Proof. Let $\ce \subset \cm$ and $\seqi{A}\subset \cm$ be a decomposition of $X$.

For each $i \in I$, let $\ce_{i} = \bracs{E \cap A_i|E \in \ce}$. By Lemma 22.5.4, there exists an essential upper bound $S_{i} \in \cm$ of $\ce_{i}$ contained in $A_{i}$ with respect to the restricted measure $\mu|_{A_i}$. In other words,

1. (i)

   $S_{i}$ is an essential upper bound of $\ce_{i}$.

2. (ii)

   For any essential upper bound $T \in \cm$ of $\ce_{i}$ with $T \subset A_{i}$, $\mu(S_{i} \setminus T) = 0$.

Now, let $S = \bigsqcup_{i \in I}S_{i}$, then since $X = \bigsqcup_{i \in I}A_{i}$ and $S_{i} \subset A_{i}$ for all $i \in I$, $S \cap A_{i} = S_{i} \in \cm$ for all $i \in I$, so $S \in \cm$. For any $E \in \ce$, $E \cap A_{i} \in \ce_{i}$. Passing through the decomposition, (i) implies that,

and $S$ is an essential upper bound of $\ce$.

Finally, let $T \in \cm$ be an essential upper bound of $\ce$, then $T \cap A_{i}$ is an essential upper bound of $\ce_{i}$ for all $i \in I$. The decomposition and (ii) then shows that

therefore $S$ is an essential supremum of $\ce$.$\square$