Proof. For each $ i < \nu $ , let $ D_i = \bigcup _{j \neq i} P_j$ . Then $ P_i $ and $ D_i $ are disjoint $ \boldsymbol {\Gamma }$ -sets. Applying the separation property we get $ C^{\prime }_i \in \boldsymbol {\Delta }_{\boldsymbol {\Gamma }}(X) $ such that $ P_i \subseteq C^{\prime }_i $ and $ C^{\prime }_i \cap P_j = \emptyset $ for every $ j < \nu $ different from $ i $ . Since our hypotheses imply that $ \boldsymbol {\Delta }_{\boldsymbol {\Gamma }}(X)$ is a $ \nu ^+ $ -algebra, it is enough to let $ C_0 = (C \cap C^{\prime }_0) \cup \left (C \setminus \bigcup _{ j < \nu } C^{\prime }_j \right ) $ and $ C_i = (C \cap C^{\prime }_i) \setminus \bigcup _{j < i} C^{\prime }_i $ , for $ 0 < i < \nu $ .

Proof. The dual results can be obtained by taking complements and using De Morgan’s rules, so we only consider the case of the classes $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X) $ . By Theorem 2.1, we can assume that .

We argue by induction on $ 1 \leq \xi < \lambda ^+ $ . For the basic case $\xi =1$ , observe that every $ A \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ 1}(X) $ can be written as $ \bigcup _{i < \mu } A_i $ , where

$$\begin{align*}A_i = \bigcup \{ \boldsymbol{N}_s \cap X \mid \operatorname{lh}(s) = i, \boldsymbol{N}_s \cap X \subseteq A \}. \end{align*}$$

Each $ A_i $ belongs to $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ 1}(X) $ because

$$\begin{align*}X \setminus A_i = \bigcup \{ \boldsymbol{N}_t \cap X \mid \operatorname{lh}(t) = i , \boldsymbol{N}_t \cap X \not\subseteq A \}. \end{align*}$$

Clearly, $ A_i \subseteq A_j $ whenever $ i \leq j $ . If instead we want the sets $ A_i $ to be pairwise disjoint, we replace each $ A_i $ with $ A_i \setminus \bigcup _{j < i} A_i $ : such sets are still clopen because $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ 1}(X) $ is a $ \mu $ -algebra (because $ X $ , being a member of $ \mathscr {M}_\lambda $ , is $ \mu $ -additive). This proves (1). Part (2) needs not to be considered in the case $ \xi = 1 $ . As for (3), it suffices to prove that $ \bigcap _{j < \nu } A_j \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ 2}(X) $ . Using (1), write every $ A_j $ as $ A_j = \bigcup _{i< \mu } A_{j,i} $ with $ A_{j,i} \in \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ 1}(X) $ . Then,

If $ \lambda $ is regular, then our assumption $ 2^{< \lambda } = \lambda $ entails $ \lambda ^{< \lambda } = \lambda $ , and thus because $ \mu = \lambda $ . If $ \lambda $ is singular, then $ 2^{< \lambda } $ entails that $ \lambda $ is strong limit, and thus for $ \kappa = \max \{ \mu ,\nu \} < \lambda $ . Therefore in all cases , and thus $ \bigcap _{j < \nu } A_j \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ 2}(X) $ , as desired.

Let now $\xi> 1 $ . Part (1) is relevant only when $ \xi $ is successor, so assume that $ \xi = \xi '+ 1$ . Write $ A = \bigcup _{j < \lambda } B_j$ with $ B_j \in \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi ' }(X) $ . Fix a strictly increasing sequence $ (\lambda _i)_{i < \mu } $ cofinal in $ \lambda $ . Then each $ A_i = \bigcup _{j < \lambda _i } B_j $ belongs to $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi '+1}(X) = \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi }(X) $ because (3) holds at level $ \xi ' $ by inductive hypothesis, and clearly $ A = \bigcup _{i < \mu } A_i $ . Moreover, if $ i \leq j < \mu $ then $ A_i \subseteq A_j $ by construction. If instead we want the sets $ A_i $ to be pairwise disjoint, then we again replace each $ A_i $ with $ A_i \setminus \bigcup _{j < i} A_j $ : since $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi }(X) $ is a $ \mu $ -algebra (because $ \xi $ is a successor ordinal), this works.

