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Theorem 4.8.2 (Miller[84]) Let \( G \) be a Polish group and \( H \) a Borel subgroup. Suppose the \( \sigma \) -algebra of invariant Borel sets is countably generated. Then \( H \) is closed.
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Proof of 4.8.2. Let \( X = G/H \), the set of right cosets, and \( q : G \rightarrow \) \( G/H \) the quotient map. Equip \( G/H \) with the largest \( \sigma \) -algebra making \( q \) Borel measurable. By our hypothesis, \( X \) is a countably generated measurable space with singletons as atoms. Consider the action \( \left( {g,{g}^{\prime }H}\right) \rightarrow g \cdot {g}^{\prime }H \) of \( G \) on \( X \) . Let \( x = H \) . Then the stabilizer\n\n\[ \n{G}_{x} = \{ g \in G : g \cdot x = x\} = H.\n\]\n\nSince \( g \rightarrow g \cdot x \) is Borel, the result follows from 4.8.5.
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Yes
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Proposition 4.8.3 Let \( X \) be a Polish space and \( G \) a group of homeomorphisms of \( X \) such that for every pair \( U, V \) of nonempty open sets there is a \( g \in G \) with \( g\left( U\right) \cap V \neq \varnothing \) . Suppose \( A \) is a \( G \) -invariant Borel set; i.e., \( g\left( A\right) = A \) for all \( g \in G \) . Then either \( A \) or \( {A}^{c} \) is meager in \( X \) .
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Proof. Suppose neither \( A \) nor \( {A}^{c} \) is meager in \( X \) . Then there exist nonempty open sets \( U, V \) such that \( A \) and \( {A}^{c} \) are comeager in \( U \) and \( V \) respectively. By our hypothesis, there is a \( g \in G \) such that \( g\left( U\right) \cap V \neq \varnothing \) . Let \( W = g\left( U\right) \bigcap V \) . It follows that \( W \) is meager. This contradicts the Baire category theorem.
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Yes
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Theorem 4.8.4 (Miller[84]) Let \( \\left( {G, \\cdot }\\right) \) be a Polish group, \( X \) a second countable \( {T}_{1} \) space, and \( \\left( {g, x}\\right) \\rightarrow g \\cdot x \) an action of \( G \) on \( X \) . Suppose that for a given \( x \), the map \( g \\rightarrow g \\cdot x \) is Borel. Then the stabilizer \( {G}_{x} \) is closed.
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Proof. Let \( H = \\operatorname{cl}\\left( {G}_{x}\\right) \) . It is fairly easy to see that we can replace \( G \) by \( H \) . Hence, without loss of generality we assume that \( {G}_{x} \) is dense in \( G \) . \n\nSince \( X \) is second countable and \( {T}_{1},{G}_{x} \) is Borel. Therefore, by 3.5.13, we shall be done if we show that \( {G}_{x} \) is nonmeager. Suppose not. We shall get a contradiction. Take a countable base \( \\left( {U}_{n}\\right) \) for \( X \) . Let \( f\\left( g\\right) = g \\cdot x \) . As \( f \) is Borel, \( {f}^{-1}\\left( {U}_{n}\\right) = {A}_{n} \), say, is Borel. For every \( h \\in {G}_{x},{A}_{n} \\cdot h = {A}_{n} \) . Since \( X \) is \( {T}_{1} \), for any two \( g, h \) we have \n\n\[ \n g \\cdot x = h \\cdot x \\Leftrightarrow \\forall n\\left( {g \\in {A}_{n} \\Leftrightarrow h \\in {A}_{n}}\\right) . \n\] \n\nHence, for any \( g \\in G \)\n\n\[ \ng{G}_{x} = \\bigcap \\left\\{ {{A}_{n} : g \\in {A}_{n}}\\right\\} \n\] \n\nApplying 4.8.3 to the group of homeomorphisms of \( G \) induced by right multiplication by elements of \( {G}_{x} \), we see that \( {A}_{n} \) is either meager or comeager. Since \( {G}_{x} \) is meager, there exists \( n \) such that \( g \\in {A}_{n} \) and \( {A}_{n} \) is meager. Hence, \n\n\[ \nG = \\bigcup \\left\\{ {{A}_{n} : {A}_{n}\\text{ meager }}\\right\\} . \n\] \n\nThis contradicts the Baire category theorem, and our result is proved.
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Yes
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Theorem 4.8.6 Let \( G \) be a Polish group, \( X \) a Polish space, and \( a\left( {g, x}\right) = \) \( g \cdot x \) an action of \( G \) on \( X \) . Assume that \( g \cdot x \) is continuous in \( x \) for all \( g \) and Borel in \( g \) for all \( x \) . Then the action is continuous.
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Proof. By 3.1.30, the action \( a : G \times X \rightarrow X \) is Borel. Let \( \left( {V}_{n}\right) \) be a countable base for \( X \) . Put \( {C}_{n} = {a}^{-1}\left( {V}_{n}\right) \) . Then \( {C}_{n} \) is Borel with open sections. By 4.7.2, write\n\n\[ \n{C}_{n} = \mathop{\bigcup }\limits_{m}\left( {{B}_{nm} \times {W}_{nm}}\right)\n\]\n\nthe \( {B}_{nm} \)’s Borel, the \( {W}_{nm} \)’s open. By 3.5.1, \( {B}_{nm} \) has the Baire property. Let \( {I}_{nm} \) be a meager set in \( G \) such that \( {B}_{nm}\Delta {I}_{nm} \) is open. Put \( I = \mathop{\bigcup }\limits_{{nm}}{I}_{nm} \) . Then \( I \) is meager in \( G \) and \( a \mid \left( {G \smallsetminus I}\right) \times X \) is continuous.\n\nNow take a sequence \( \left( {{g}_{k},{x}_{k}}\right) \) in \( G \times X \) converging to \( \left( {g, x}\right) \), say. We need to show that \( {g}_{k} \cdot {x}_{k} \rightarrow g \cdot x \) . Let\n\n\[ \nJ = \mathop{\bigcup }\limits_{k}I \cdot {g}_{k}^{-1}\bigcup I \cdot {g}^{-1}.\n\]\n\nSince \( G \) is a topological group, \( J \) is meager in \( G \) . By the Baire category theorem, \( G \neq J \) . Take any \( h \in G \smallsetminus J \) . Then \( h \cdot g, h \cdot {g}_{k} \in G \smallsetminus I \) . As \( {g}_{k} \rightarrow g \) , \( h \cdot {g}_{k} \rightarrow h \cdot g \) . Since \( a \mid \left( {G \smallsetminus I}\right) \times X \) is continuous, \( \left( {h \cdot {g}_{k}}\right) \cdot {x}_{k} \rightarrow \left( {h \cdot g}\right) \cdot x \) . Since the action is continuous in the second variable,\n\n\[ \n{g}_{k} \cdot {x}_{k} = {h}^{-1} \cdot \left( {\left( {h \cdot {g}_{k}}\right) \cdot {x}_{k}}\right) \rightarrow {h}^{-1} \cdot \left( {\left( {h \cdot g}\right) \cdot x}\right) = g \cdot x.\n\]
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Yes
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Lemma 4.8.8 If \( \left( {G, \cdot }\right) \) is a group with a Polish topology such that the group operation \( \left( {g, h}\right) \rightarrow g \cdot h \) is Borel, then \( g \rightarrow {g}^{-1} \) is continuous.
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Proof. Since \( \left( {g, h}\right) \rightarrow g \cdot h \) is Borel, the graph\n\n\[ \n\{ \left( {g, h}\right) : g \cdot h = e\}\n\]\n\nof \( g \rightarrow {g}^{-1} \) is Borel. Hence, by 4.5.2, \( g \rightarrow {g}^{-1} \) is Borel measurable. An imitation of the proof of 3.5.9 shows that \( g \rightarrow {g}^{-1} \) is continuous.
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No
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Proposition 4.8.9 If \( \left( {G, \cdot }\right) \) is a group with a Polish topology such that the group operation is separately continuous in each variable, then \( G \) is a topological group.
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Proof. In view of 4.8.8, we have only to show that the group operation is jointly continuous. This we get immediately by applying 4.8.6 to \( X = G \) and action \( g \cdot x \) the group operation.
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Yes
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Theorem 4.8.10 (S. Solecki and S. M. Srivastava[109]) Let \( \left( {G, \cdot }\right) \) be a group with a Polish topology such that \( h \rightarrow g \cdot h \) is continuous for every \( g \in G \), and \( g \rightarrow g \cdot h \) Borel for all \( h \) . Then \( G \) is a topological group.
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Proof. By 4.8.9, we only have to show that the group operation \( g \cdot h \) is jointly continuous. A close examination of the proof of 4.8.6 shows that this follows from the following result.\n\nLemma 4.8.11 Let \( G \) satisfy the hypothesis of our theorem. Then for every meager set \( I \) and every \( g \) ,\n\n\[ \n{Ig} = \{ h \cdot g : h \in I\} \n\]\n\n is meager.\n\nProof.\n\nClaim. If \( I \) is meager in \( G \), so is \( {I}^{-1} = \left\{ {h \in G : {h}^{-1} \in I}\right\} \) .\n\nAssuming the claim, we prove the lemma as follows. Let \( I \) be meager in \( G \) and \( g \in G \) . By the claim, \( {I}^{-1} \) is meager. Since the group operation is continuous in the second varible, \( J = {g}^{-1} \cdot {I}^{-1} \) is meager. As \( I \cdot g = {J}^{-1} \) , it is meager by our claim.\n\nProof of the claim. Let \( I \) be meager. Since every meager set is contained in a meager \( {F}_{\sigma } \), without any loss of generality we assume that \( I \) is Borel. By 3.1.30, the group operation \( \left( {g, h}\right) \rightarrow g \cdot h \) is a Borel map. Since the graph of \( g \rightarrow {g}^{-1} \) is Borel, \( g \rightarrow {g}^{-1} \) is Borel measurable (4.5.2). Hence, \( \left( {g, h}\right) \rightarrow {g}^{-1} \cdot h \) is Borel measurable. Let\n\n\[ \n\widehat{I} = \left\{ {\left( {h, g}\right) : {g}^{-1} \cdot h \in I}\right\} .\n\]\n\nSince \( \widehat{I} \) is a Borel set, it has the Baire property. Now, for every \( g \in G \) ,\n\n\[ \n{\widehat{I}}^{g} = \left\{ {h \in G : {g}^{-1} \cdot h \in I}\right\} = g \cdot I.\n\]\n\nHence, by our hypothesis, \( {\widehat{I}}^{g} \) is meager for every \( g \) . Therefore, by the Kura-towski - Ulam theorem (3.5.16), the set \( \left\{ {h : {\widehat{I}}_{h}}\right. \) is meager \( \} \) is comeager and hence nonempty by the Baire category theorem. In particular, there exists \( h \in G \) such that \( {\widehat{I}}_{h} = h \cdot {I}^{-1} \) is meager. It follows that \( {I}^{-1} = {h}^{-1}\left( {h{I}^{-1}}\right) \) is meager.
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Yes
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Lemma 4.8.11 Let \( G \) satisfy the hypothesis of our theorem. Then for every meager set \( I \) and every \( g \) , \[ {Ig} = \{ h \cdot g : h \in I\} \] is meager.
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Proof. Claim. If \( I \) is meager in \( G \), so is \( {I}^{-1} = \left\{ {h \in G : {h}^{-1} \in I}\right\} \) . Assuming the claim, we prove the lemma as follows. Let \( I \) be meager in \( G \) and \( g \in G \) . By the claim, \( {I}^{-1} \) is meager. Since the group operation is continuous in the second varible, \( J = {g}^{-1} \cdot {I}^{-1} \) is meager. As \( I \cdot g = {J}^{-1} \) , it is meager by our claim. Proof of the claim. Let \( I \) be meager. Since every meager set is contained in a meager \( {F}_{\sigma } \), without any loss of generality we assume that \( I \) is Borel. By 3.1.30, the group operation \( \left( {g, h}\right) \rightarrow g \cdot h \) is a Borel map. Since the graph of \( g \rightarrow {g}^{-1} \) is Borel, \( g \rightarrow {g}^{-1} \) is Borel measurable (4.5.2). Hence, \( \left( {g, h}\right) \rightarrow {g}^{-1} \cdot h \) is Borel measurable. Let \[ \widehat{I} = \left\{ {\left( {h, g}\right) : {g}^{-1} \cdot h \in I}\right\} . \] Since \( \widehat{I} \) is a Borel set, it has the Baire property. Now, for every \( g \in G \) , \[ {\widehat{I}}^{g} = \left\{ {h \in G : {g}^{-1} \cdot h \in I}\right\} = g \cdot I. \] Hence, by our hypothesis, \( {\widehat{I}}^{g} \) is meager for every \( g \) . Therefore, by the Kura-towski - Ulam theorem (3.5.16), the set \( \left\{ {h : {\widehat{I}}_{h}}\right. \) is meager \( \} \) is comeager and hence nonempty by the Baire category theorem. In particular, there exists \( h \in G \) such that \( {\widehat{I}}_{h} = h \cdot {I}^{-1} \) is meager. It follows that \( {I}^{-1} = {h}^{-1}\left( {h{I}^{-1}}\right) \) is meager.
