathbb{P}(F)=0$.  Clearly,
$$G^c = A^c \cap N^c = \underbrace{A^c}_W \backslash \underbrace{(N \cap A^c)}_{U}$$
As $$\underbrace{N \cap A^c}_{U} \subseteq \underbrace{F \cap A^c}_{V} \subseteq \underbrace{A^c}_{W}$$ it follows that
$$G^c = \underbrace{\big(A^c \backslash (F \cap A^c)\big)}_{W \backslash V} \cup \underbrace{\big( (F \cap A^c) \backslash (N \cap A^c) \big)}_{V \backslash U}; \tag{1}$$
here we have used the general fact that
$$U \subseteq V \subseteq W \implies W \backslash U = (W \backslash V) \cup (V \backslash U).$$ Consequently, we have shown that
$$G^c = \tilde{A} \cup \tilde{N} \tag{2}$$
for
$$\tilde{A} := A^c \backslash (F \cap A^c)\quad \text{and} \quad \tilde{N}:= (F \cap A^c) \backslash (N \cap A^c).$$
As $\tilde{N} \subseteq F \in \mathcal{F}$ and $\mathbb{P}(F)=0$, we have $\tilde{N} \in \mathcal{N}$. Morover, $A \in \mathcal{F}$ and $F \in \mathcal{F}$ implies $\tilde{A} \in \mathcal{F}$. Consequently, $(2)$ shows that $G^c \in \mathcal{G}$.