If $ \xi $ is limit, we instead need to prove part (2). By definition, $A= \bigcup _{j<\lambda } B_{j}$ , where $B_{j} \in \bigcup _{ \xi ' < \xi } \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi ' }(X) $ . For all $i<\operatorname {cof}(\xi )$ , set $A_{i} = \bigcup \{ B_{j} \mid B_{j} \in \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi _i }(X) \}$ . Then $ A_i \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _i+1}(X) \subseteq \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi _i+2}(X) $ . Moreover, $ A_i \subseteq A_j $ if $ i \leq j $ , and $ A = \bigcup _{i < \operatorname {cof}(\xi )} A_i $ . If instead we want the sets $ A_i $ to be pairwise disjoint, we once again replace each $ A_i $ with $ A_i \setminus \bigcup _{j < i} A_j $ , which belongs to $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi _i+2}(X) $ because the sets $ A_i $ are in $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _i+1}(X) $ .

Finally, we prove (3). Again, we only need to check that $\bigcap _{j<\nu } A_j \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi +1}$ . For all $i<\nu $ , applyFootnote ⁴ (1) or (2), depending on whether $\xi $ is a successor or a limit ordinal, to get $A_j=\bigcup _{i<\kappa } B_{j,i}$ with $B_{j,i} \in \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi }(X)$ and $\kappa =\mu $ if $\xi $ is successor or $\kappa =\operatorname {cof}(\xi )$ if $ \xi $ is limit. Arguing as in the case $ \xi = 1 $ , we have , and we have to check that in all cases . If $ \lambda $ is regular, then because $ \kappa \leq \lambda $ . If instead $ \lambda $ is singular, then $ \kappa < \lambda $ : therefore because $ \lambda $ is strong limit.

Proof. The case $ \mu = \omega $ has already been treatedFootnote ⁵ in [Reference Dimonte and Motto Ros7, Proposition 4.4.4]. We show that the same argument can be adapted to deal with the remaining cases, so from now on assume $ \mu> \omega $ . By Proposition 3.3, it is enough to show that $\lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }$ has the ordinal $\lambda $ -uniformization property. Fix any $ X \in \mathscr {M}_\lambda $ : since we assumed $ \mu> \omega $ , by Theorem 2.1 we can suppose that . We distinguish three cases.

First suppose that $\xi =1$ . For any $i<\mu $ , we say that a set $A \subseteq X$ is i-clopen if there is such that $A=\bigcup \left \{\boldsymbol {N}_s \cap X \mid s \in S\right \}$ . It is easy to check that i-clopen sets are closed under complements, and arbitrary unions and intersections. Moreover, if $j \leq i$ then every j-clopen set is also i-clopen. Fix any $R \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ 1}(X \times \lambda )$ . For every $i<\mu $ and $\gamma <\lambda $ , let

and notice that $ \bigcup _{i < \mu } \bigcup _{\gamma < \lambda } R_i^\gamma $ coincides with the projection of $ R $ on the first coordinate. Each $R_i^\gamma $ is i-clopen, hence so is $R_i=\bigcup _{\gamma <\lambda } R^\gamma _i$ . The set

$$\begin{align*}Q_i^\gamma = R^\gamma_i \setminus \left(\bigcup\nolimits_{i' < i} R_{i'} \cup \bigcup\nolimits_{\gamma' < \gamma} R^{\gamma'}_i \right) \end{align*}$$

is i-clopen too, hence

$$\begin{align*}\begin{aligned} Q^* & =\left\{(x, \gamma, i) \in X \times \lambda \times \mu \mid x \in Q_i^\gamma\right\} \\ & =\bigcup_{i < \mu} \bigcup_{\gamma<\lambda} \left(Q_i^\gamma \times\{\gamma\} \times\{i\} \right) \in \lambda^+\text{-}\boldsymbol{\Sigma}^{ 0 }_{ 1}(X \times \lambda \times \mu ). \end{aligned} \end{align*}$$

Finally, the set

$$\begin{align*}R^* =\{ (x, \gamma) \in X \times \lambda \mid \exists i < \mu \, [(x, \gamma, i) \in Q^*] \} \end{align*}$$

is open and uniformizes R.