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No
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Lemma 4.9.3 Let \( X \) be a Polish space, \( A \subseteq X \) coanalytic, and \( \varphi \) a norm on \( A \) . Then \( \varphi \) is a \( {\mathbf{\Pi }}_{1}^{1} \) -norm if and only if both \( { \leq }_{\varphi }^{ * },{ < }_{\varphi }^{ * } \) are coanalytic.
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Proof. We first prove the \
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No
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Example 4.9.4 Let \( X = {2}^{\mathbb{N} \times \mathbb{N}} \) and \( A = {WO} \). For \( x \in {WO} \), Let \( \left| x\right| < {\omega }_{1} \) be the order type of \( x \).
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For \( x \in {2}^{\mathbb{N} \times \mathbb{N}} \), define\n\n\[ m{ < }_{x}n \Leftrightarrow x\left( {m, n}\right) = 1\& x\left( {n, m}\right) = 0. \]\n\nFor \( x, y \) in \( {2}^{\mathbb{N} \times \mathbb{N}} \), set\n\n\[ x{ \leq }_{\left| \cdot \right| }^{{\sum }_{1}^{1}}y \Leftrightarrow \exists z \in {\mathbb{N}}^{\mathbb{N}}\forall m\forall n\left\lbrack {m{ < }_{x}n \Leftrightarrow z\left( m\right) { < }_{y}z\left( n\right) }\right\rbrack \]\n\nand\n\n\[ x{ < }_{\left| \cdot \right| }^{{\sum }_{1}^{1}}y\; \Leftrightarrow \;\exists k\exists z \in {\mathbb{N}}^{\mathbb{N}}\forall m\forall n\lbrack z\left( m\right) { < }_{y}k \]\n\n\[ \left. {\& \left( {m{ < }_{x}n \Leftrightarrow z\left( m\right) { < }_{y}z\left( n\right) }\right) }\right\rbrack . \]\n\nThus, \( x{ \leq }_{\left| \cdot \right| }^{{\sum }_{1}^{1}}y \) if and only if there is an order-preserving map from \( x \) to \( y \), and \( x{ < }_{\left| \cdot \right| }^{{\sum }_{1}^{1}}y \) if and only if there is an order-preserving map from \( x \) into an initial segment of \( y \). The sets \( { \leq }_{\left| \cdot \right| }^{{\sum }_{1}^{1}} \) and \( { < }_{\left| \cdot \right| }^{{\sum }_{1}^{1}} \) are clearly \( {\mathbf{\sum }}_{1}^{1} \). Further, for \( y \in {WO} \), \n\n\[ x{ \leq }_{\left| \cdot \right| }^{{\sum }_{1}^{1}}y \Leftrightarrow x \in {WO}\& \left| x\right| \leq \left| y\right| \]\n\nand\n\n\[ x{ < }_{\left| \cdot \right| }^{{\sum }_{1}^{1}}y \Leftrightarrow x \in {WO}\& \left| x\right| < \left| y\right| . \]\n\nTherefore, by 4.9.2, \( \left| \cdot \right| \) is a \( {\mathbf{\Pi }}_{1}^{1} \) -norm on \( {WO} \), which we shall call the canonical norm on \( {WO} \).
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Yes
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Theorem 4.9.8 (Boundedness theorem for \( {\mathbf{\Pi }}_{1}^{1} \) -norms) Suppose \( A \) is a \( {\mathbf{\Pi }}_{1}^{1} \) set in a Polish space \( X \) and \( \varphi \) a norm on \( A \) as defined in 4.9.1. Then for every \( {\mathbf{\sum }}_{1}^{1} \) set \( B \subseteq A,\sup \{ \varphi \left( x\right) : x \in B\} < {\omega }_{1} \) .
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Proof. Suppose \( \sup \{ \varphi \left( y\right) : y \in B\} = {\omega }_{1} \) . Take any \( {\mathbf{\Pi }}_{1}^{1} \) set \( C \) that is not \( {\mathbf{\sum }}_{1}^{1} \) . Fix a Borel function \( g \) such that\n\n\[ x \in C \Leftrightarrow g\left( x\right) \in {WO}. \]\n\nThen,\n\n\[ x \in C\; \Leftrightarrow \;\exists y\left( {y \in B\& \left| {g\left( x\right) }\right| \leq \varphi \left( y\right) }\right) \]\n\n\[ \Leftrightarrow \exists y\left( {y \in B\& g\left( x\right) { \leq }_{\left| .\right| }^{{\sum }_{1}^{1}}f\left( y\right) }\right) , \]\n\nwhere \( f \) is as in 4.9.1. This contradicts the fact that \( C \) is not \( {\mathbf{\sum }}_{1}^{1} \) . Hence,\n\n\[ \sup \{ \varphi \left( x\right) : x \in B\} < {\omega }_{1}. \]\n\nIf \( A \) is Borel, then taking \( B \) to be \( A \), we see that \( \sup \{ \varphi \left( x\right) : x \in A\} < {\omega }_{1} \) . On the other hand, if \( \sup \{ \varphi \left( x\right) : x \in A\} < {\omega }_{1} \), then \( A \) is a union of countably many Borel sets of the form \( \{ x \in A : \varphi \left( x\right) = \xi \} ,\xi < {\omega }_{1} \) . So \( A \) is Borel.
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Yes
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Example 4.9.11 (A. Maitra and C. Ryll-Nardzewski[76]) Let \( X, Y \) be uncountable Polish spaces. Let \( U \subseteq X \times X \) be universal analytic and \( C \subseteq Y \) an uncountable coanalytic set not containing a perfect set. We mentioned earlier that Gödel's axiom of constructibility implies the existence of such a set. The set \( C \) does not contain any uncountable Borel set. Take \( A = Y \smallsetminus C \) . Then \( U \) and \( A \) are not Borel isomorphic.
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Here is a proof. Suppose they are Borel isomorphic. Take a Borel isomorphism \( f : U \rightarrow A \) . By 3.3.5, there exist Borel sets \( {B}_{1} \supseteq U,{B}_{2} \supseteq A \) and a Borel isomorphism \( g : {B}_{1} \rightarrow {B}_{2} \) extending \( f \) . Let \( \varphi \) be a \( {\mathbf{\Pi }}_{1}^{1} \) norm on \( {U}^{c} \) as defined in 4.9.8. It is easy to verify that for uncountably many \( \xi < {\omega }_{1},\left\{ {\left( {x, y}\right) \in {U}^{c} : \varphi \left( {x, y}\right) = }\right. \) \( \xi \} \) is uncountable. By 4.9.8, \( \sup \left\{ {\varphi \left( {x, y}\right) : \left( {x, y}\right) \in {B}_{1}^{c}}\right\} < {\omega }_{1} \) . Therefore, \( {B}_{1} \smallsetminus U \) contains an uncountable Borel set. It follows that \( C \) contains an uncountable Borel set, which is not the case. Hence, \( U \) and \( A \) are not Borel isomorphic.
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Yes
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Theorem 4.9.14 (The reduction principle for coanalytic sets) (Kuratowski) Let \( \\left( {A}_{n}\\right) \) be sequence of \( {\\mathbf{\\Pi }}_{1}^{1} \) sets in a Polish space \( X \) . Then there is a sequence \( \\left( {A}_{n}^{ * }\\right) \) of \( {\\mathbf{\\Pi }}_{1}^{1} \) sets such that they are pairwise disjoint, \( {A}_{n}^{ * } \\subseteq {A}_{n} \) , and \( \\mathop{\\bigcup }\\limits_{n}{A}_{n}^{ * } = \\mathop{\\bigcup }\\limits_{n}{A}_{n} \) .
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Proof. Consider \( A \\subseteq X \\times \\mathbb{N} \) given by\n\n\[ \n\\left( {x, n}\\right) \\in A \\Leftrightarrow x \\in {A}_{n} \n\] \n\nClearly, \( A \) is \( {\\mathbf{\\Pi }}_{1}^{1} \) with projection \( \\mathop{\\bigcup }\\limits_{n}{A}_{n} \) . Let \( \\varphi \) be a \( {\\mathbf{\\Pi }}_{1}^{1} \)-norm on \( A \) . Define \( {A}^{ * } \\subseteq X \\times \\mathbb{N} \) by\n\n\[ \n\\left( {x, n}\\right) \\in {A}^{ * }\\; \\Leftrightarrow \\;\\left( {x, n}\\right) \\in A\\& \\forall m\\left\\lbrack {\\left( {x, n}\\right) { \\leq }_{\\varphi }^{ * }\\left( {x, m}\\right) }\\right\\rbrack \n\] \n\n\[ \n\\text{&}\\forall m\\left\\lbrack {\\left( {x, n}\\right) { < }_{\\varphi }^{ * }\\left( {x, m}\\right) \\text{or}n \\leq m}\\right\\rbrack \\text{.} \n\] \n\nThus, for each \( x \) in the projecton of \( A \) we first look at the set of integers \( n \) with \( \\left( {x, n}\\right) \\in A \) such that \( \\varphi \\left( {x, n}\\right) \) is the minimum. Then we choose the least among these integers. Note that \( {A}^{ * } \) is \( {\\mathbf{\\Pi }}_{1}^{1},{A}^{ * } \\subseteq A \), and for every \( x \\in \\mathop{\\bigcup }\\limits_{n}{A}_{n} \) there is exactly one \( n \) such that \( \\left( {x, n}\\right) \\in {A}^{ * } \) . Let\n\n\[ \n{A}_{n}^{ * } = \\{ x : \\left( {x, n}\\right) \\in A * \\} . \n\] \n\nClearly, \( {A}_{n}^{ * } \) is \( {\\mathbf{\\Pi }}_{1}^{1} \) . It is easy to check that the \( {A}_{n}^{ * } \\)’s are pairwise disjoint and \( \\mathop{\\bigcup }\\limits_{n}{A}_{n}^{ * } = \\mathop{\\bigcup }\\limits_{n}{A}_{n} \) .
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Yes
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Corollary 4.9.15 Let \( X \) be Polish and \( {A}_{0},{A}_{1} \) coanalytic subsets of \( X \) . Then there exist pairwise disjoint coanalytic sets \( {A}_{0}^{ * },{A}_{1}^{ * } \) contained in \( {A}_{0} \) , \( {A}_{1} \) respectively such that \( {A}_{0}^{ * }\bigcup {A}_{1}^{ * } = {A}_{0}\bigcup {A}_{1} \) .
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Proof. In the above theorem, take \( {A}_{n} = \varnothing \) for \( n > 1 \) .
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Yes
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Theorem 4.9.19 Let \( X \) be a Polish space. Then there exist sets \( C \in \) \( {\mathbf{\Pi }}_{1}^{1}\left( {\mathbb{N}}^{\mathbb{N}}\right) \) and \( V \in {\mathbf{\Pi }}_{1}^{1}\left( {{\mathbb{N}}^{\mathbb{N}} \times X}\right), U \in {\mathbf{\sum }}_{1}^{1}\left( {{\mathbb{N}}^{\mathbb{N}} \times X}\right) \) such that for every \( \alpha \in C,{U}_{\alpha } = {V}_{\alpha } \) and\n\n\[ \n{\mathbf{\Delta }}_{1}^{1}\left( X\right) = \left\{ {{U}_{\alpha } : \alpha \in C}\right\} \n\] \n\nIn particular, there are a coanalytic set and an analytic set contained in \( {\mathbb{N}}^{\mathbb{N}} \times X \) that are universal for \( {\mathbf{\Delta }}_{1}^{1}\left( X\right) \) .
