The trace of the $\sigma$-algebra of all $\mu^*$-measurable sets

Question

Let $(X, \mathcal X, \mu)$ be a measure space. Let $\mu^*$ be the outer measure induced from $\mu$. Let $\mathcal{M}$ be the $\sigma$-algebra of all $\mu^*$-measurable sets. Let $(X, \overline{\mathcal X}, \overline \mu)$ be the completion of $(X, \mathcal X, \mu)$.

Let $(X_n)$ be a sequence of pairwise disjoint sets in $\mathcal X$. Let $\mathcal X_n$ be the sub $\sigma$-algebra that $\mathcal X$ induces on $X_n$, i.e., $\mathcal X_n := \{A \cap X_n : A \in \mathcal X\}$. Let $\mu_n$ be the restriction of $\mu$ to $\mathcal X_n$, i.e., $\mu_n (A) := \mu (A)$ for all $A \in \mathcal X_n$. We construct the corresponding objects $(\mu^*_n, \mathcal M_n, \overline{\mathcal X_n}, \overline{\mu_n})$ from the measure space $(X_n, \mathcal X_n, \mu_n)$. Then we have

Lemma $\mu_n^* (A) = \mu^* (A)$ for all $A \subset X_n$.

The proof of the lemma is given at the end. Now I would like to prove the following analogue, i.e.

Theorem $\mathcal M_n = \{A \cap X_n : A \in \mathcal M\}$.

Could you have a check on my attempt?

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Proof We have $\mathcal{M}$ is the collection of all those sets $A \subseteq X$ such that $$ \mu^*(E)=\mu^*(E\cap A) +\mu^*(E\cap (X \setminus A)) \quad \forall E\subset X. $$

• Let $A \in \mathcal M$. Let's prove that $A_n := A \cap X_n \in \mathcal M_n$. Let $E\subset X_n$. Then $E \subset X$. So $$ \begin{align} \mu^*(E ) &= \mu^*(E\cap A) +\mu^*(E\cap (X \setminus A)) \\ &= \mu^*(E\cap A_n) +\mu^*(E\cap (X_n \setminus A)) \text{ because } E \subset X_n\\ &= \mu^*(E\cap A_n) +\mu^*(E\cap ((X_n \setminus A) \cup (X_n \setminus X_n))) \\ &= \mu^*(E\cap A_n) +\mu^*(E\cap (X_n \setminus (A \cap X_n))) \\ &= \mu^*(E\cap A_n) +\mu^*(E\cap (X_n \setminus A_n). \end{align} $$ It follows from the Lemma that $\mu_n^*(E ) = \mu_n^*(E\cap A_n) +\mu_n^*(E\cap (X_n \setminus A_n)$.

• Let $A' \in \mathcal M_n$. Let's prove that there is $A \in \mathcal M$ such that $A' = A \cap X_n$. It suffices to prove that $A' \in \mathcal M$. Let $E \subset X$. I proved previously that $\mathcal X \subset \mathcal M$. Then $X_n \in \mathcal M$ and thus $$ \mu^*(E) = \mu^*(E\cap X_n) +\mu^*(E\cap (X \setminus X_n)). $$ On the other hand, $(E \cap X_n) \subset X_n$. So $$ \begin{align} \mu^* (E \cap X_n ) &= \mu^*((E\cap X_n) \cap A') +\mu^*((E\cap X_n) \cap (X_n \setminus A')) \\ &= \mu^*(E\cap A') +\mu^*((E\cap X_n) \cap (X \setminus A')). \end{align} $$ It follows that $$ \begin{align} \mu^*(E) &= \mu^*(E\cap A') +\mu^*((E\cap X_n) \cap (X \setminus A')) +\mu^*(E\cap (X \setminus X_n)) \\ &= \mu^*(E\cap A') +\mu^*(E \cap (X_n \setminus A')) +\mu^*(E\cap (X \setminus X_n)) \\ &\ge \mu^*(E\cap A') +\mu^*( [E \cap (X_n \setminus A')] \cup [E\cap (X \setminus X_n)]) \\ &= \mu^*(E\cap A') +\mu^*( E \cap [(X_n \setminus A') \cup (X \setminus X_n)]) \\ &= \mu^*(E\cap A') +\mu^*( E \cap (X \setminus A')). \end{align} $$

$\tag*{$\blacksquare$}$

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Proof of Lemma Fix $A \subset X_n$. First, $$ \begin{align} \mu_n^*(A) &= \inf \left\{ \sum_{m=1}^{\infty} \mu_n (B_m) : (B_m) \subset \mathcal X_n , A \subset \bigcup_m B_m \right\} \\ &= \inf \left\{ \sum_{m=1}^{\infty} \mu (B_m) : (B_m) \subset \mathcal X_n , A \subset \bigcup_m B_m \right\} \text{ because } \mu_n = \mu |_{\mathcal X_n}\\ &\ge \inf \left\{ \sum_{m=1}^{\infty} \mu (B_m) : (B_m) \subset \mathcal X , A \subset \bigcup_m B_m \right\} \\ &= \mu^* (A). \end{align} $$

Second, $$ \begin{align} \mu^*(A) &= \inf \left\{ \sum_{m=1}^{\infty} \mu (B_m) : (B_m) \subset \mathcal X , A \subset \bigcup_m B_m \right\} \\ &\ge \inf \left\{ \sum_{m=1}^{\infty} \mu (B_m \cap X_n) : (B_m) \subset \mathcal X , A \subset \bigcup_m B_m \right\}\\ &\ge \inf \left\{ \sum_{m=1}^{\infty} \mu_n (B_m) : (B_m) \subset \mathcal X_n , A \subset \bigcup_m B_m \right\} \\ &= \mu_n^* (A). \end{align} $$

$\tag*{$\blacksquare$}$