Proof. First suppose that $Y$ is finite. For each $y \in Y$, let $P(y)$ be an essential supremum of $\bracs{f_A^{-1}(y)|A \in \cf}$. By Lemma 22.5.6, for any $x, y \in Y$ with $x \ne y$, $\mu(P(x) \cap P(y)) = 0$. After modification by null sets, assume without loss of generality that $X = \bigsqcup_{y \in Y}P(y)$.

For each $x \in X$, let $f(x) \in Y$ be the unique element of $Y$ such that $x \in P(f(x))$, then:

1. (1)

   For each $y \in Y$, $f^{-1}(y) = P(y)$, so $f \in \mathcal{L}^{0}(X; Y)$ is measurable.

2. (2)

   Let $A \in \cf$, then for each $y \in Y$, $\mu(f_{A}^{-1}(y) \setminus P(y)) = 0$. On the other hand,

   \begin{align*}\mu\braks{(P(y) \cap A) \setminus f_A^{-1}(y)}&\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap f_{A}^{-1}(z) ) \\&\le \sum_{z \in Y \setminus \bracs{y}}\mu(P(y) \cap P(z)) \\&+ \sum_{z \in Y \setminus \bracs{y}}\mu(f_{A}^{-1}(z) \setminus P(z)) \\&= 0\end{align*}

   
   

   so $f|_{A} = f_{A}$ almost everywhere on $A$.

3. (U)

   For all $A \in \cf$, $f|_{A} = g|_{A}$ almost everywhere. Since $\mu$ is semifinite, $f = g$ almost everywhere.

Therefore $f$ is the desired function.

Now suppose that $Y$ is an arbitrary separable metrisable space. By Lemma 25.6.4, there exists $\seq{I_n}\subset Y^{Y}$ such that:

1. (i)

   $I_{n} \to \text{Id}$ pointwise as $n \to \infty$.

2. (ii)

   For each $n \in \natp$, $I_{n}(Y)$ is finite and Borel measurable.

For each $n \in \natp$, let $f_{A, n}= I_{n} \circ f_{A}$, then

1. (a)

   For each $A \in \cf$, $f_{A, n}\in \mathcal{L}^{0}(A; Y)$.

2. (b)

   For each $A, B \in \cf$, $f_{A, n}|_{A \cap B}= f_{B, n}|_{A \cap B}$ almost everywhere.

3. (c)

   For each $A \in \cf$, $f_{A, n}(A) \subset I_{n}(Y)$.

By the finite case, there exists $f_{n}: X \to Y$ such that:

1. (1)

   $f_{n} \in \mathcal{L}^{0}(X; Y)$.

2. (2)

   For each $A \in \cf$, $f_{n}|_{A} = f_{A, n}$ almost everywhere.

Since $Y$ is Polish, Proposition 25.5.4 implies that $\bracsn{\limv{n}f_n \text{ exists}}\in \cm$. For each $A \in \cf$, $f_{A, n}\to f_{n}$ pointwise by (i). For each$n \in \natp$, $f_{A, n}= f_{n}|_{A}$ almost everywhere on $A$ by (2). Thus

As $\mu$ is semifinite, $\mu\bracsn{\limv{n}f_n \text{ does not exist}}= 0$. By (1) and Proposition 25.5.4, there exists $f \in \mathcal{L}^{0}(X; Y)$ such that $f = \limv{n}f_{n}$ almost everywhere. In which case,

1. (1)

   $f \in \mathcal{L}^{0}(X; Y)$.

2. (2)

   For each $A \in \cf$, $f|_{A} = \limv{n}f_{n}|_{A} = \limv{n}f_{A, n}= f_{A}$ almost everywhere.

3. (U)

   For all $A \in \cf$, $f|_{A} = g|_{A}$ almost everywhere. Since $\mu$ is semifinite, $f = g$ almost everywhere.

$\square$