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Proof. Let \( {W}_{0},{W}_{1} \) be coanalytic subsets of \( {\mathbb{N}}^{\mathbb{N}} \times X \) such that for every pair \( \left( {{C}_{0},{C}_{1}}\right) \) of sets in \( {\mathbf{\Pi }}_{1}^{1}\left( X\right) \) there is an \( \alpha \) with \( {C}_{i} = {\left( {W}_{i}\right) }_{\alpha }, i = 0 \) or 1 . By the reduction principle for coanalytic sets, (4.9.15), there are pairwise disjoint coanalytic sets \( {V}_{i} \subseteq {W}_{i}, i = 0 \) or 1, such that \( {V}_{0}\bigcup {V}_{1} = {W}_{0}\bigcup {W}_{1} \) . Define\n\n\[ \nC = \left\{ {\alpha : \forall x\left( {\left( {\alpha, x}\right) \in {V}_{0}\bigcup {V}_{1}}\right) }\right\} \n\] \n\nSo, \( C \) is coanalytic. Take \( V = {V}_{0} \) and \( U = {V}_{1}^{c} \) . A routine argument shows that \( C, U \), and \( V \) have the desired properties. So, we have proved the first part of the result.\n\nTo see the second part, note that \( V\bigcap \left( {C \times X}\right) \) is a coanalytic set universal for \( {\mathbf{\Delta }}_{1}^{1}\left( X\right) \), and its complement is an analytic set universal for \( {\mathbf{\Delta }}_{1}^{1}\left( X\right) \) .
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Yes
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Example 4.10.1 Let \( \mu \) be a finite Borel measure on a Polish space \( X \) and \( {\mu }^{ * } \) the associated outer measure. Thus, for any \( A \subseteq X \) , \[ {\mu }^{ * }\left( A\right) = \inf \{ \mu \left( B\right) : B \supseteq A, B\text{ Borel }\} . \]
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It is easy to check that \( {\mu }^{ * } \) is a capacity on \( X \) .
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No
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Proposition 4.10.4 Let \( I \) be a capacity on a Polish space \( X \) and that \( {I}^{ * } : \mathcal{P}\left( X\right) \rightarrow \left\lbrack {0,\infty }\right\rbrack \) be defined by\n\n\[ \n{I}^{ * }\left( A\right) = \inf \{ I\left( B\right) : B \supseteq A, B\text{ Borel }\} .\n\]\n\nThen \( {I}^{ * } \) is a capacity on \( X \) .
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Proof. Clearly, \( {I}^{ * } \) is monotone. Further, \( {I}^{ * } \) and \( I \) coincide on Borel sets. As \( I \) is a capacity, it follows that \( {I}^{ * }\left( K\right) < \infty \) for every compact \( K \) and that \( {I}^{ * } \) is right-continuous over compacta.\n\nTo show that \( {I}^{ * } \) is going up, take any nondecreasing sequence \( \left( {A}_{n}\right) \) of subsets of \( X \) . Set \( A = \mathop{\bigcup }\limits_{n}{A}_{n} \) . Note that for every \( C \subseteq X \), there is a Borel \( D \supseteq C \) such that \( {I}^{ * }\left( C\right) = I\left( D\right) \) . Hence, for every \( n \) there is a Borel \( {B}_{n} \supseteq {A}_{n} \) such that \( I\left( {B}_{n}\right) = {I}^{ * }\left( {A}_{n}\right) \) . Replacing \( {B}_{n} \) by \( \mathop{\bigcap }\limits_{{m > n}}{B}_{m} \), we may assume that \( \left( {B}_{n}\right) \) is nondecreasing. Set \( B = \mathop{\bigcup }\limits_{n}{B}_{n} \) . Clearly,\n\n\[ \nI\left( B\right) \geq {I}^{ * }\left( A\right) \geq {I}^{ * }\left( {A}_{n}\right) = I\left( {B}_{n}\right)\n\]\n\nfor every \( n \) . Since \( I \) is going up, \( \lim I\left( {B}_{n}\right) = I\left( B\right) \) . It follows that \( \left. {\lim {I}^{ * }\left( {A}_{n}\right) = {I}^{ * }A}\right) \) .
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Yes
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Proposition 4.10.4 Let \( I \) be a capacity on a Polish space \( X \) and that \( {I}^{ * } : \mathcal{P}\left( X\right) \rightarrow \left\lbrack {0,\infty }\right\rbrack \) be defined by\n\n\[ \n{I}^{ * }\left( A\right) = \inf \{ I\left( B\right) : B \supseteq A, B\text{ Borel }\} .\n\]\n\nThen \( {I}^{ * } \) is a capacity on \( X \) .
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Proof. Clearly, \( {I}^{ * } \) is monotone. Further, \( {I}^{ * } \) and \( I \) coincide on Borel sets. As \( I \) is a capacity, it follows that \( {I}^{ * }\left( K\right) < \infty \) for every compact \( K \) and that \( {I}^{ * } \) is right-continuous over compacta.\n\nTo show that \( {I}^{ * } \) is going up, take any nondecreasing sequence \( \left( {A}_{n}\right) \) of subsets of \( X \) . Set \( A = \mathop{\bigcup }\limits_{n}{A}_{n} \) . Note that for every \( C \subseteq X \), there is a Borel \( D \supseteq C \) such that \( {I}^{ * }\left( C\right) = I\left( D\right) \) . Hence, for every \( n \) there is a Borel \( {B}_{n} \supseteq {A}_{n} \) such that \( I\left( {B}_{n}\right) = {I}^{ * }\left( {A}_{n}\right) \) . Replacing \( {B}_{n} \) by \( \mathop{\bigcap }\limits_{{m > n}}{B}_{m} \), we may assume that \( \left( {B}_{n}\right) \) is nondecreasing. Set \( B = \mathop{\bigcup }\limits_{n}{B}_{n} \) . Clearly,\n\n\[ \nI\left( B\right) \geq {I}^{ * }\left( A\right) \geq {I}^{ * }\left( {A}_{n}\right) = I\left( {B}_{n}\right)\n\]\n\nfor every \( n \) . Since \( I \) is going up, \( \lim I\left( {B}_{n}\right) = I\left( B\right) \) . It follows that \( \left. {\lim {I}^{ * }\left( {A}_{n}\right) = {I}^{ * }A}\right) \) .
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Yes
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Proposition 4.10.10 Let \( I \) be a capacity on a Polish space \( X \) and \( A \subseteq X \) universally capacitable. Then\n\n\[ I\left( A\right) = {I}^{ * }\left( A\right) \]\n\nwhere \( {I}^{ * } \) is as defined in 4.10.4.
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Proof. By 4.10.4, \( {I}^{ * } \) is a capacity. Now note the following.\n\n\[ {I}^{ * }\left( A\right) = \sup \left\{ {{I}^{ * }\left( K\right) : K \subseteq A\text{ compact }}\right\} \;\text{ (as }A\text{ is }{I}^{ * } - \text{ capacitable) }\n\n= \;\sup \{ I\left( K\right) : K \subseteq A\text{ compact}\} \n\n= I\left( A\right) \]
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Yes
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Proposition 4.10.11 \( {\mathbb{N}}^{\mathbb{N}} \) is universally capacitable.
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Proof. For any \( s = \left( {{n}_{0},{n}_{1},\ldots ,{n}_{k - 1}}\right) \in {\mathbb{N}}^{ < \mathbb{N}} \), set\n\n\[ \n{\sum }^{ * }\left( s\right) = \left\{ {\alpha \in {\mathbb{N}}^{\mathbb{N}} : \left( {\forall i < k}\right) \left( {\alpha \left( i\right) \leq {n}_{i}}\right) }\right\} \n\]\n\nTake any capacity \( I \) on \( {\mathbb{N}}^{\mathbb{N}} \) and a real number \( t \) such that \( I\left( {\mathbb{N}}^{\mathbb{N}}\right) > t \) . To prove our result, we shall show that there is a compact set \( K \) such that \( I\left( K\right) \geq t \) .\n\nSince the sequence \( \left( {{\sum }^{ * }\left( n\right) }\right) \) increases to \( {\mathbb{N}}^{\mathbb{N}} \), there is a natural number \( {n}_{0} \) such that \( I\left( {{\sum }^{ * }\left( {n}_{0}\right) }\right) > t \) . Again, since \( \left( {{\sum }^{ * }\left( {{n}_{0}n}\right) }\right) \) increases to \( {\sum }^{ * }\left( {n}_{0}\right) \), there is a natural number \( {n}_{1} \) such that \( I\left( {{\sum }^{ * }\left( {{n}_{0}{n}_{1}}\right) }\right) > t \) . Proceeding similarly, we get a sequence \( {n}_{0},{n}_{1},{n}_{2},\ldots \) of natural numbers such that\n\n\[ \nI\left( {{\sum }^{ * }\left( {{n}_{0}{n}_{1}\ldots {n}_{k - 1}}\right) }\right) > t \n\]\n\nfor every \( k \) . Now consider\n\n\[ \nK = \left\{ {\alpha \in {\mathbb{N}}^{\mathbb{N}} : \alpha \left( i\right) \leq {n}_{i}\text{ for every }i}\right\} .\n\]\n\nClearly, \( K \) is compact. We claim that \( I\left( K\right) \geq t \) . Suppose not. Since \( I \) is right-continuous over compacta, there is an open set \( U \supseteq K \) such that \( I\left( U\right) < t \) . It is not very hard to show that \( U \supseteq {\sum }^{ * }\left( {{n}_{0}{n}_{1}\ldots {n}_{k - 1}}\right) \) ) for some \( k \) . Since \( I \) is monotone, we have arrived at a contradiction.
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Yes
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Theorem 4.10.12 (Choquet capacitability theorem [30], [107]) Every analytic subset of a Polish space is universally capacitable.
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Proof. Let \( X \) be a Polish space and \( A \subseteq X \) analytic. Let \( I \) be any capacity on \( X \) . Suppose \( I\left( A\right) > t \) . Let \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) be a continuous map with range \( A \) . By 4.10.11, there is a compact \( K \subseteq {\mathbb{N}}^{\mathbb{N}} \) such that \( {I}_{f}\left( K\right) > t \) . Plainly \( I\left( {f\left( K\right) }\right) > t \), and our result is proved.
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No
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Proposition 4.10.13 Let \( X \) be a Polish space and \( I \) the separation capacity on \( X \times X \) as defined in 4.10.2. Assume that a rectangle \( {A}_{1} \times {A}_{2} \) be universally capacitable. If \( I\left( {{A}_{1} \times {A}_{2}}\right) = 0 \), then there is a Borel rectangle \( B = {B}_{1} \times {B}_{2} \) containing \( {A}_{1} \times {A}_{2} \) of \( I \) -capacity 0 .
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Proof of 4.10.13. Set \( {C}_{0} = {A}_{1} \times {A}_{2} \) . By 4.10.10, there is a Borel \( {C}_{1} \supseteq {C}_{0} \) such that \( I\left( {C}_{1}\right) = 0 \) . Set \( {C}_{2} = R\left\lbrack {C}_{1}\right\rbrack \) . (Recall that \( \mathrm{R}\left\lbrack \mathrm{A}\right\rbrack \) denotes the smallest rectangle containing \( A \) .) Clearly \( I\left( {C}_{2}\right) = 0 \) . Since \( {C}_{2} \) is analytic, by 4.10.12, it is universally capacitable. By 4.10.10, there is a Borel \( {C}_{3} \supseteq {C}_{2} \) such that \( I\left( {C}_{3}\right) = 0 \) . Set \( {C}_{4} = R\left\lbrack {C}_{3}\right\rbrack \) . Proceeding similarly, we get a nondecreasing sequence \( \left( {C}_{n}\right) \) of subsets of \( X \times X \) such that \( {C}_{n} \) is a rectangle for even \( n \) and \( {C}_{n} \) ’s are Borel for odd \( n \) . Further \( I\left( {C}_{n}\right) = 0 \) for all \( n \) . Take \( B = \mathop{\bigcup }\limits_{n}{C}_{n} \) .