Next we consider the case where $\xi =\xi '+1>1$ is a successor ordinal. Consider $R \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X \times \lambda )$ , and write it as $R=\bigcup _{i <\lambda } R_i$ with $ R_i \in \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi ' }(X \times \lambda )$ . Let

$$\begin{align*}Q=\left\{(x, \gamma, i) \in X \times \lambda \times \lambda \mid(x, \gamma) \in R_i\right\}, \end{align*}$$

and notice that $Q \in \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi ' }(X \times \lambda \times \lambda )$ because $\lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi ' }$ is $\lambda $ -reasonable. Endow $\lambda \times \lambda $ with the Gödel well-ordering $\preceq $ , and define

$$ \begin{align*} Q^* =\{ (x, \gamma, i) \in X \times \lambda \times \lambda \mid (x, \gamma, i) \in Q \wedge \forall (\gamma',i') \prec (\gamma,i) \left[\left(x, \gamma', i'\right) \notin Q\right]\}. \end{align*} $$

There are less than $\lambda $ -many pairs $(\gamma ',i') \prec (\gamma ,i)$ , and since the intersection of less than $ \lambda $ -many sets in $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi '}(X \times \lambda \times \lambda ) $ is in $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi '+1}(X \times \lambda \times \lambda ) = \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi }(X \times \lambda \times \lambda )$ by Lemma 3.5(3), we get that $Q^* \in \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi }(X \times \lambda \times \lambda ) \subseteq \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X \times \lambda \times \lambda )$ . Therefore the set

$$\begin{align*}R^* =\{ (x, \gamma) \in X \times \lambda \mid \exists i < \mu \, [(x, \gamma, i) \in Q^*] \} \end{align*}$$

belongs to $\lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X \times \lambda )$ and uniformizes R.

Finally, assume that $\xi $ is a limit ordinal, and let $R \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X \times \lambda )$ . By Lemma 3.5(2), there are an increasing sequence $ (\xi _i)_{i < \operatorname {cof}(\xi )} $ of ordinals smaller than $ \xi $ and sets $ R_i \in \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi _i}(X \times \lambda ) $ such that $R=\bigcup _{i<\operatorname {cof}(\xi )} R_i$ . For $i<\operatorname {cof}(\xi )$ , let $Q^i \subseteq X \times \lambda \times \operatorname {cof}(\xi )$ be defined by

$$\begin{align*}Q^i=\left\{\left(x, \gamma, i'\right) \in X \times \lambda \times \operatorname{cof}(\xi) \mid i' \leq i \wedge (x, \gamma) \in R_{i'}\right\}, \end{align*}$$

and set $ Q=\bigcup _{i<\operatorname {cof}(\xi )} Q^i=\left \{(x, \gamma , i) \in X \times \lambda \times \operatorname {cof}(\xi ) \mid (x, \gamma ) \in R_i\right \} $ . Since each pointclass $\lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _i} $ is $ \lambda $ -reasonable, then $ Q^i \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _i}(X) $ , and thus $ Q \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X) $ . Let

$$ \begin{align*} Q^* =\big\{ (x, \gamma, i) \in X \times \lambda \times \operatorname{cof}(\xi) \mid {} & (x, \gamma, i) \in Q \wedge \forall i'<i \, \forall \gamma'<\lambda \, [\left(x, \gamma', i'\right) \notin Q^i] \\ & {} \wedge \forall \gamma'<\gamma \, \left[\left(x, \gamma', i\right) \notin Q^i\right] \big\}. \end{align*} $$

For any fixed pair $(\gamma , i) \in \lambda \times \operatorname {cof}(\xi )$ , the set

$$\begin{align*}\left\{ x \in X \mid \forall i'<i \, \forall \gamma'<\lambda \left[\left(x, \gamma', i'\right) \notin Q^i\right] \wedge \forall \gamma'<\gamma \,\left[\left(x, \gamma', i\right) \notin Q^i\right] \right\} \end{align*}$$

belongs to $ \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi _i }(X) $ . Since $ \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi _i }(X) \subseteq \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X) $ and $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi } $ is $ \lambda $ -reasonable, $Q^* \in \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X \times \lambda \times \operatorname {cof}(\xi ))$ . As before, it follows that the set $R^*$ consisting of those $(x, \gamma ) \in X \times \lambda $ such that $(x, \gamma , i) \in Q^*$ for some $i<\operatorname {cof}(\xi )$ is the desired uniformization of R in $\lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi }(X \times \lambda )$ .

