In proving the Fubini's theorem, I've come across below statement for which I'm not sure if it's indeed correct. Could you please confirm the validity of this statement and if my proof is correct?
Let
$(X, \mathcal A)$ and $(Y, \mathcal B)$ be measure spaces.
$[n] \triangleq \{1, \ldots, n\}$ and $A \times B \in \mathcal A \times \mathcal B$.
$\{X_k \times Y_k\}_{k=1}^n$ a finite collection of sets in $\mathcal A \times \mathcal B$.
$(X_k \times Y_k) \cap (X_h \times Y_h) = \emptyset$, either $X_k = X_h$ or $X_k \cap X_h = \emptyset$, and either $Y_k = Y_h$ or $Y_k \cap Y_h = \emptyset$ for all $k \neq h$.
Then there is a finite collection $\Delta = \{A_k \times B_k\}_{k=1}^m$ of sets in $\mathcal A \times \mathcal B$ such that
$\{X_k \times Y_k\}_{k=1}^n \subseteq \Delta$.
$\bigcup \Delta = (A \times B) \bigcup \cup_{k=1}^n X_k \times Y_k$.
$(A_k \times B_k) \cap (A_h \times B_h) = \emptyset$, either $A_k = A_h$ or $A_k \cap A_h = \emptyset$, and either $B_k = B_h$ or $B_k \cap B_h = \emptyset$ for all $k \neq h$.
Proof:
We define $A_k \triangleq A \cap X_k$ and $B_k \triangleq B \cap Y_k$ for $k \in [n]$. Also, $A' \triangleq A \setminus \cup_{k=1}^n A_k$ and $B' \triangleq B \setminus \cup_{k=1}^n B_k$. Then $A_k, A' \in \mathcal A$ and $B_k, B' \in \mathcal B$ for $k \in [n]$. Moreover, $$A \times B = (A' \cup A_1 \cup \cdots \cup A_n) \times (B' \cup B_1 \cup \cdots \cup B_n).$$
Clearly, $(A_i \times B_j) \cap (X_k \times Y_k) \neq \emptyset$ if and only if $A_i \cap X_k \neq \emptyset$ and $B_j \cap Y_k \neq \emptyset$ if and only if $A_i \subseteq X_k$ and $B_j \subseteq Y_k$. Then it's either $(A_i \times B_j) \cap (X_k \times Y_k) = \emptyset$ or $(A_i \times B_j) \subseteq (X_k \times Y_k)$ for all $i,j,k \in [n]$. Let $$I \triangleq \{(i, j) \in [n]^2 \mid \forall k \in [n], (A_i \times B_j) \cap (X_k \times Y_k) = \emptyset\}$$ and $$\Delta \triangleq \{X_k \times Y_k, A' \times B_k, A_k \times B'\}_{k=1}^n \cup \{A' \times B'\} \cup \{A_i \times B_j\}_{(i, j) \in I}.$$
Then $\Delta$ satisfies the required conditions.
Update: After looking closer to the problem, I found that the "theorem" can be simplified. The price to pay for this simplification is that we no longer have $\{X_k \times Y_k\}_{k=1}^n \subseteq \Delta$.
Let
$(X, \mathcal A)$ and $(Y, \mathcal B)$ be measure spaces.
$[n] \triangleq \{1, \ldots, n\}$ and $A \times B \in \mathcal A \times \mathcal B$.
$\{X_k \times Y_k\}_{k=1}^n$ a finite collection of sets in $\mathcal A \times \mathcal B$ such that $X_k \cap X_h = \emptyset$ for all $k \neq h$.
Then there is a finite collection $\Delta = \{A_k \times B_k\}_{k=1}^m$ of sets in $\mathcal A \times \mathcal B$ such that
$\bigcup \Delta = (A \times B) \bigcup \cup_{k=1}^n X_k \times Y_k$.
$A_k \cap A_h = \emptyset$ for all $k \neq h$.
Update 2: I apply above theorem to build a countable partition of a measurable set from one of its countable cover containing measurable subsets.
Let $\Delta = \{X_k \times Y_k\}_k$ be a sequence in $\mathcal A \times \mathcal B$. We define a sequence $(\Lambda_k)$ of collections with $\Lambda_k \subseteq \mathcal A \times \mathcal B$ as follows. First, $\Lambda_1 = \{X_1 \times Y_1\}$. We construct $\Lambda_{k+1}$ by applying above theorem to the finite collection $\Lambda_k$ and an element $X_{k+1} \times Y_{k+1}$. It follows that $\Lambda_{k} \subseteq \Lambda_{k+1}$ and that $\Lambda_k$ contains only pairwise disjoints sets. Let $\Lambda = \bigcup_k \Lambda_k$. Then $\bigcup \Lambda = \bigcup \Delta$ and that $\Lambda$ contains only pairwise disjoint sets.
We notice that $\Lambda_1$ contains only $1$ element and thus satisfies the condition of the theorem. Hence the recursive construction by the theorem is valid.