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Yes
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Theorem 4.11.1 (Second separation theorem for analytic sets) (Kuratowski) Let \( X \) be a Polish space and \( A, B \) two analytic subsets. There exist disjoint coanalytic sets \( C \) and \( D \) such that\n\n\[ A \smallsetminus B \subseteq C \\text{and} B \smallsetminus A \subseteq D. \]
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Proof. By 4.1.20, there exist Borel maps \( f : X \rightarrow {LO}, g : X \rightarrow {LO} \) such that \( {f}^{-1}\left( {WO}\right) = {A}^{c} \) and \( {g}^{-1}\left( {WO}\right) = {B}^{c} \). \n\nFor \( \alpha ,\beta \) in \( {LO} \), define\n\n\[ \alpha \preccurlyeq \beta \; \Leftrightarrow \;\exists f \in {\\mathbb{N}}^{\\mathbb{N}}(f \\mid D\\left( \\alpha \\right) \\text{ is one-to-one }\n\n\\left. {\\& \\forall m\\forall n\\left( {\\alpha \\left( {m, n}\\right) = 1 \\Leftrightarrow \\beta \\left( {f\\left( m\\right), f\\left( n\\right) }\\right) = 1}\\right) }\\right) .\n\n(Recall that for any \( \\alpha \\in {\\mathbb{N}}^{\\mathbb{N}}, n \\in D\\left( \\alpha \\right) \\Leftrightarrow \\alpha \\left( {n, n}\\right) = 1 \).) So \( \\preccurlyeq \) is an analytic subset of \( {\\mathbb{N}}^{\\mathbb{N}} \\times {\\mathbb{N}}^{\\mathbb{N}} \). Let\n\n\[ C = {B}^{c}\\bigcap \\{ x \\in X : f\\left( x\\right) \\preccurlyeq g\\left( x\\right) {\\} }^{c}\n\nand\n\n\[ D = {A}^{c}\\bigcap \\{ x \\in X : g\\left( x\\right) \\preccurlyeq f\\left( x\\right) {\\} }^{c}.\n\nClearly, \( C \) and \( D \) are coanalytic. We claim that \( C \) and \( D \) are disjoint. Suppose not. Take any \( x \\in C \\cap D \). Then both \( f\\left( x\\right) \) and \( g\\left( x\\right) \) are in \( {WO} \). Therefore, either \( \\left| {f\\left( x\\right) }\\right| \\leq \\left| {g\\left( x\\right) }\\right| \) or \( \\left| {g\\left( x\\right) }\\right| \\leq \\left| {f\\left( x\\right) }\\right| \). Since \( x \\in C \\cap D \), this is impossible.\n\nFinally, we show that \( A \\smallsetminus B \\subseteq C \). Let \( x \\in A \\smallsetminus B \). Then, of course, \( x \\in {B}^{c} \). As \( f\\left( x\\right) \\notin {WO} \) and \( g\\left( x\\right) \\in {WO} \), there is no order-preserving one-to-one map from \( D\\left( {f\\left( x\\right) }\\right) \) into \( D\\left( {g\\left( x\\right) }\\right) \). So, \( x \\in C \). Similarly it follows that \( B \\smallsetminus A \\subseteq D \).
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Yes
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Corollary 4.11.3 Suppose \( X \) is a Polish space and \( \left( {A}_{n}\right) \) a sequence of analytic subsets of \( X \) . Then there exists a sequence \( \left( {C}_{n}\right) \) of pairwise disjoint coanalytic sets such that
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Proof. By the second separation theorem, for each \( n \) there exist pairwise disjoint coanalytic sets \( {C}_{n}^{\prime } \) and \( {D}_{n}^{\prime } \) such that\n\n\[ {A}_{n} \smallsetminus \mathop{\bigcup }\limits_{{m \neq n}}{A}_{m} \subseteq {C}_{n}^{\prime }\text{ and }\mathop{\bigcup }\limits_{{m \neq n}}{A}_{m} \smallsetminus {A}_{n} \subseteq {D}_{n}^{\prime } \]\n\nTake\n\n\[ {C}_{n} = {C}_{n}^{\prime } \cap \mathop{\bigcap }\limits_{{m \neq n}}{D}_{m}^{\prime } \]
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Yes
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Proposition 4.11.4 Suppose \( X \) is a Polish space and \( \left( {A}_{n}\right) \) a sequence of analytic subsets of \( X \) . Then there exists a sequence \( \left( {C}_{n}\right) \) of coanalytic subsets of \( X \) such that\n\n\[ \n{A}_{n} \smallsetminus \lim \sup {A}_{m} \subseteq {C}_{n} \n\]\n\n(1)\n\nand\n\n\[ \n\lim \sup {C}_{n} = \varnothing \text{.} \n\]\n\n(2)
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Proof. For each \( n \), set \( {\beta }_{n} = {\beta }_{{A}_{n}} \), where \( {\beta }_{{A}_{n}} \) is as defined in 4.11.2. Let\n\n\[ \n{Q}_{nm} = \left\{ {x \in X : {\beta }_{n}\left( x\right) \leq {\beta }_{m}\left( x\right) }\right\} \n\]\n\n\( {Q}_{nm} \) is analytic by 4.11.2. Take\n\n\[ \n{C}_{n} = {\left\lbrack \mathop{\limsup }\limits_{m}\left\{ {Q}_{nm}\right\} \right\rbrack }^{c}. \n\]\n\nThen \( {C}_{n} \) is coanalytic and\n\n\[ \nx \notin {C}_{n} \Leftrightarrow \exists \eta \subseteq \mathbb{N}\left( {\eta \text{ infinite,}\& x \in \mathop{\bigcap }\limits_{{m \in \eta }}{Q}_{nm}}\right) . \n\]\n\n\( \left( *\right) \)\n\nProof of (1): Let \( x \in {A}_{n} \smallsetminus \lim \sup {A}_{m} \) . Then \( {\beta }_{n}\left( x\right) = {\omega }_{1} \) . Let \( \eta \) be any infinite subset of \( \mathbb{N} \) . Find \( m \in \eta \) such that \( x \notin {A}_{m} \) . Then \( {\beta }_{m}\left( x\right) < {\omega }_{1} = \) \( {\beta }_{n}\left( x\right) \) . So, \( x \notin {Q}_{nm} \) . By \( \left( \star \right) x \in {C}_{n} \) .\n\nProof of (2): Suppose \( \lim \sup {C}_{n} \neq \varnothing \) . Take any \( x \in \lim \sup {C}_{n} \) . Choose an infinite subset \( \eta \) of \( \mathbb{N} \) such that \( x \in {C}_{n} \) for all \( n \in \eta \) . Choose \( {n}_{0} \in \eta \) such that \( {\beta }_{{n}_{0}}\left( x\right) = \min \left\{ {{\beta }_{n}\left( x\right) : n \in \eta }\right\} \) . So, \( x \notin {C}_{{n}_{0}} \) by \( \left( \star \right) \) . This is a contradiction. Hence, \( \lim \sup {C}_{n} = \varnothing \) .
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Yes
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Theorem 4.12.3 (Lusin[71]) If \( X, Y \) are Polish and \( B \) a Borel subset of \( X \times Y \) such that for every \( x \in X \) the section \( {B}_{x} \) is countable, then \( {\pi }_{X}\left( B\right) \) is Borel.
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Proof. Let \( E \subseteq {\mathbb{N}}^{\mathbb{N}} \) be a closed set and \( f : E \rightarrow X \times Y \) a one-to-one continuous map from \( E \) onto \( B \) . Consider \( g = {\pi }_{X} \circ f \) . For every \( x \in {\pi }_{X}\left( B\right) \) , \( {g}^{-1}\left( x\right) \) is a countable closed subset of \( E \) . Hence, by the Baire category theorem, \( {g}^{-1}\left( x\right) \) has an isolated point. Let \( {g}_{s} = g \mid \sum \left( s\right), s \in {\mathbb{N}}^{ < \mathbb{N}} \) . As\n\n\[ \n{\pi }_{X}\left( B\right) = \mathop{\bigcup }\limits_{s}{Z}_{{g}_{s}} \n\]\n\nit is coanalytic by 4.12.1. The result follows from Souslin's theorem.
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Yes
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Theorem 4.12.4 Suppose \( X, Y \) are Polish spaces and \( f : X \rightarrow Y \) is a countable-to-one Borel map. Then \( f\left( B\right) \) is Borel for every Borel set \( B \) in \( X \) .
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Proof. The result follows from 4.12.3 and the identity\n\n\[ f\left( B\right) = {\pi }_{Y}\left( {\operatorname{graph}\left( f\right) \bigcap \left( {B \times Y}\right) }\right) .\n\]
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Yes
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Theorem 4.12.5 (Purves [93]) Let \( X \) be a standard Borel space, \( Y \) Polish, and \( f : X \rightarrow Y \) a bimeasurable map. Then\n\n\[ \left\{ {y \in Y : {f}^{-1}\left( y\right) \text{ is uncountable }}\right\} \] \n\nis countable.
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Proof of 4.12.5. Assume that \( {f}^{-1}\left( y\right) \) is uncountable for uncountably many \( y \) . We shall show that there is a Borel \( B \subseteq X \) such that \( f\left( B\right) \) is not Borel.\n\nCase 1: \( f \) is continuous.\n\nFix a countable base \( \left( {U}_{n}\right) \) for the topology of \( X \) . Let \( G = \operatorname{graph}\left( f\right) \) . For each \( n \), let\n\n\[ {E}_{n} = \left\{ {y \in Y : {U}_{n}\bigcap {G}^{y}\text{ is countable }}\right\} \]\n\nand\n\n\[ A = G \smallsetminus \mathop{\bigcup }\limits_{n}\left( {{U}_{n} \times {E}_{n}}\right) \]\n\nBy 4.3.7, \( {E}_{n} \) is coanalytic. Hence, \( A \) is analytic. Further, \( {\pi }_{Y}\left( A\right) \) is uncountable and \( {A}^{y} \) is perfect for every \( y \in {\pi }_{Y}\left( A\right) \) . By 4.12.6, there is a homeomorph of the Cantor set \( C \) contained in \( {\pi }_{Y}\left( A\right) \) and a one-to-one Borel map \( g : {2}^{\mathbb{N}} \times C \rightarrow A \) such that \( {\pi }_{Y}\left( {g\left( {\alpha, y}\right) }\right) = y \) . Let \( D \) be a Borel subset of \( {2}^{\mathbb{N}} \times C \) such that \( {\pi }_{C}\left( D\right) \) is not Borel and let \( B = {\pi }_{X}\left( {g\left( D\right) }\right) \) . Since \( {\pi }_{X} \circ g \) is one-to-one, \( B \) is Borel by 4.5.4. Since \( f\left( B\right) = {\pi }_{C}\left( D\right) \), the result follows in this case.\n\nThe general case follows from case 1 by replacing \( X \) by \( \operatorname{graph}\left( f\right) \) and \( f \) by \( {\pi }_{Y} \mid \operatorname{graph}\left( f\right) \) .
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Yes
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Lemma 4.12.6 Let \( X \) be a standard Borel space, \( Y \) Polish, and \( A \subseteq X \times \) \( Y \) analytic with \( {\pi }_{X}\left( A\right) \) uncountable. Suppose that for every \( x \in {\pi }_{X}\left( A\right) \) , the section \( {A}_{x} \) is perfect. Then there is a \( C \subseteq {\pi }_{X}\left( A\right) \) homeomorphic to the Cantor set and a one-to-one Borel map \( f : C \times {2}^{\mathbb{N}} \rightarrow A \) such that \( {\pi }_{X}\left( {f\left( {x,\alpha }\right) }\right) = x \) for every \( x \) and every \( \alpha \) .
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Proof of 4.12.6.\n\nFix a compatible complete metric on \( Y \) and a countable base \( \left( {U}_{n}\right) \) for the topology of \( Y \) . For each \( s \in {2}^{ < \mathbb{N}} \), we define a map \( {n}_{s}\left( x\right) : {\pi }_{X}\left( A\right) \rightarrow \mathbb{N} \) satifying the following conditions.\n\n(i) \( x \rightarrow {n}_{s}\left( x\right) \) is \( \sigma \left( {\mathbf{\sum }}_{1}^{1}\right) \) -measurable,\n\n(ii) diameter \( \left( {U}_{{n}_{s}\left( x\right) }\right) < \frac{1}{{2}^{\left| s\right| }} \) ,\n\n(iii) \( {U}_{{n}_{s}\left( x\right) } \cap {A}_{x} \neq \varnothing \) for all \( x \in {\pi }_{X}\left( A\right) \) ,\n\n(iv) \( \operatorname{cl}\left( {U}_{{n}_{s{}^{ \frown }\epsilon }\left( x\right) }\right) \subseteq {U}_{{n}_{s}\left( x\right) },\epsilon = 0 \) or \( 1 \), and\n\n(v)\n\n\[ \operatorname{cl}\left( {U}_{{n}_{{s}^{\prime }0}\left( x\right) }\right) \cap \operatorname{cl}\left( {U}_{{n}_{{s}^{\prime }1}\left( x\right) }\right) = \varnothing . \]\n\nSuch a system of functions is defined by induction on \( \left| s\right| \) . This is a fairly routine exercise, which we leave for the reader. Now fix a continuous probability measure \( P \) on \( X \) such that \( P\left( {{\pi }_{X}\left( A\right) }\right) = 1 \) . Since every set in \( \sigma \left( {\mathbf{\sum }}_{1}^{1}\right) \) is \( P \) -measurable and since \( {\pi }_{X}\left( A\right) \) is uncountable, there is a homeomorph \( C \) of the Cantor set contained in \( {\pi }_{X}\left( A\right) \) such that \( {n}_{s} \mid C \) is Borel measurable for all \( s \in {2}^{ < \mathbb{N}} \) . Take \( x \in C \) and \( \alpha \in {2}^{\mathbb{N}} \) . Note that \( \mathop{\bigcap }\limits_{k}{U}_{{n}_{\alpha \mid k\left( x\right) }} \) is a singleton, say \( \{ y\} \) . Put \( f\left( {x,\alpha }\right) = \left( {x,y}\right) \).