Proof. Let $n \in \omega $ and $(A_j)_{j \leq n}$ be a finite $\lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi }$ -partition of X such that $ f $ is constant with value $ y_j $ on each $ A_j $ . Using Lemma 3.5(2) if $ \mu> \omega $ or [Reference Dimonte and Motto Ros7, Proposition 4.2.1] if $ \mu = \omega $ , we can find a sequence of ordinals $(\xi _i)_{i<\operatorname {cof}(\xi )}$ cofinal in $\xi $ and sets $B^j_i \in \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi _i}(X) $ such that $ A_j=\bigcup _{i<\operatorname {cof}(\xi )}B^j_i$ for every $ j \leq n $ , and moreover $ B^j_i\subseteq B^j_{i'}$ for every $i \leq i' < \operatorname {cof}(\xi ) $ . Fix $\bar y \in Y $ , and for $ i <\operatorname {cof}(\xi )$ let

$$\begin{align*}f_i(x)= \begin{cases} y_j, & \text{ if } x \in B^j_i \text{ for some } j \leq n, \\ \bar y , & \text{ otherwise}. \end{cases} \end{align*}$$

Notice that $ f $ is well-defined because $ B^j_i \subseteq A_j $ and $A_j \cap A_{j'} = \emptyset $ for every $ j \neq j' $ . Since $ X \setminus \bigcup _{j \leq n} B^j_i \in \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi _i}(X) $ , each $f_i$ is constant on a finite $\lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi _i}$ -partition, and hence $\lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _i}$ -measurable. It remains to show that $f=\lim _{i<\operatorname {cof}(\xi )}f_i$ . Given $x \in X$ , let $j \leq n$ and $ i < \operatorname {cof}(\xi ) $ be such that $x \in B^j_i$ . Since the sequence $(B^j_i)_{i<\operatorname {cof}(\xi )}$ is increasing, $ x \in B^j_{i'} $ for every $ i' \geq i $ , and thus $ f_{i'}(x) = y_j = f(x) $ .

Proof. Let $ \mathcal {T}_Y $ be the tree of $ Y $ . For each $ s \in \mathcal {T}_Y $ , let $ \boldsymbol {N}_s(Y) = \boldsymbol {N}_s \cap Y $ and fix any $ y_s \in \boldsymbol {N}_s(Y) $ . Let $ \{ P^s_\alpha \mid \alpha < \lambda \} $ be a family of nonempty $ \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi }(X) $ -sets such that $ f^{-1}(\boldsymbol {N}_s(Y)) = \bigcup _{\alpha < \lambda } P^s_\alpha $ .

Since $ |\mathcal {T}_Y| \leq \lambda $ , we can clearly work with the directed set $ D = ([\mathcal {T}_Y \times \lambda ]^{< \aleph _0},{\subseteq }) $ instead of $ \mathrm {Fin}_\lambda $ . Fix a nonempty $ d \in D $ . Let $ S_d = \{ s \in \mathcal {T}_Y \mid \exists \alpha < \lambda \, [(s,\alpha ) \in d] \} $ , and let $ \beta _0, \dotsc , \beta _k $ enumerate the set $ \{ \operatorname {lh} (s) \mid s \in S_d \} $ in increasing order, for the appropriate $ k \in \omega $ . To simplify the notation, for $ s \in \mathcal {T}_Y $ we let $ \operatorname {\mathrm {lev}} (s) = j $ if and only if $ \operatorname {lh}(s) = \beta _j $ , and for every $ i \leq k $ we let

$$\begin{align*}s|_i = \begin{cases} s \restriction \beta_i & \text{if } \operatorname{lh}(s) \geq \beta_i \\ s & \text{otherwise}. \end{cases} \end{align*}$$

