# $\sigma$-algebra generated by a union of a collection of null-sets and another $\sigma$-algebra

I honestly don't even know how to approach this assignment. I will be extremely grateful for any tips.

Let $(\Omega,\Sigma,\mu)$ be a measure space. Call a subset $N$ of $\Omega$ a $(\Omega,\Sigma)$-null set if there exists a set $N' \in \Sigma$ with $N \subset N'$ and $\mu(N')=0$. Denote by $\mathcal{N}$ collection of all $(\mu,\Sigma)$-null sets. Let $\Sigma^*$ be the collection of subsets $E$ of $\Omega$ for which there exists $F,G \in \Sigma$ s.t. $F \subset E \subset G$ and $\mu(G\setminus F)=0$. For $E \in \Sigma^*$ and $F,G$ as above we define $\mu^*(E)=\mu(F)$.

Show that $\Sigma^*$ is a $\sigma$-algebra and that $\Sigma^* = \sigma(\mathcal{N} \cup \Sigma)$

Show that $\mu^*$ restricted to $\Sigma$ coincides with $\mu$ and that $\mu^*(E)$ doesn't depend on the specific choice of $F$ in the definition.

Show that the collection of $(\mu^*,\Sigma^*)$-null sets is $\mathcal{N}$.

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So start from the beginning: can you show that $\Omega\in\sigma^*$? –  Davide Giraudo Sep 17 '13 at 19:48
I see that $N' \subset \Sigma$ but I can't show from the definition of $\Sigma^*$ that it contains $\mathcal{N}$. I just don't see it. –  Piotr Sokol Sep 17 '13 at 20:33
$\Sigma^*$ contains $\mathcal N$ because you can take $G:=N'$ and $F:=\emptyset$. –  Davide Giraudo Sep 17 '13 at 20:35
To me this implies that $\Sigma^*$ generated by is trivial, since $G \supset E$ and $\mu(G \setminus F) =0$ and $\mu(F) = 0$. For the second point I should use the fact that if two measures agree on a pi-system then they agree on the sigma-algebra generated by that pi-system. –  Piotr Sokol Sep 17 '13 at 21:37