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No
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Lemma 5.1.2 Suppose \( Y \) is metrizable, \( G : X \rightarrow Y \) strongly \( \mathcal{A} \) - measurable, and \( \mathcal{A} \) closed under countable unions. Then \( G \) is \( \mathcal{A} \) -measurable.
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Proof. Let \( U \) be open in \( Y \) . Since \( Y \) is metrizable, \( U \) is an \( {F}_{\sigma } \) set in \( Y \) . Let \( U = \mathop{\bigcup }\limits_{n}{C}_{n},{C}_{n} \) closed. Then\n\n\[ \n{G}^{-1}\left( U\right) = \mathop{\bigcup }\limits_{n}{G}^{-1}\left( {C}_{n}\right) \n\]\n\nSince \( G \) is strongly \( \mathcal{A} \) -measurable and \( \mathcal{A} \) closed under countable unions, \( {G}^{-1}\left( U\right) \in \mathcal{A} \) .
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Yes
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Lemma 5.1.4 Suppose \( \left( {X,\mathcal{A}}\right) \) is a measurable space, \( Y \) a Polish space, and \( G : X \rightarrow Y \) a closed-valued measurable multifunction. Then \( \operatorname{gr}\left( G\right) \in \) \( \mathcal{A} \otimes {\mathcal{B}}_{Y} \)
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Proof. Let \( \left( {U}_{n}\right) \) be a countable base for \( Y \) . Note that\n\n\[ y \notin G\left( x\right) \Leftrightarrow \exists n\left\lbrack {G\left( x\right) \bigcap {U}_{n} = \varnothing \& y \in {U}_{n}}\right\rbrack . \]\n\nTherefore,\n\n\[ \left( {X \times Y}\right) \smallsetminus \operatorname{gr}\left( G\right) = \mathop{\bigcup }\limits_{n}\left\lbrack {{\left( {G}^{-1}\left( {U}_{n}\right) \right) }^{c} \times {U}_{n}}\right\rbrack \]\n\nand the result follows.
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Yes
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Proposition 5.1.9 Suppose \( X \) is a Polish space and \( \mathbf{\Pi } \) a Borel equivalence relation on \( X \) . Then the following statements are equivalent.\n\n(i) \( \Pi \) has a Borel section.\n\n(ii) II admits a Borel cross section.
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Proof. If \( f \) is a Borel section of \( \mathbf{\Pi } \), then the corresponding cross section is clearly Borel. On the other hand, let \( S \) be a Borel cross section of \( \mathbf{\Pi } \) . Let \( f\left( x\right) \) be the unique point of \( S \) equivalent to \( x \) . It is clearly a section of \( \mathbf{\Pi } \) . Note that\n\n\[ y = f\left( x\right) \Leftrightarrow x \coprod y\& y \in S. \]\n\nTherefore, as \( \mathbf{\Pi } \) and \( S \) are Borel, the graph of \( f \) is Borel. Hence, \( f \) is Borel measurable by 4.5.2.
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Yes
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Proposition 5.1.11 Every closed equivalence relation \( \mathbf{\Pi } \) on a Polish space \( X \) is countably separated.
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Proof. Take any countable base \( \left( {U}_{n}\right) \) for the topology of \( X \) . For every \( x, y \) in \( X \) such that \( \left( {x, y}\right) \notin \mathbf{\Pi } \), there exist basic open sets \( {U}_{n} \) and \( {U}_{m} \) containing \( x \) and \( y \) respectively with \( {U}_{n} \times {U}_{m} \subseteq \left( {X \times Y}\right) \smallsetminus \Pi \) . In particular, \( {U}_{n}^{ * } \) is disjoint from \( {U}_{m} \) . Since \( {U}_{n}^{ * } \) is the projection onto the first coordinate axis of \( {\pi }_{X}\left( {\mathbf{\Pi } \cap \left( {X \times {U}_{n}}\right) }\right) \), which is Borel, \( {U}_{n}^{ * } \) is analytic. Thus \( {U}_{n}^{ * } \) is an invariant analytic set disjoint from \( {U}_{m} \) . Hence, by 4.4.5, there exists an invariant Borel set \( {B}_{n} \) containing \( {U}_{n}^{ * } \) and disjoint from \( {U}_{m} \) . It is now fairly easy to see that\n\n\[ \nX \times Y \smallsetminus \mathbf{\Pi } = \mathop{\bigcup }\limits_{n}\left( {{B}_{n} \times {B}_{n}^{c}}\right) \n\]\n\nThe result follows from 5.1.10.
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Yes
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Proposition 5.1.12 Every Borel measurable partition of a Polish space into \( {G}_{\delta } \) sets is countably separated.
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Proof. Let \( X \) be a Polish space and \( \mathbf{\Pi } \) a Borel measurable partition of \( X \) into \( {G}_{\delta } \) sets. Take \( Y = F\left( X\right) \), the Effros Borel space of \( X \) . Then \( Y \) is standard Borel (3.3.10). For \( x \in X \), let \( \left\lbrack x\right\rbrack \) be the equivalence class containing \( x \) and \( p\left( x\right) = \operatorname{cl}\left( \left\lbrack x\right\rbrack \right) \) . For any open \( U \subseteq X \) ,\n\n\[ \n\{ x \in X : p\left( x\right) \bigcap U \neq \varnothing \} = {U}^{ * },\n\]\n\nwhich is Borel, since \( \mathbf{\Pi } \) is measurable. Therefore, \( p : X \rightarrow Y \) is Borel measurable (5.1.1). We now show that:\n\n\[ \nx \equiv y \Leftrightarrow p\left( x\right) = p\left( y\right) .\n\]\n\n\( \left( *\right) \)\n\nClearly, \( x \equiv y \Rightarrow p\left( x\right) = p\left( y\right) \) . Suppose \( x ≢ y \) but \( p\left( x\right) = p\left( y\right) \) . Then \( \left\lbrack x\right\rbrack \) and \( \left\lbrack y\right\rbrack \) are two disjoint dense \( {G}_{\delta } \) sets in \( p\left( x\right) \) . This contradicts the Baire category theorem, and we have proved \( \left( \star \right) \) . Thus, \( p : X \rightarrow Y \) witnesses the fact that \( \mathbf{\Pi } \) is countably separated.
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Yes
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Lemma 5.1.16 Let \( \Pi \) be a Borel partition of a Polish space \( X \) . The following statements are equivalent.\n\n(i) \( \Pi \) is countably separated.\n\n(ii) The \( \sigma \) -algebra \( {\mathcal{B}}^{ * } \) of \( \mathbf{\Pi } \) -invariant Borel sets is countably generated.
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Proof. (i) implies (ii): Let \( \mathbf{\Pi } \) be countably separated. Take a Polish space \( Y \) and \( f : X \rightarrow Y \) a Borel map such that\n\n\[ x \coprod {x}^{\prime } \Leftrightarrow f\left( x\right) = f\left( {x}^{\prime }\right) .\n\]\n\nWe show that \( {\mathcal{B}}^{ * } = {f}^{-1}\left( {\mathcal{B}}_{Y}\right) \), which will then show that \( \mathbf{\Pi } \) satisfies (ii). Clearly, \( {\mathcal{B}}^{ * } \supseteq {f}^{-1}\left( {\mathcal{B}}_{Y}\right) \) . To prove the reverse inclusion, let \( B \subseteq X \) be an invariant Borel set. Then \( f\left( B\right) \) and \( f\left( {B}^{c}\right) \) are disjoint analytic subsets of \( Y \) . By the first separation principle for analytic sets (4.4.1), there is a Borel set \( C \) such that\n\n\[ f\left( B\right) \subseteq C\text{ and }C\bigcap f\left( {B}^{c}\right) = \varnothing .\n\]\n\nTherefore, \( B = {f}^{-1}\left( C\right) \in {f}^{-1}\left( {\mathcal{B}}_{Y}\right) \) . Hence,(i) implies (ii).\n\n(ii) implies (i): Let \( {\mathcal{B}}^{ * } \) be countably generated. Take any countable generator \( \left( {A}_{n}\right) \) of \( {\mathcal{B}}^{ * } \) . Note that the atoms of \( {\mathcal{B}}^{ * } \) are precisely the \( \mathbf{\Pi } \) - equivalence classes. Therefore, for any \( x,{x}^{\prime } \) in \( X \) ,\n\n\[ x \coprod {x}^{\prime } \Leftrightarrow \forall n\left( {x \in {A}_{n} \Leftrightarrow {x}^{\prime } \in {A}_{n}}\right) .\n\]\n\nFrom this and 5.1.10, it follows that (ii) implies (i).
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Yes
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Theorem 5.2.1 (Kuratowski and Ryll-Nardzewski [63]) Every \( {\mathcal{L}}_{\sigma } \) - measurable, closed-valued multifunction \( F : X \rightarrow Y \) admits an \( {\mathcal{L}}_{\sigma } \) - measurable selection.
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Proof of 5.2.1. Inductively we define a sequence \( \left( {s}_{n}\right) \) of \( {\mathcal{L}}_{\sigma } \) -measurable maps from \( X \) to \( Y \) such that for every \( x \in X \) and every \( n \in \mathbb{N} \) ,\n\n(i) \( d\left( {{s}_{n}\left( x\right), F\left( x\right) }\right) < {2}^{-n} \), and\n\n(ii) \( d\left( {{s}_{n}\left( x\right) ,{s}_{n + 1}\left( x\right) }\right) < {2}^{-n} \) .\n\nTo define \( \left( {s}_{n}\right) \) we take a countable dense set \( \left( {r}_{n}\right) \) in \( Y \) . Define \( {s}_{0} \equiv {r}_{0} \) . Let \( n > 0 \) . Suppose that for every \( m < n,{s}_{m} \) satisfying conditions (i) and (ii) have been defined. Let\n\n\[ {E}_{k} = \left\{ {x \in X : d\left( {{s}_{n - 1}\left( x\right) ,{r}_{k}}\right) < {2}^{-n + 1}\& d\left( {{r}_{k}, F\left( x\right) }\right) < {2}^{-n}}\right\} .\n\]\n\nSo,\n\n\[ {E}_{k} = {s}_{n - 1}^{-1}\left( {B\left( {{r}_{k},{2}^{-n + 1}}\right) }\right) \bigcap {F}^{-1}\left( {B\left( {{r}_{k},{2}^{-n}}\right) }\right) ,\n\]\n\nwhere \( B\left( {y, r}\right) \) denotes the open ball in \( Y \) with center \( y \) and radius \( r \) . It follows that \( {E}_{k} \in {\mathcal{L}}_{\sigma } \) .\n\nFurther, \( \mathop{\bigcup }\limits_{k}{E}_{k} = X \) . To see this, take any \( x \in X \) . As \( d\left( {{s}_{n - 1}\left( x\right), F\left( x\right) }\right) < \) \( {2}^{-n + 1} \), there is a \( y \) in \( F\left( X\right) \) such that \( d\left( {y,{s}_{n - 1}\left( x\right) }\right) < {2}^{-n + 1} \) . Since \( \left( {r}_{k}\right) \) is dense, there exists an \( l \) such that \( d\left( {{r}_{l}, y}\right) < {2}^{-n} \) and \( d\left( {{r}_{l},{s}_{n - 1}\left( x\right) }\right) < {2}^{-n + 1} \) . Then \( x \in {E}_{l} \) .\n\nBy 5.2.2, there exist pairwise disjoint sets \( {D}_{k} \subseteq {E}_{k} \) in \( {\mathcal{L}}_{\sigma } \).
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Yes
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Lemma 5.2.2 Suppose \( {A}_{n} \in {\mathcal{L}}_{\sigma } \). Then there exist \( {B}_{n} \subseteq {A}_{n} \) such that the \( {B}_{n} \)’s are pairwise disjoint elements of \( {\mathcal{L}}_{\sigma } \) and \( \mathop{\bigcup }\limits_{n}{A}_{n} = \mathop{\bigcup }\limits_{n}{B}_{n} \).