Finally, for any $ j \leq k $ we let $ S_d^j = S_d^{j,0} \cup S_d^{j,1} $ , where

$$ \begin{align*} S_d^{j,0} & = \{ s|_j \mid s \in S_d \wedge \operatorname{\mathrm{lev}}(s) \geq j \} \qquad \text{and} \\ S_d^{j,1} & = \{ s \in S_d \mid \operatorname{\mathrm{lev}}(s) < j \wedge s \text{ maximal in } S_d \}. \end{align*} $$

Clearly, $ S^j_d $ consists of pairwise incomparable sequences because of the maximality requirement in the definition of $ S_d^{j,1} $ . Moreover, $ S_d^j $ is finite because so is $ d $ . Notice also that if $ s \in S^j_d $ , then $ s|_i \in S^i_d $ for every $ i \leq j $ . Finally, by definition $ S^k_d $ coincides with the set of all maximal elements of $ S_d $ .

We build a collection of finite $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi } $ -partitions $ \mathcal {C}_j = \{ C^j_s \mid s \in S^j_d \} $ of $ X $ satisfying the following two conditions:

1. (1) For every $ s \in S^{j,0}_d $ , $ P^t_\alpha \subseteq C^j_s $ for every $ (t,\alpha ) \in d $ with $ t \supseteq s $ .

2. (2) For every $ 0 < j \leq k $ and $ s \in S^j_d $ , $ C^j_s \subseteq C^{j-1}_{s|_{j-1}} $ .

In particular, $ \mathcal {C}_j $ refines $ \mathcal {C}_{j-1} $ . Notice that condition (2) is equivalent to: $ C^j_s \subseteq C^i_{s|_i} $ for every $ i \leq j \leq k $ and $ s \in S^j_d $ .

The construction is by recursion on $ j \leq k $ . If $ j = 0 $ , then $ S^j_d = S^{j,0}_d = \{ s |_ 0 \mid s \in S_d \} $ . For every $ s \in S^{j,0}_d $ , let $ P_s = \bigcup \{ P^t_\alpha \mid (t,\alpha ) \in d \wedge t \supseteq s \} $ . The finite family $ \{ P_s \mid s \in S^{j,0}_d \} $ consists of pairwise disjoint $ \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi } $ -sets because $ P^t_\alpha \subseteq f^{-1}(\boldsymbol {N}_t(Y)) \subseteq f^{-1}(\boldsymbol {N}_{t|_0}(Y)) $ and $ \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi }(X) $ is closed under finite unions. Using Corollary 3.7, let $ \mathcal {C}_0 = \{ C^0_s \mid s \in S^{j,0}_d = S^j_d \} $ be any (finite) $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi } $ -partition of $ X $ separating the sets $ P_s $ from each other. It is clear that (1) holds by construction, while (2) needs not to be checked in this case.

Assume now that $ j> 0 $ , and that $ \mathcal {C}_{j-1} $ has already been defined. Fix any $ s \in S^j_d $ . We distinguish two cases. If $ s \in S^{j,1}_d $ , then $ s = s|_{j-1} \in S^{j-1}_d $ and we can set $ C^j_s = C^{j-1}_s $ . With this choice, (2) is trivially satisfied. The remaining case is when $ s \in S^{j,0}_d $ . Let $ \hat {S}^j_d = \{ s|_{j-1} \mid s \in S^{j,0}_d \} $ , and notice that $ \hat {S}^j_d \subseteq S^{j-1,0}_d \subseteq S^{j-1}_d $ . For each $ \hat {s} \in \hat {S}^j_d $ , we repeat the argument from the basic case $ j = 0 $ but working within $ C^{j-1}_{\hat {s}} $ and considering only those $ s \in S^{j,0}_d $ such that $ s|_{j-1} = \hat {s} $ . More precisely, for each such $ s $ let $ P_s = \bigcup \{ P^t_\alpha \mid (t,\alpha ) \in d \wedge t \supseteq s \} $ . By (1) applied to $ \hat {s} $ , we get $ P_s \subseteq C^{j-1}_{\hat {s}} $ . Moreover, the sets $ P_s $ are pairwise disjoint and they belong to $ \lambda ^+\text {-}\boldsymbol {\Pi }^{ 0 }_{ \xi }(X) $ . So by Corollary 3.7 we can find a $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi } $ -partition $ \{ C^j_s \mid s \in S^{j,0}_d, s|_{j-1} = \hat {s} \} $ of $ C^{j-1}_{\hat {s}}$ separating the sets $ P_s $ from each other. It is clear that both (1) and (2) are satisfied by construction.

