Proof. Let $\fF \subset \mathcal{L}^{0}_{\cf}(X; Y)$ be a Cauchy filter. By Theorem 26.7.4, for each $A \in \cf$, there exists an almost everywhere unique $f_{A} \in \mathcal{L}^{0}(A; Y)$ such that $\fF$ converges to $f_{A}$ when restricted to $A$. Thus for any $A, B \in \cf$, $f_{A \cup B}|_{A \cap B}= f_{A}|_{A \cap B}= f_{B}|_{B \cap A}$ almost everywhere. By the gluing lemma for measurable functions, there exists $f \in \mathcal{L}^{0}(X; Y)$ such that $f|_{A} = f_{A}$ for all $A \in \cf$. Thus $\fF \to f$ locally in measure, and $\mathcal{L}^{0}(X; Y)$ is complete.$\square$