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Proof. Write\n\n\[ \n{A}_{n} = \mathop{\bigcup }\limits_{m}{C}_{nm} \n\] \n\n\( {C}_{nm} \in \mathcal{L} \). Enumerate \( \left\{ {{C}_{nm} : n, m \in \mathbb{N}}\right\} \) in a single sequence, say \( \left( {D}_{i}\right) \). Set\n\n\[ \n{E}_{i} = {D}_{i} \smallsetminus \mathop{\bigcup }\limits_{{j < i}}{D}_{j} \n\] \n\n\nClearly, \( {E}_{i} \in \mathcal{L} \). Take\n\n\[ \n{B}_{n} = \bigcup \left\{ {{E}_{i} : {E}_{i} \subseteq {A}_{n}\& \left( {\forall m < n}\right) \left( {{E}_{i} \nsubseteq {A}_{m}}\right) }\right\} .\n\]
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Yes
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Lemma 5.2.3 Suppose \( {f}_{n} : X \rightarrow Y \) is a sequence of \( {\mathcal{L}}_{\sigma } \) -measurable functions converging uniformly to \( f : X \rightarrow Y \) . Then \( f \) is \( {\mathcal{L}}_{\sigma } \) -measurable.
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Proof. Replacing \( \left( {f}_{n}\right) \) by a subsequence if necessary, we assume that\n\n\[ \forall x\forall n\left( {d\left( {f\left( x\right) ,{f}_{n}\left( x\right) }\right) < 1/\left( {n + 1}\right) }\right) .\n\]\n\nLet \( F \) be a closed set in \( Y \) and\n\n\[ {F}_{n} = \operatorname{cl}\left( {\{ y \in Y : d\left( {y, F}\right) < 1/\left( {n + 1}\right) \} }\right) .\n\]\n\nThen\n\n\[ f\left( x\right) \in F \Leftrightarrow \forall n{f}_{n}\left( x\right) \in {F}_{n} \]\n\ni.e., \( {f}^{-1}\left( F\right) = \mathop{\bigcap }\limits_{n}{f}_{n}^{-1}\left( {F}_{n}\right) \in {\mathcal{L}}_{\delta } \), and our result is proved.
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Yes
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Corollary 5.2.4 Let \( X \) be a Polish space and \( F\left( X\right) \) the space of nonempty closed subsets of \( X \) with Effros Borel structure. Then there is a measurable \( s : F\left( X\right) \rightarrow X \) such that \( s\left( F\right) \in F \) for all \( F \in F\left( X\right) \) .
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Proof. Apply 5.2.1 to the multifunction \( G : F\left( X\right) \rightarrow X \), where \( G\left( F\right) = \) \( F \), with \( \mathcal{L} \) the Effros Borel \( \sigma \) -algebra on \( F\left( X\right) \) .
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Yes
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Corollary 5.2.5 Let \( \\left( {T,\\mathcal{T}}\\right) \) be a measurable space and \( Y \) a separable metric space. Then every \( \\mathcal{T} \) -measurable, compact-valued multifunction \( F \) : \( T \\rightarrow Y \) admits a \( \\mathcal{T} \) -measurable selection.
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Proof. Let \( X \) be the completion of \( Y \) . Then \( F \) as a multifunction from \( T \) to \( X \) is closed-valued and \( \\mathcal{T} \) -measurable. Apply 5.2.1 now.
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No
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Corollary 5.2.6 Suppose \( Y \) is a compact metric space, \( X \) a metric space, and \( f : Y \rightarrow X \) a continuous onto map. Then there is a Borel map \( s : X \rightarrow Y \) of class 2 such that \( f \circ s \) is the identity map on \( X \) .
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Proof. Let \( F\left( x\right) = {f}^{-1}\left( x\right), x \in X \), and \( \mathcal{L} = {\mathbf{\Delta }}_{2}^{0} \) . For any closed set \( C \) in \( Y \) ,\n\n\[ \n{F}^{-1}\left( C\right) = {\pi }_{X}\left( {\operatorname{graph}\left( f\right) \bigcap \left( {X \times C}\right) }\right) .\n\]\n\nTherefore, by \( {2.3.24},{F}^{-1}\left( C\right) \) is closed. Hence, \( F \) is \( {\mathcal{L}}_{\sigma } \) -measurable. Now apply 5.2.1.
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Yes
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Proposition 5.2.7 (A. Maitra and B. V. Rao[77]) Let \( T \) be a nonempty set, \( \mathcal{L} \) an algebra on \( T \), and \( X \) a Polish space. Suppose \( F : T \rightarrow X \) is a closed-valued \( {\mathcal{L}}_{\sigma } \) -measurable multifunction. Then there is a sequence \( \left( {f}_{n}\right) \) of \( {\mathcal{L}}_{\sigma } \) -measurable selections of \( F \) such that\n\n\[ F\left( t\right) = \operatorname{cl}\left( \left\{ {{f}_{n}\left( t\right) : n \in \mathbb{N}}\right\} \right) ,\;t \in T. \]
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Proof. Fix a countable base \( \left\{ {{U}_{n} : n \in \mathbb{N}}\right\} \) for the topology of \( X \) and fix also an \( {\mathcal{L}}_{\sigma } \) -measurable selection \( f \) for \( F \) . For each \( n,{T}_{n} = {F}^{-1}\left( {U}_{n}\right) \in {\mathcal{L}}_{\sigma } \) . Write \( {T}_{n} = \mathop{\bigcup }\limits_{m}{T}_{nm},{T}_{nm} \in \mathcal{L} \) . Define \( {F}_{nm} : {T}_{nm} \rightarrow X \) by\n\n\[ {F}_{nm}\left( t\right) = \operatorname{cl}\left( {F\left( t\right) \bigcap {U}_{n}}\right) ,\;t \in {T}_{nm}. \]\n\nBy 5.2.1, there is an \( {\mathcal{L}}_{\sigma } \mid {T}_{nm} \) -measurable selection \( {h}_{nm} \) for \( {F}_{nm} \) . Define\n\n\[ {f}_{nm}\left( t\right) = \left\{ \begin{array}{ll} {h}_{nm}\left( t\right) & \text{ if }t \in {T}_{nm}, \\ f\left( t\right) & \text{ otherwise. } \end{array}\right. \]\n\nThen each \( {f}_{nm} \) is \( {\mathcal{L}}_{\sigma } \) -measurable. Further,\n\n\[ F\left( t\right) = \operatorname{cl}\left\{ {{f}_{nm}\left( t\right) : n, m \in \mathbb{N}}\right\} ,\;t \in T. \]
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Yes
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Theorem 5.2.8 (Srivastava[115]) Let \( T,\mathcal{L}, X \), and \( F \) be as in 5.2.7. Then there is a map \( f : T \times {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) such that\n\n(i) for every \( \alpha \in {\mathbb{N}}^{\mathbb{N}}, t \rightarrow f\left( {t,\alpha }\right) \) is \( {\mathcal{L}}_{\sigma } \) -measurable, and\n\n(ii) for every \( t \in T, f\left( {t, \cdot }\right) \) is a continuous map from \( {\mathbb{N}}^{\mathbb{N}} \) onto \( F\left( t\right) \) .
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Proof of 5.2.8 Fix a complete compatible metric \( d \) on \( X \) . Applying 5.2.9 and 5.2.7 repeatedly, for each \( s \in {\mathbb{N}}^{ < \mathbb{N}} \), we get an \( {\mathcal{L}}_{\sigma } \) -measurable selection \( {f}_{s} : T \rightarrow X \) for \( F \) satisfying the following condition: For every \( s \in {\mathbb{N}}^{ < \mathbb{N}} \) and every \( t \in T,\left\{ {{f}_{{s}^{ \frown }n}\left( t\right) : n \in \mathbb{N}}\right\} \) is dense in \( F\left( t\right) \bigcap B\left( {{f}_{s}\left( t\right) ,1/{2}^{-\left| s\right| }}\right) \) ). Note that for every \( \alpha \in {\mathbb{N}}^{\mathbb{N}} \) and every \( t \in T \), the sequence \( \left( {{f}_{\alpha \mid n}\left( t\right) }\right) \) is Cauchy and hence convergent. Take \( f : T \times {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) defined by\n\n\[ f\left( {t,\alpha }\right) = \mathop{\lim }\limits_{n}{f}_{\alpha \mid n}\left( t\right) \]
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Yes
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Theorem 5.2.10 (S. Bhattacharya and S. M. Srivastava [12]) Let \( F \) : \( X \rightarrow Y \) be closed-valued and strongly \( {\mathcal{L}}_{\sigma } \) -measurable. Suppose \( Z \) is a separable metric space and \( g : Y \rightarrow Z \) a Borel map of class 2 . Then there is an \( {\mathcal{L}}_{\sigma } \) -measurable selection \( f \) of \( F \) such that \( g \circ f \) is \( {\mathcal{L}}_{\sigma } \) -measurable.
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Proof. Let \( \left( {U}_{n}\right) \) be a countable base for the topology of \( Z \) . Write \( {g}^{-1}\left( {U}_{n}\right) = \mathop{\bigcup }\limits_{m}{H}_{nm} \), the \( {H}_{nm} \) ’s closed. Also, take a countable base \( \left( {W}_{n}\right) \) for \( Y \) and write \( {W}_{n} = \mathop{\bigcup }\limits_{m}{C}_{nm} \), the \( {C}_{nm} \) ’s closed. Let \( \mathcal{B} \) be the smallest family of subsets of \( Y \) closed under finite intersections, containing each \( {H}_{nm} \) and each \( {C}_{nm} \) . Let \( {\mathcal{T}}^{\prime } \) be the topology on \( Y \) with \( \mathcal{B} \) a base. Note that \( {\mathcal{T}}^{\prime } \) is finer than \( \mathcal{T} \), the original topology of \( Y \) . By Observations 1 and 2 of Section 2, Chapter 3, we see that \( {\mathcal{T}}^{\prime } \) is a Polish topology on \( Y \) . Note that with \( Y \) equipped with the topology \( {\mathcal{T}}^{\prime }, g \) is continuous and \( F{\mathcal{L}}_{\sigma } \) - measurable. By 5.2.1, there is an \( {\mathcal{L}}_{\sigma } \) -measurable selection \( f \) of \( F \) . This \( f \) works.
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Yes
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Theorem 5.2.11 Let \( X, Y \) be compact metric spaces, \( f : X \rightarrow Y \) a continuous onto map. Suppose \( A \subseteq Y \) and \( 1 \leq \alpha < {\omega }_{1} \). Then\n\n\[ \n{f}^{-1}\left( A\right) \in {\mathbf{\Pi }}_{\alpha }^{0}\left( X\right) \Leftrightarrow A \in {\mathbf{\Pi }}_{\alpha }^{0}\left( Y\right)\n\]
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Proof of 5.2.11 We need to prove the \
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No
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Lemma 5.2.12 Let \( X, Y \), and \( f \) be as in 5.2.11. Suppose \( 1 \leq \alpha < {\omega }_{1}, Z \) is a separable metric space, and \( g : X \rightarrow Z \) is a Borel map of class \( \alpha \) . Then there is a class 2 map \( s : Y \rightarrow X \) such that \( g \circ s \) is of class \( \alpha \) and \( f\left( {s\left( y\right) }\right) = y \) for all \( y \) .
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Proof. Let \( F\left( y\right) = {f}^{-1}\left( y\right), y \in Y \) . Then \( F : Y \rightarrow X \) is an upper-semicontinuous closed-valued multifunction. By 5.2.1 there is a selection \( s \) of \( F \) that is Borel of class 2 . This \( s \) works if either \( \alpha = 1 \) (i.e., if \( g \) is continuous) or if \( \alpha \geq {\omega }_{0} \) (in this case \( g \circ s \) is of class \( 1 + \alpha = \alpha \) ). So, we need to prove the result for \( 2 \leq \alpha < {\omega }_{0} \) only. We prove this by induction on \( \alpha \) .\n\nFor \( \alpha = 2 \) we get this by 5.2.10. Let \( n \geq 2 \), and the result is true for \( \alpha = n \) . Let \( g : X \rightarrow Z \) be of class \( n + 1 \) . By 3.6.15, there is a sequence \( \left( {g}_{n}\right) \) of Borel measurable functions from \( X \) to \( Z \) of class \( n \) converging pointwise to \( g \) . By 3.6.5, \( h = \left( {g}_{n}\right) : X \rightarrow {Z}^{\mathbb{N}} \) is of class \( n \) . So, by the induction hypothesis, there is a selection \( s \) of \( F \) of class 2 such that \( h \circ s \) is of class \( n \) . In particular, each \( {g}_{n} \circ s \) is of class \( n \) . As \( {g}_{n} \circ s \rightarrow g \circ s \) pointwise, \( g \circ s \) is of class \( \left( {n + 1}\right) \) by 3.6.5.