For each $ d \in D $ , let $ f_d \colon X \to Y $ be the unique function which is locally constant on the finite $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi } $ -partition $ \mathcal {C}_k $ and assumes value $ y_s $ on each $ C^k_s \in \mathcal {C}_k $ .

To conclude the proof, we just need to show that $ \lim _{d \in D} f_d = f $ . Fix $ x \in X $ and any $ s \in \mathcal {T}_Y $ such that $ f(x) \in \boldsymbol {N}_s(Y) $ . Let $ \alpha < \lambda $ be such that $ x \in P^s_\alpha $ , and set $ d = \{ (s,\alpha ) \} $ . Then Claim 4.3.1 entails that for every $ d' \supseteq d $ there is $ t \supseteq s $ such that $ f_{d'}(x) = y_t \in \boldsymbol {N}_t(Y) $ , and since $ \boldsymbol {N}_t(Y) \subseteq \boldsymbol {N}_s(Y) $ we are done.

Proof. By Theorems 2.1 and 2.3, we can assume that . Therefore it is enough to combine Proposition 4.3 with Lemma 4.1.

Proof. (1} $\Rightarrow $ (2) First assume that either $ \mu> \omega $ , or else $ \mu = \omega $ and $ \dim (Y) = 0 $ . Then we may assume, without loss of generality, that by Theorems 2.1 and 2.3. We show by induction on $ \xi < \lambda ^+ $ that $ \mathcal {M}_{\xi +1}(X,Y) $ is contained in the $ \mathcal {D}_0 $ -closure of $ \mathcal {M}_1(X,Y) $ . The basic case $ \xi = 0 $ is trivial, so assume that $\xi>0$ and fix any $ f \in \mathcal {M}_{\xi +1}(X,Y) $ . If $\xi $ is a limit ordinal, then since $ \mathcal {M}_{<\xi }(X,Y) = \bigcup _{\xi ' < \xi } \mathcal {M}_{\xi '+1}(X,Y)$ and $ \operatorname {cof}(\xi ) \in \mathcal {D}_0 $ (because $ \operatorname {cof}(\xi ) \leq |\xi | \leq \lambda $ is regular), it is enough to use Corollary 4.4 and the inductive hypothesis. If instead $\xi = \xi '+1 $ is a successor ordinal, then we use Proposition 4.3 and the inductive hypothesis applied to $ \xi ' $ : this works because if a function is locally constant on a finite $ \lambda ^+\text {-}\boldsymbol {\Delta }^{ 0 }_{ \xi } $ -partition of $ X $ , then it is trivially $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi '+1} $ -measurable.

It remains to consider the case where $ \mu = \omega $ but $ \dim (Y) \neq 0 $ . First we prove the result for all $ \lambda ^+ $ -Borel measurable functions with finite range, which are precisely the functions which are locally constant on a finite $ \lambda ^+\text {-}\mathbf {Bor} $ -partition.

