I've just started with an advanced course in measure theory and I'm having trouble working with $\sigma$-algebras. Here's my problem:
Let $(S,\Sigma, \mu)$ be a measure space. Call $N \subset S$ a $(\mu,\Sigma)$-null set if there exists a set $N' \in \Sigma$ with $N \subset N'$ and $\mu(N')=0$. Denote by $\mathcal{N}$ the collection of all $(\mu,\Sigma)$-null sets. Let $\Sigma^*$ be the collection of subsets $E$ of $S$ for which there exist $F,G \in \Sigma$ such that $F \subset E \subset G$ and $\mu(G\backslash F)=0$. For $E \in \Sigma^*$ and $F,G$ as above we define $\mu^*(E)=\mu(F)$.
Prove that $\Sigma^* = \sigma(\mathcal{N} \cup \Sigma)$.
I'm having trouble proving the $\subset$ inclusion. Let $E \in \Sigma^*$, then there exist $F,G\in \Sigma$ with $F \subset E \subset G$ and $\mu(G\setminus F)=0$. From here I just can't see how to show that $E$ is in the $smallest$ $\sigma$-algebra containing $\mathcal{N} \cup \Sigma$.
Any tips in how I should reason?