Suppose we have a $\sigma$-finite measure (i.e. $E$ is a countable union of finite measure sets) space $(E,\mathcal{E},\mu)$. Say a set is $N$ is null if $N \subseteq B \in \mathcal{E}$ with $\mu(B) = 0$. We can then define a new $\sigma$-algebra $\mathcal{E}^\mu = \{A\cup N: A \in \mathcal{E}, N \text{ is null}\}$. How can I show that this is exactly the $\sigma$-algebra $\mathcal{M}$ generated by the set of all $\mu^*$-measurable sets where $\mu^*$ is the outer measure?
So far I have managed to show that $\mathcal{E}^\mu \subseteq \mathcal{M}$. For the other direction I pick any $M \in \mathcal{M}$ and try to show it can be written as something in $\mathcal{E}^\mu$. To do this I let $A \in \mathcal{E}$ be a maximal set in $\mathcal{E}$ contained in $M$. This exists because of Zorn's Lemma. Then we have $M = A \cup L$ where $A,L \in E$ are disjoint.
By maximality of $A$ I know that $L$ does not contain any elements of $\mathcal{E}$ that are not the empty set and we also have $L \in \mathcal{M}$. I want to prove that $L$ is null and I am trying to do this by showing that $\mu^*(L) = 0$ but without any luck. I still haven't used the fact that $E$ is $\sigma$-finite.
How should I proceed?
EDIT: I believe I have a solution, can someone please check and see.
Note that we have that $L \in M$. We know $E$ is $\sigma$-finite so we have $L \subseteq \bigcup\limits_{n=1}^\infty B_i$ for some $B_i \in \mathcal{E}$ with $\mu(B_i)$ finite for all $i$ and also $B_i \cap L \neq \emptyset$ (otherwise we just throw away this $B_i$ and proceed onto the next one). This gives us $\mu^*(B_i) = \mu^*(B_i\cap L)+\mu^*(B_i\cap L^C)$. We show that $\mu(B_i) = \mu^*(B_i) = \mu^*(B_i \cap L^C)$ (first equality is because $B_i \in \mathcal{E}$).
Note that as $B_i$ covers $B_i \cap L^C$ we get $\mu^*(B_i\cap L^C) \leq \mu(B_i)$. We show equality by contradiction. Suppose not, then we can find $C_1,C_2,\cdots$ s.t. $B_i \cap L^C \subseteq \bigcup\limits_{j=1}^\infty C_j$ (some $C_j$ may be empty, if we have a finite cover) with $\sum\limits_{j=1}^\infty \mu(C_j) < \mu(B_i)$.
Note that this is not a cover of $B_i$, as if it was that would mean $\mu^*(B_i) \leq \mu(B_i)$, a contradiction. As $E$ is $\sigma$-finite we can extend this cover with $D_j \in \mathcal{E}$ such that $B_i \subseteq (\bigcup\limits_{j=1}^\infty C_j)\cup(\bigcup\limits_{j=1}^\infty D_j)$. Now note that we have $B_i \cap D_j \in \mathcal{E}$ too and we can take this intersection to get $B_i \subseteq (\bigcup\limits_{j=1}^\infty C_j)\cup(\bigcup\limits_{j=1}^\infty (B_i\cap D_j))$.
Call $(\bigcup\limits_{j=1}^\infty (B_i\cap D_j)) = F \in \mathcal{E} $ and note that $F$ is not empty, otherwise $(\bigcup\limits_{j=1}^\infty C_j)$ would cover all of $B_i$, and $\mu(F)$ is finite. We can set $G = F \cap (\bigcup\limits_{j=1}^\infty C_j)^C \in \mathcal{E}$ which is also non-empty, otherwise $(\bigcup\limits_{j=1}^\infty C_j)$ would cover all of $B_i$. We have $G \subseteq B_i$. Now we have $B_i \subseteq (\bigcup\limits_{j=1}^\infty C_j) \cup G$.
Note that $G$ does not contain anything in $(\bigcup\limits_{j=1}^\infty C_j)$ so it does not contain anything in $B_i \cap L^C$, so as it is in $B_i$ it must be contained within $B_i\cap L$, so contained within $L$. Hence we have found a non-empty $G \in \mathcal{E}$ that is contained in $L$, which is a contradiction to the fact that $L$ does not contain any non-empty elements of $\mathcal{E}$.
So we get $\mu^*(B_i) = \mu^*(B_i \cap L^C)$. So $\mu^*(B_i) = \mu^*(B_i\cap L)+\mu^*(B_i\cap L^C) \implies \mu^*(B_i\cap L) = 0$. This holds for all $i$. So by countable subadditivity we get $\mu^*(L) \leq \sum\limits_{i=1}^\infty \mu^*(B_i\cap L) = \sum\limits_{i=1}^\infty 0 = 0$. So $\mu^*(L) = 0$ as required and we are done!
Best Answer
First prove two simple lemmas:
and
Now, let $X\in\mathcal M$ and $n\in\Bbb N$. By Lemma 2 there exists $B_n\in\mathcal E$ such that $X^c\subset B_n$ and $\mu(B_n)\le\mu^*(X^c)+\tfrac 1 n$. Since $X\in\mathcal M$, $$ \mu^*(B_n) = \mu^*(B_n\cap X) + \mu^*(B_n\cap X^c)\ge\mu^*(B_n\cap X)+\mu^*(B_n)-\tfrac 1n. $$ Thus, $\mu^*(B_n\cap X)\le\tfrac 1 n$.
Now, define $A_n := B_n^c\subset X$ and $A := \bigcup_nA_n\in\mathcal E$. Then $\mu^*(X\backslash A_n)\le\tfrac 1n$ for each $n$ and thus (as $X\backslash A\subset X\backslash A_n$ for all $n$) $\mu^*(X\backslash A) = 0$. Hence, by Lemma 1, $N := X\backslash A$ is null and $X = A\cup N$, i.e., $X\in\mathcal E^\mu$.