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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?

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To show that $E$ is in the smallest $\sigma$-algebra containing $\mathcal{N} \cup \mathcal{\Sigma}$ it is enough to show that it is generated from elements of $\mathcal{N} \cup \mathcal{\Sigma}$ under the operations of countable union, countable intersection, and complementation. You have already found some relevant elements of $\mathcal{\Sigma}$, namely $F$ and $G$. Hint: find a null set whose union with $F$ is $E$.

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My first intuition would be to take the union of $\bigcup\{N \in S: N \subset G \setminus F\}$ with F, but the problem is this left set isn't necessarily a countable union so it doesn't have to be in $\mathcal{N}$. – BallzofFury Sep 8 '12 at 13:55
The definition of the class $\mathcal{N}$ of null sets gives another way to find null sets, namely as subsets of sets in $\Sigma$ with $\mu$-measure zero. You already found the relevant set in $\Sigma$ with $\mu$-measure zero; now find the subset that you want to add to $F$. – Trevor Wilson Sep 8 '12 at 14:02

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