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Yes
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Theorem 5.3.1 (Schäl) Suppose \( \\left( {T,\\mathcal{T}}\\right) \) is a measurable space and let \( Y \) be a separable metric space. Suppose \( G : T \\rightarrow Y \) is a \( \\mathcal{T} \) -measurable compact-valued multifunction. Let \( v \) be a real-valued function on \( \\operatorname{gr}\\left( G\\right) \) , the graph of \( G \), that is the pointwise limit of a nonincreasing sequence \( \\left( {v}_{n}\\right) \) of \( \\mathcal{T} \\otimes {\\mathcal{B}}_{Y} \\mid \\operatorname{gr}\\left( G\\right) \) -measurable functions on \( \\operatorname{gr}\\left( G\\right) \) such that for each \( n \) and each \( t \\in T,{v}_{n}\\left( {t,\\text{.}}\\right) {is}\;{continuous}\;{on}\;G\\left( t\\right) .{Let} \n\n\[ \n{v}^{ * }\\left( t\\right) = \\sup \\{ v\\left( {t, y}\\right) : y \\in G\\left( t\\right) \\} ,\\;t \\in T. \n\] \n\nThen there is a \( \\mathcal{T} \) -measurable selection \( g : T \\rightarrow Y \) for \( G \) such that \n\n\[ \n{v}^{ * }\\left( t\\right) = v\\left( {t, g\\left( t\\right) }\\right) \n\] \n\nfor every \( t \\in T \) .
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Proof of 5.3.1. (Burgess and Maitra[24]) Without any loss of generality we assume that \( Y \) is Polish. Fix a complete metric \( d \) on \( Y \) compatible with its topology. By 5.2.7, we get \( \\mathcal{T} \) -measurable selections \( {g}_{n} : T \\rightarrow Y \) of \( G \) such that \n\n\[ \nG\\left( t\\right) = \\operatorname{cl}\\left( \\left\\{ {{g}_{n}\\left( t\\right) : n \\in \\mathbb{N}}\\right\\} \\right) ,\\;t \\in T. \n\] \n\nThen \( {v}^{ * }\\left( t\\right) = \\sup \\left\\{ {v\\left( {t,{g}_{n}\\left( t\\right) }\\right) : n \\in \\mathbb{N}}\\right\\} \) . Hence, \( {v}^{ * } \) is \( \\mathcal{T} \) -measurable.\n\nWe first prove the result when \( v \) is \( \\mathcal{T} \\otimes {\\mathcal{B}}_{Y} \\mid \\operatorname{gr}\\left( G\\right) \) -measurable with \( v\\left( {t,\\text{.}}\\right) \) continuous for all \( t \) . Set \n\n\[ \nH\\left( t\\right) = \\left\\{ {y \\in G\\left( t\\right) : v\\left( {t, y}\\right) = {v}^{ * }\\left( t\\right) }\\right\\} ,\\;t \\in T. \n\] \n\nClearly, \( H \) is a compact-valued multifunction. Let \( C \) be any closed set in \( Y \) and let \n\n\[ \n{C}_{n} = \\{ y \\in Y : d\\left( {y, C}\\right) < 1/n\\} ,\\;n \\geq 1. \n\] \n\nWe easily check that \n\n\[ \n\\{ t : H\\left( t\\right) \\cap C \\neq \\varnothing \\} = \\mathop{\\bigcap }\\limits_{n}\\mathop{\\bigcup }\\limits_{i}\\left\\{ {t : v\\left( {t,{g}_{i}\\left( t\\right) }\\right) > {v}^{ * }\\left( t\\right) - 1/n\\text{ and }{g}_{i}\\left( t\\right) \\in {C}_{n}}\\right\\} . \n\] \n\nIt follows that \( H \) is \( \\mathcal{T} \) -measurable. To complete the proof in the special case, apply the Kuratowski and Ryll-Nardzewski selection theorem (5.2.1) and take any \( \\mathcal{T} \) -measurable selection \( g \) for \( H \) .\n\nWe now turn to the general case. By the above case, for each \( n \) there is a \( \\mathcal{T} \) -measurable selection \( {g}_{n} : T \\rightarrow Y \) of \( G \) such that \n\n\[ \n{v}_{n}\\left( {t,{g}_{n}\\left( t\\right) }\\right) = \\sup \\left\\{ {{v}_{n}\\left( {t, y}\\right) : y \\in G\\left( t\\right) }\\right\\} \n\] \n\nfor every \( t \\in T \) . For \( t \\in T \), set \n\n\( H\\left( t\\right) = \\left\\{ {y \\in G\\left( t\\right) : \\text{ there is a subsequence }\\left( {{g}_{{n}_{i}}\\left( t\\right) }\\right) \\text{ such that }{g}_{{n}_{i}}\\left( t\\right) \\rightarrow y}\\right\\} . \n\nSince \( G\\left( t\\right) \) is nonempty and compact, so is \( H\\left( t\\right) \) . We now show that \( H \) is \( \\mathcal{T} \) -measurable. Let \( C \) be closed in \( Y \) . Then \n\n\[ \n\\{ t \\in T : H\\left( t\\right) \\bigcap C \\neq \\varnothing \\} = \\mathop{\\bigcap }\\limits_{{k \\geq 1}}\\mathop{\\bigcup }\\limits_{{m > k}}\\left\\{ {t \\in T : d\\left( {{g}_{m}\\left( t\\right), C}\\right) < 1/k}\\right\\} . \n\] \n\nIt follows that \( H \) is \( \\mathcal{T} \) -measurable. By the Kur
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Yes
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It is not unreasonable to conjecture that 5.3.1 remains true even for \( v \) that are \( \mathcal{T}\bigotimes {\mathcal{B}}_{Y} \mid {gr}\left( G\right) \) -measurable such that \( v\left( {t,\text{.}}\right) {isupper} \) semicontinuous for every \( t \) . However, this is not true.
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Recall that in the last chapter, using Solovay's coding of Borel sets, we showed that there is a coanalytic set \( T \) and a function \( g : T \rightarrow {2}^{\mathbb{N}} \) whose graph is relatively Borel in \( T \times {2}^{\mathbb{N}} \) but that is not Borel measurable. Take \( \mathcal{T} = {\mathcal{B}}_{T}, G\left( t\right) = {2}^{\mathbb{N}} \) \( \left( {t \in T}\right) \), and \( v : T \times {2}^{\mathbb{N}} \rightarrow \mathbb{R} \) the characteristic function of \( \operatorname{graph}\left( g\right) \) .
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Yes
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Theorem 5.4.1 (Effros [40]) Every lower-semicontinuous or upper-semicontinuous partition \( \mathbf{\Pi } \) of a Polish space \( X \) into closed sets admits a Borel measurable section \( f : X \rightarrow X \) of class 2. In particular, they admit a \( {G}_{\delta } \) cross section.
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Proof. In 5.2.1, take \( Y = X,\mathcal{L} \) the family of invariant sets that are simultaneously \( {F}_{\sigma } \) and \( {G}_{\delta } \), and \( F\left( x\right) = \left\lbrack x\right\rbrack \), the equivalence class containing \( x \) . So, there is an \( {\mathcal{L}}_{\sigma } \) -measurable selection \( f : X \rightarrow X \) of \( F \) . This means that \( f \) is a Borel measurable section of \( \mathbf{\Pi } \) of class 2 . The corresponding cross section \( S = \{ x \in X : x = f\left( x\right) \} \) is a \( {G}_{\delta } \) cross section of \( \Pi \) .
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Yes
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Theorem 5.4.2 (Effros - Mackey cross section theorem) Suppose \( H \) is a closed subgroup of a Polish group \( G \) and \( \mathbf{\Pi } \) the partition of \( G \) consisting of all the right cosets of \( H \) . Then \( \mathbf{\Pi } \) admits a Borel measurable section of class 2. In particular, it admits a \( {G}_{\delta } \) cross section.
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Proof. Note that for any open set \( U \) in \( G \) ,\n\n\[ \n{U}^{ * } = \bigcup \{ g \cdot U : g \in H\} . \n\]\n\nSo, \( {U}^{ * } \) is open. Thus \( \mathbf{\Pi } \) is lower semicontinuous. The result follows from Effros's cross section theorem (5.4.1).
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No
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Theorem 5.4.3 Every Borel measurable partition \( \mathbf{\Pi } \) of a Polish space \( X \) into closed sets admits a Borel measurable section \( f : X \rightarrow X \) . In particular, it admits a Borel cross section.
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Proof. Let \( \mathcal{A} \) be the \( \sigma \) -algebra of all invariant Borel subsets of \( X \) and \( F : X \rightarrow X \) the multifunction that assigns to each \( x \in X \) the member of \( \mathbf{\Pi } \) containing \( x \) . By our assumptions, \( F \) is \( \mathcal{A} \) -measurable. By 5.2.1, we get a measurable selection \( f \) for \( F \) . Note that \( f \) is a section of \( \mathbf{\Pi } \) . The corresponding cross section \( S = \{ x \in X : x = f\left( x\right) \} \) of \( \mathbf{\Pi } \) is clearly a Borel cross section of \( \Pi \) .
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Yes
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Theorem 5.4.4 The classification space \( \operatorname{irr}\left( n\right) / \sim \) is standard Borel.
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Proof. Fix any irreducible \( A \) . Then the \( \sim \) -equivalence class \( \left\lbrack A\right\rbrack \) containing \( A \) equals\n\n\[ \n{\pi }_{1}\left\{ {\left( {B, U}\right) \in \operatorname{irr}\left( n\right) \times U\left( n\right) : A = {UB}{U}^{ * }}\right\} , \n\]\n\nwhere \( {\pi }_{1} : \operatorname{irr}\left( n\right) \times U\left( n\right) \rightarrow \operatorname{irr}\left( n\right) \) is the projection map to the first coordinate space. (Recall that \( U\left( n\right) \) denotes the set of all \( n \times n \) unitary matrices.) As the set\n\n\[ \n\left\{ {\left( {B, U}\right) \in \operatorname{irr}\left( n\right) \times U\left( n\right) : A = {UB}{U}^{ * }}\right\} \n\]\n\nis closed and \( U\left( n\right) \) compact, \( \left\lbrack A\right\rbrack \) is closed by 2.3.24.\n\nNow let \( \mathcal{O} \) be any open set in \( \operatorname{irr}\left( n\right) \) . Its saturation is\n\n\[ \n\mathop{\bigcup }\limits_{{U \in U\left( n\right) }}\left\{ {A \in \operatorname{irr}\left( n\right) : {UA}{U}^{ * } \in \mathcal{O}}\right\} \n\]\n\nwhich is open. Thus \( \sim \) is a lower-semicontinuous partition of \( \operatorname{irr}\left( n\right) \) into closed sets. By 5.4.3, let \( C \) be a Borel cross section of \( \sim \) . Then \( q \mid C \) is a one-to-one Borel map from \( C \) onto \( \operatorname{irr}\left( n\right) / \sim \), where \( q : \operatorname{irr}\left( n\right) \rightarrow \operatorname{irr}\left( n\right) / \sim \) is the canonical quotient map. By the Borel isomorphism theorem (3.3.13), \( q \) is a Borel isomorphism, and our result is proved.
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Yes
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Theorem 5.4.5 (Miller[84]) Let \( \\left( {G, \\cdot }\\right) \) be a Polish group, \( X \) a Polish space, and \( a\\left( {g, x}\\right) = g \\cdot x \) an action of \( G \) on \( X \) . Suppose for a given \( x \\in X \) that \( g \\rightarrow g \\cdot x \) is Borel. Then the orbit\n\n\\[ \n\\{ g \\cdot x : g \\in G\\} \n\\]\n\nof \( x \) is Borel.
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Proof. Let \( H = {G}_{x} \) be the stabilizer of \( x \) . By 4.8.4, \( H \) is closed in \( G \) . Let \( S \) be a Borel cross section of the partition \( \\mathbf{\\Pi } \) consisting of the left cosets of \( H \) . The map \( g \\rightarrow g \\cdot x \) restricted to \( S \) is one-to-one, Borel, and onto the orbit of \( x \) . The result follows from 4.5.4.
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Yes
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Proposition 5.5.1 Let \( X, Y \) be Polish spaces, \( B \subseteq X \times Y \) Borel, and \( C \) an analytic uniformization of \( B \) . Then \( C \) is Borel.
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Proof. We show that \( C \) is also coanalytic. The result will then follow from Souslin’s theorem. That \( C \) is coanalytic follows from the following relation:\n\n\[ \left( {x, y}\right) \in C \Leftrightarrow \left( {x, y}\right) \in B\& \forall z\left( {\left( {x, z}\right) \in C \Rightarrow y = z}\right) . \]
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No
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Theorem 5.5.2 (Von Neumann[124]) Let \( X \) and \( Y \) be Polish spaces, \( A \subseteq \) \( X \times Y \) analytic, and \( \mathcal{A} = \sigma \left( {{\mathbf{\sum }}_{1}^{1}\left( X\right) }\right) \), the \( \sigma \) -algebra generated by the analytic subsets of \( X \) . Then there is an \( \mathcal{A} \) -measurable section \( u : {\pi }_{X}\left( A\right) \rightarrow Y \) of A.