Consider now an arbitrary $ \lambda ^+ $ -Borel measurable function $ f \colon X \to Y $ . Let $ \tau $ be the topology of $ Y $ . By [Reference Dimonte and Motto Ros7, Corollary 4.3.6], there is a topology $ \tau ' \supseteq \tau $ on $ Y $ such that $ Y' = (Y,\tau ') \in \mathscr{M}_\lambda $ , $ \dim (Y') = 0 $ , and $ \lambda ^+\text {-}\mathbf {Bor}(Y') = \lambda ^+\text {-}\mathbf {Bor}(Y) $ . It follows that $ f $ , viewed as a function from $ X $ to $ Y' $ , is still $ \lambda ^+ $ -Borel measurable, and thus it is $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi +1} $ -measurable for some $ \xi < \lambda ^+ $ . Since both $ X $ and $ Y' $ can now be construed as subspaces of by Theorem 2.3, we can apply Proposition 4.3 and get a family of functions $ f_d \colon X \to Y '$ which are locally constant on a finite $ \lambda ^+\text {-}\mathbf {Bor} $ -partition and such that $ f = \lim _{d \in \mathrm {Fin}_\lambda } f_d $ , where the limit is computed with respect to $ Y' $ . Since $ \tau ' \supseteq \tau $ , we still have $ f = \lim _{d \in \mathrm {Fin}_\lambda } f_d $ if the limit is computed with respect to $ Y $ , and clearly the functions $ f_d $ remain locally constant on the same finite $ \lambda ^+\text {-}\mathbf {Bor} $ -partition if we step back from $ Y' $ to $ Y $ . Therefore we are done by Claim 4.5.1.

(2) $\Rightarrow $ (3) Trivial, as $ \mathcal {D}_0 \subseteq \mathcal {D}_\lambda $ .

(3) $\Rightarrow $ (1) Let $f=\lim _{d \in D}f_d$ with $(f_d)_{d \in D}$ a family of $ \lambda ^+ $ -Borel measurable functions. Assume first that $ \mu> \omega $ , so that $ Y $ is zero-dimensional. Then for every clopen $U \subseteq Y$ ,

(4.1) $$ \begin{align} f^{-1}(U)=\bigcup_{d \in D} \bigcap_{d' \geq d }f^{-1}_{d'}(U). \end{align} $$

Since $|D|\leq \lambda $ and $f^{-1}_{d'}(U) \in \lambda ^+\text {-}\mathbf {Bor}(X)$ , this proves that f is $ \lambda ^+ $ -Borel measurable too. If instead $ \mu = \omega $ , given any open set $ U \subseteq Y $ we consider an open covering $ \{ U_n \mid n \in \omega \} $ of $ U $ such that $ \operatorname {\mathrm {cl}}(U_n) \subseteq U $ for every $ n \in \omega $ . Then,

(4.2) $$ \begin{align} f^{-1}(U)=\bigcup_{n \in \omega} \bigcup_{d \in D} \bigcap_{d' \geq d}f^{-1}_{d'}(\operatorname{\mathrm{cl}}(U_n)) , \end{align} $$

hence $ f^{-1}(U) \in \lambda ^+\text {-}\mathbf {Bor}(X) $ again.

Proof. Assume first that $ \operatorname {cof}(\xi ) < \operatorname {cof}(\kappa ) $ . Suppose that $f=\lim _{\alpha <\kappa }f_\alpha $ for some sequence $ (f_\alpha )_{\alpha < \kappa } $ of functions in $ \mathcal {M}_{<\xi }(X,Y)$ . Let $(\xi _i)_{i <\operatorname {cof}(\xi )}$ be a sequence of ordinals cofinal in $\xi $ . Then for every $\alpha <\kappa $ there exists $i<\operatorname {cof}(\xi )$ such that $f_\alpha $ is $\lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _i}$ -measurable. Since $\operatorname {cof}(\xi )< \operatorname {cof}(\kappa )$ , there exists some $\bar \imath <\operatorname {cof}(\xi )$ such that

$$\begin{align*}A = \{ \alpha < \kappa \mid f_\alpha \text{ is } \lambda^+\text{-}\boldsymbol{\Sigma}^{ 0 }_{ \xi_{\bar \imath}} \text{ -measurable} \} \end{align*}$$

is unbounded in $ \kappa $ , so that $ f = \lim _{\alpha \in A} f_\alpha $ . Being a limit of $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _{\bar \imath }}$ -measurable functions over an index set of size at most $ \lambda $ , we get that $ f $ is $ \lambda ^+\text {-}\boldsymbol {\Sigma }^{ 0 }_{ \xi _{\bar \imath }+1} $ -measurable (by the computations (4.1) and (4.2) in Theorem 4.5), and thus $ f \in \mathcal {M}_{< \xi }(X,Y) $ because $ \xi $ is limit.

