How to show that $\mathcal S^* \subseteq \mathcal S (\mathcal A) \cup \mathcal N$?

Question

Let $\mathcal A$ be an algebra of subsets of a set $X$ and $\mu : \mathcal A \longrightarrow [0,\infty]$ be a measure on $\mathcal A.$ Let $\mu^*$ be the outer measure induced by $\mu$ and $\mathcal S^*$ be the collection of all $\mu^*$-measurable subsets of $X.$ Then $\mathcal S^*$ is a $\sigma$-algebra and $\mu^*$ restricted to $\mathcal S^*$ is a measure on $\mathcal S^*.$ Let $\mathcal N = \{E \subseteq X\ |\ \mu^*(E) = 0 \}.$ Then $\mathcal A, \mathcal N \subseteq S^*.$ Let $\mathcal S (\mathcal A)$ be the $\sigma$-algebra generated by $\mathcal A.$ Then clearly $\mathcal S (\mathcal A) \subseteq \mathcal S^*.$ Now how do I prove that $\mathcal S^* = \mathcal S (\mathcal A) \cup \mathcal N := \{ E \cup N\ |\ E \in \mathcal S (\mathcal A), N \in \mathcal N \}$? I have proved that $\mathcal S (\mathcal A) \cup \mathcal N \subseteq \mathcal S^*.$ How do I prove the other part of the inclusion i.e. $\mathcal S^* \subseteq \mathcal S (\mathcal A) \cup \mathcal N$?

Any help will be highly appreciated. Thanks in advance.

EDIT $:$ I think that there exists some $N \in \mathcal N$ such that $E \cap N^c \in \mathcal S (\mathcal A)$ and then we are through because $E \cap N$ being a subset of $N$ is of measure $0$ by monotonicity of $\mu^*.$ But how do I find such an $N \in \mathcal N$?

Answer 1

$\mathcal S^* = \mathcal S (\mathcal A) \cup \mathcal N$ is not true in general, But if $\left.\mu^*\right|_\mathcal {S(\mathcal A)}$ is $\sigma$-finite then $\mathcal S^* = \mathcal S (\mathcal A) \cup \mathcal N$.

First let's prove the following :$ \ \ $ $ \forall B \subset X, \exists E \in \mathcal S (\mathcal A):\mu^*(E)=\mu^*(B)$.

From the definition of $ \ \mu^* (B)$, there exists $ \ (A_{i,n})_{i\geq1,n\geq1} \subset \mathcal A \ \ $ such that for all $n \geq1 \ \ B \subset \overline{A}_{n} =\bigcup\limits_{i=1}^{\infty}A_{i,n}$ and $\lim\limits_{n \to \infty}\sum\limits_{i=1}^{\infty} \mu \left (A_{i,n} \right )=\mu^*(B)$. $ \ $ Let $E=\bigcap\limits_{n=1}^{\infty}\overline{A}_{n} \ $ then we have $B \subset E$ and $E\in\mathcal S (\mathcal A) \ $ since $ \overline{A}_{n}\in\mathcal S (\mathcal A), n \geq1 $.

From $B \subset E \subset \overline{A}_{n} =\bigcup\limits_{i=1}^{\infty}A_{i,n}$ we get $\mu^*(B) \leq\mu^*(E) \leq\mu^*(\overline{A}_{n})\leq\sum\limits_{i=1}^{\infty} \mu \left (A_{i,n} \right ) \ $, thus $ \mu^*(E)=\mu^*(B)$.

Now we return to the main problem where $\left.\mu^*\right|_\mathcal {S(\mathcal A)}$ is $\sigma$-finite (i.e there exists $\ (X_{n})_{n\geq1}\subset \mathcal {S(\mathcal A)} $ such that $X=\bigcup\limits_{n=1}^{\infty}X_{n}$ and $\mu^*(X_{n})<\infty$ ), in this case it suffices to prove that $ \{B\in \mathcal S^* |\mu^*(B)<\infty\}\subset \mathcal S (\mathcal A) \cup \mathcal N $ . Let $E \in \mathcal S (\mathcal A)$ such that $B\subset E$ and $\mu^*(E)=\mu^*(B)$, since $B\in \mathcal S^*$ then $ \ \mu^*(E)=\mu^*(E\cap B)+\mu^*(E\cap B^{c})=\mu^*(B)+\mu^*(E\cap B^{c}) $, since $ \ \mu^*(B)<\infty$ then $\mu^*(E\cap B^{c})=0$. Hence $E\cap B^{c} \in \mathcal N$, we have $B^c=E^c \cup(E \cap B^c)\in \mathcal S (\mathcal A) \cup \mathcal N$, then $B\in \mathcal S (\mathcal A) \cup \mathcal N. \square $

For a counter example let $X$ be an uncountable set, $\mathcal A=\{A\subset X |$ $ A $ is countable or $ A^c $ is countable $ \} $ and $\mu$ be the counting measure whiche is not $\sigma$-finite.

It's easy to see that $\mathcal A$ is $\sigma$-algebra (i.e $ \mathcal S (\mathcal A)=\mathcal A $), $ \ $ $\mathcal N=\{\emptyset \}$ and $\mathcal S^*=P(X) $. Hence $\mathcal S (\mathcal A) \cup \mathcal N\subsetneq S^*$.