Let $(X, \mathcal X, \mu)$ be a measure space. 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 $(X_n, \overline{\mathcal X_n}, \overline{\mu_n})$ from the measure space $(X_n, \mathcal X_n, \mu_n)$.
I would like to prove the following analogue, i.e.
Theorem $\overline{\mathcal X_n} = \{A \cap X_n : A \in \mathcal{\overline X}\}$.
Could you have a check on my attempt?
Proof Let $\mathcal N$ be the collection of all subsets of $\mu$-null subsets of $X$, i.e., $$ \mathcal N := \{A \subset X :\exists N \in \mathcal X \text{ such that } A \subset N \text{ and } \mu (N)=0\}. $$
We define $\mathcal N_n$ similarly, i.e., $$ \mathcal N_n := \{A \subset X_n :\exists N \in \mathcal X_n \text{ such that } A \subset N \text{ and } \mu_n (N)=0\}. $$
Let $A \in \mathcal{\overline X}$. Let's prove that $A_n := A \cap X_n \in \overline{\mathcal X_n}$. We have $A \in \overline{\mathcal X}$ if and only if $A = B \cup C$ for some $B \in \mathcal X$ and $C \in \mathcal N$. Let $B_n := B \cap X_n \in \mathcal X_n$ and $C_n := C \cap X_n$. Then $A_n = B_n \cup C_n$. There is $N \in \mathcal X$ such that $\mu(N)=0$ and $C \subset N$. Then $C_n \subset N_n := N \cap X_n \in \mathcal X_n$. Clearly, $\mu_n (N_n)=0$. It follows that $C_n \in \mathcal N_n$ and thus $A_n \in \overline{\mathcal X_n}$.
Let $A' \in \overline{\mathcal X_n}$. We will prove that there is $A \in \overline{\mathcal X}$ such that $A'=A \cap X_n$. It suffices to show that $A' \in \overline{\mathcal X}$. We have $A' = B \cup C$ for some $B \in \mathcal X_n$ and $C \in \mathcal N_n$. Clearly, $B \in \mathcal X$. There is $N \in \mathcal X_n$ such that $\mu_n (N)=0$ and $C \subset N$. Clearly, $N \in \mathcal X$ and $\mu(N)=0$. Thus $C \in \mathcal N$. It follows that $A' \in \overline{\mathcal X}$.