Assume now that $ \operatorname {cof}(\kappa ) < \operatorname {cof}(\xi ) $ . By Remark 2.4, if $ f $ is a $ \kappa $ -limit of functions in $ \mathcal {M}_{<\xi }(X,Y) $ , then $ f = \lim _{\alpha < \operatorname {cof}(\kappa )} f_\alpha $ for some sequence $ (f_\alpha )_{\alpha < \operatorname {cof}(\kappa )} $ of functions in the same class $ \mathcal {M}_{<\xi }(X,Y) $ . By the assumption on $ \operatorname {cof}(\kappa ) $ , there is $ 1 \leq \xi ' < \xi $ such that $ f_\alpha \in \mathcal {M}_{\xi '}(X,Y) $ for all $ \alpha < \operatorname {cof}(\kappa ) $ . Using again the computations (4.1) and (4.2) in Theorem 4.5, we get $ f \in \mathcal {M}_{\xi '+1}(X,Y) \subseteq \mathcal {M}_{<\xi }(X,Y) $ , as desired.

Proof. If $ D = \mathrm {Fin}_\lambda $ , then we let $ \iota ( d) = \operatorname {\mathrm {supp}}(d) $ . If $ D = \kappa $ for some regular $ \kappa < \lambda $ , we let $ \iota ( d) = d(\kappa ) $ . Finally, if $ D = \lambda $ then we let $ \iota (d) = \max \operatorname {\mathrm {supp}}(d) $ . It is easy to check that in all three cases $ \iota $ is as required.

Proof. It is obvious that the collection of full sets with constant $ \rho $ is closed under arbitrary unions and consists of open sets, so it is enough to show that it is also closed under complements. Let $ A \subseteq X $ be full with constant $ \rho $ , and let $ y \in X \setminus A $ . Assume towards a contradiction that $ B_d(y,\rho ) \cap A \neq \emptyset $ , as witnessed by $ x $ . Then $ B_d(x,\rho ) = B_d(y,\rho ) $ and hence $ y \in A $ by fullness, a contradiction.

Finally, let $ (A_\alpha )_{\alpha < \nu } $ for $ \nu < \mu $ be a family of full sets, and let $ \rho _\alpha \in \mathbb {G}^+ $ be such that $ A_\alpha $ is full with constant $ \rho _\alpha $ . Since $ \mu $ is regular and $ \mathbb {G} $ has degree $ \mu $ , there is $ \rho \in \mathbb {G}^+ $ such that $ \rho \leq \rho _\alpha $ for all $ \alpha < \nu $ . Then each $ A_\alpha $ is full with constant $ \rho $ , and thus so is $ \bigcup _{\alpha < \nu } A_\alpha $ .

Proof. The first part of the lemma is obvious, so we only prove the second one. Let $ r' \in \mathbb {G}^+ $ be such that $ \operatorname {diam}(\operatorname {\mathrm {ran}}(f)) \leq r' $ , and set $ R = r' \cdot \rho ^{-1} $ . Pick any $ x,x' \in X $ . If $ d_X(x,x') < \rho $ , then $ d_Y(f(x),f(x')) = 0 \leq R \cdot d_X(x,x') $ because $ f $ is full with constant $ \rho $ . If instead $ d_X(x,x') \geq \rho $ , then

$$\begin{align*}d_Y(f(x),f(x')) \leq r' = r' \cdot \rho^{-1} \cdot \rho \leq R \cdot d_X(x,x'). \end{align*}$$

Thus $ f $ is Lipschitz with constant $ R $ .

Proof. Suppose that $ h $ is Lipschitz with constant $ R \in \mathbb {G}^+ $ . It is enough to show that the preimage via h of a full set $A \subseteq X$ with constant $\rho \in \mathbb {G}^+$ is a full set with constant $\rho ' = \rho \cdot R^{-1}$ . Indeed, let $z \in Z $ be such that $h(z) \in A $ and let $z' \in Z$ be such that $d_Z(z, z') < \rho ' $ . Then $d_X(h(z), h(z')) \leq R \cdot d_Z(z, z') < R \cdot \rho \cdot R^{-1}= \rho $ , hence $h(z') \in A$ . This shows that $B_{d_Z}(z , \rho ' ) \subseteq h^{-1}(A)$ .