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Proof. Let \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow A \) be a continuous surjection. Consider\n\n\[ B = \left\{ {\left( {x,\alpha }\right) \in X \times {\mathbb{N}}^{\mathbb{N}} : {\pi }_{X}\left( {f\left( \alpha \right) }\right) = x}\right\} . \]\n\nThen \( B \) is a closed set with \( {\pi }_{X}\left( B\right) = {\pi }_{X}\left( A\right) \) . For \( x \in {\pi }_{X}\left( A\right) \), define \( g\left( x\right) \) to be the lexicographic minimum of \( {B}_{x} \) ; i.e.,\n\n\[ g\left( x\right) = \alpha \Leftrightarrow \left( {x,\alpha }\right) \in B \]\n\n\[ \& \forall \beta \{ \left( {x,\beta }\right) \in B \Rightarrow \]\n\n\[ \exists n\left\lbrack {\alpha \left( n\right) < \beta \left( n\right) \text{ and }\forall m < n\left( {\alpha \left( m\right) = \beta \left( m\right) }\right) }\right\rbrack \} . \]\n\nBy induction on \( \left| s\right| \), we prove that \( {g}^{-1}\left( {\sum \left( s\right) }\right) \in \mathcal{A} \) for every \( s \in {\mathbb{N}}^{ < \mathbb{N}} \) . Since \( \left\{ {\sum \left( s\right) : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) is a base for \( {\mathbb{N}}^{\mathbb{N}} \), it follows that \( g \) is \( \mathcal{A} \) -measurable. Suppose \( {g}^{-1}\left( {\sum \left( t\right) }\right) \in \mathcal{A} \) and \( s = t \hat{} k, k \in \mathbb{N} \) . Then for any \( x \),\n\n\[ x \in {g}^{-1}\left( {\sum \left( s\right) }\right) \Leftrightarrow x \in {g}^{-1}\left( {\sum \left( t\right) }\right) \]\n\n\[ \& \exists \alpha \left( {\left( {x,\alpha }\right) \in B\& s \prec \alpha }\right) \]\n\n\[ \text{&}\forall l < k\neg \exists \beta \left( {\left( {x,\beta }\right) \in B\& t \hat{} l \prec \beta }\right) \text{.} \]\n\nHence, \( {g}^{-1}\left( {\sum \left( s\right) }\right) \in \mathcal{A} \) . Now, define \( u\left( x\right) = {\pi }_{Y}\left( {f\left( {g\left( x\right) }\right) }\right), x \in {\pi }_{X}\left( A\right) \) . Then \( u \) is an \( \mathcal{A} \) -measurable section of \( A \) .
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Yes
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Theorem 5.5.3 Every analytic subset \( A \) of the product of Polish spaces \( X, Y \) admits a section \( u \) that is universally measurable as well as Baire measurable.
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Proof. The result follows from 5.5.2, 4.3.1, and 4.3.2.
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No
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Proposition 5.5.4 In 5.5.3, further assume that \( A \) is Borel. Then the graph of the section \( u \) is coanalytic.
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Proof. Note that\n\n\[ \begin{matrix} u\left( x\right) = y & \Leftrightarrow & \left( {x, y}\right) \in A\;\& \;\left( {\forall \alpha \in {\mathbb{N}}^{\mathbb{N}}}\right) \left( {\forall \beta \in {\mathbb{N}}^{\mathbb{N}}}\right) (\lbrack \left( {x,\alpha }\right) \in B \end{matrix}\n\n\[ \left. {\& \left( {x,\beta }\right) \in B\& f\left( \alpha \right) = \left( {x, y}\right) \rbrack \; \Rightarrow \alpha { \leq }_{\text{lex }}\beta }\right) ,\n\nwhere \( { \leq }_{\text{lex }} \) is the lexicographic ordering on \( B \) .
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No
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Theorem 5.5.7 Let \( \left( {X,\mathcal{E}}\right) \) be a measurable space with \( \mathcal{E} \) closed under the Souslin operation, \( Y \) a Polish space, and \( A \in \mathcal{E}\bigotimes {\mathcal{B}}_{Y} \) . Then \( {\pi }_{X}\left( A\right) \in \mathcal{E} \) , and there is an \( \mathcal{E} \) -measurable section of \( A \) .
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Proof. By 3.1.7, there exists a countable sub \( \sigma \) -algebra \( \mathcal{D} \) of \( \mathcal{E} \) such that \( A \in \mathcal{D}\bigotimes {\mathcal{B}}_{Y} \) . Let \( \left( {B}_{n}\right) \) be a countable generator of \( \mathcal{D} \) and \( \chi : X \rightarrow \mathcal{C} \) the map defined by\n\n\[ \chi \left( x\right) = \left( {{\chi }_{{B}_{0}}\left( x\right) ,{\chi }_{{B}_{1}}\left( x\right) ,{\chi }_{{B}_{2}}\left( x\right) ,\ldots }\right) ,\;x \in X. \]\n\nLet \( Z = \chi \left( X\right) \) . Then \( \chi \) is a bimeasurable map from \( \left( {X,\mathcal{D}}\right) \) onto \( \left( {Z,{\mathcal{B}}_{Z}}\right) \) .\n\nLet\n\n\[ B = \{ \left( {\chi \left( x\right), y}\right) \in Z \times Y : \left( {x, y}\right) \in A\} . \]\n\n\( B \) is Borel in \( Z \times Y \) . Take a Borel set \( C \) in \( \mathcal{C} \times Y \) such that \( B = C\bigcap \left( {Z \times Y}\right) \) .\n\nLet \( E = {\pi }_{\mathcal{C}}\left( C\right) \) . Then \( E \) is analytic, and therefore it is the result of the Souslin operation on a system \( \left\{ {{E}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) of Borel subsets of \( \mathcal{C} \) . Note that\n\n\[ {\pi }_{X}\left( A\right) = {\chi }^{-1}\left( E\right) = \mathcal{A}\left( {{\chi }^{-1}\left( \left\{ {E}_{s}\right\} \right) }\right) . \]\n\nSince \( \mathcal{E} \) is closed under the Souslin operation, \( {\pi }_{X}\left( A\right) \in \mathcal{E} \) .\n\nBy 5.5.2, there is a \( \sigma \left( {{\mathbf{\sum }}_{1}^{1}\left( \mathcal{C}\right) }\right) \) -measurable section \( v : E \rightarrow Y \) of \( C \) . Take \( f = v \circ \chi \) . Then \( f \) is an \( \mathcal{E} \) -measurable section of \( A \) .
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Yes
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Theorem 5.5.7 Let \( \\left( {X,\\mathcal{E}}\\right) \) be a measurable space with \( \\mathcal{E} \) closed under the Souslin operation, \( Y \) a Polish space, and \( A \\in \\mathcal{E}\\bigotimes {\\mathcal{B}}_{Y} \) . Then \( {\\pi }_{X}\\left( A\\right) \\in \\mathcal{E} \) , and there is an \( \\mathcal{E} \) -measurable section of \( A \) .
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Proof. By 3.1.7, there exists a countable sub \( \\sigma \) -algebra \( \\mathcal{D} \) of \( \\mathcal{E} \) such that \( A \\in \\mathcal{D}\\bigotimes {\\mathcal{B}}_{Y} \) . Let \( \\left( {B}_{n}\\right) \) be a countable generator of \( \\mathcal{D} \) and \( \\chi : X \\rightarrow \\mathcal{C} \) the map defined by\n\n\[ \n\\chi \\left( x\\right) = \\left( {{\\chi }_{{B}_{0}}\\left( x\\right) ,{\\chi }_{{B}_{1}}\\left( x\\right) ,{\\chi }_{{B}_{2}}\\left( x\\right) ,\\ldots }\\right) ,\\;x \\in X.\n\]\n\nLet \( Z = \\chi \\left( X\\right) \) . Then \( \\chi \) is a bimeasurable map from \( \\left( {X,\\mathcal{D}}\\right) \) onto \( \\left( {Z,{\\mathcal{B}}_{Z}}\\right) \) .\n\nLet\n\n\[ \nB = \\{ \\left( {\\chi \\left( x\\right), y}\\right) \\in Z \\times Y : \\left( {x, y}\\right) \\in A\\} .\n\]\n\n\( B \) is Borel in \( Z \\times Y \) . Take a Borel set \( C \) in \( \\mathcal{C} \\times Y \) such that \( B = C\\bigcap \\left( {Z \\times Y}\\right) \) .\n\nLet \( E = {\\pi }_{\\mathcal{C}}\\left( C\\right) \) . Then \( E \) is analytic, and therefore it is the result of the Souslin operation on a system \( \\left\\{ {{E}_{s} : s \\in {\\mathbb{N}}^{ < \\mathbb{N}}}\\right\\} \) of Borel subsets of \( \\mathcal{C} \) . Note that\n\n\[ \n{\\pi }_{X}\\left( A\\right) = {\\chi }^{-1}\\left( E\\right) = \\mathcal{A}\\left( {{\\chi }^{-1}\\left( \\left\\{ {E}_{s}\\right\\} \\right) }\\right) .\n\]\n\nSince \( \\mathcal{E} \) is closed under the Souslin operation, \( {\\pi }_{X}\\left( A\\right) \\in \\mathcal{E} \) .\n\nBy 5.5.2, there is a \( \\sigma \\left( {{\\mathbf{\\sum }}_{1}^{1}\\left( \\mathcal{C}\\right) }\\right) \) -measurable section \( v : E \\rightarrow Y \) of \( C \) . Take \( f = v \\circ \\chi \) . Then \( f \) is an \( \\mathcal{E} \) -measurable section of \( A \) .
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Yes
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Corollary 5.5.8 Let \( \left( {X,\mathcal{A}, P}\right) \) be a complete probability space, \( Y \) a Polish space, and \( B \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) . Then \( {\pi }_{X}\left( B\right) \in \mathcal{A} \), and \( B \) admits an \( \mathcal{A} \) -measurable section.
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Proof. Since \( \mathcal{A} \) is closed under the Souslin operation, the result follows from 5.5.7.
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No
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Theorem 5.7.1 (Novikov [90]) Let \( X, Y \) be Polish spaces and \( \mathcal{A} \) a countably generated sub \( \sigma \) -algebra of \( {\mathcal{B}}_{X} \) . Suppose \( B \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) is such that the sections \( {B}_{x} \) are compact. Then \( {\pi }_{X}\left( B\right) \in \mathcal{A} \), and \( B \) admits an \( \mathcal{A} \) -measurable section.
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Proof. Since the projection of a Borel set with compact sections is Borel (4.7.11), \( {\pi }_{X}\left( B\right) \) is Borel. Since \( {\pi }_{X}\left( B\right) \) is a union of atoms of \( \mathcal{A} \), by the Blackwell - Mackey theorem (4.5.7), it is in \( \mathcal{A} \) .\n\nLet \( U \) be an open set in \( Y \) . Write \( U = \mathop{\bigcup }\limits_{n}{F}_{n} \), the \( {F}_{n} \) ’s closed. Then\n\n\[{\pi }_{X}\left( {B\bigcap \left( {X \times U}\right) }\right) = \mathop{\bigcup }\limits_{n}{\pi }_{X}\left( {B\bigcap \left( {X \times {F}_{n}}\right) }\right) .\n\]\n\nHence, by 4.7.11 and 4.5.7, \( {\pi }_{X}\left( {B\bigcap \left( {X \times U}\right) }\right) \in \mathcal{A} \) . It follows that the multifunction \( x \rightarrow {B}_{x} \) defined on \( {\pi }_{X}\left( B\right) \) is \( \mathcal{A} \) -measurable. The result follows from the selection theorem of Kuratowski and Ryll-Nardzewski (5.2.1). ∎
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Yes
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Theorem 5.7.2 (Lusin) Let \( X, Y \) be Polish spaces and \( B \subseteq X \times Y \) Borel with sections \( {B}_{x} \) countable. Then \( B \) admits a Borel uniformization.
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Proof. By 3.3.17, there is a closed set \( E \) in \( {\mathbb{N}}^{\mathbb{N}} \) and a one-to-one continuous map \( f : E \rightarrow X \times Y \) with range \( B \) . Set\n\n\[ H = \left\{ {\left( {x,\alpha }\right) \in X \times E : {\pi }_{X}\left( {f\left( \alpha \right) }\right) = x}\right\} . \]\n\nThen \( H \) is a closed set in \( X \times {\mathbb{N}}^{\mathbb{N}} \) with sections \( {H}_{x} \) countable. Further, \( {\pi }_{X}\left( B\right) = {\pi }_{X}\left( H\right) \) . Fix a countable base \( \left( {V}_{n}\right) \) for \( {\mathbb{N}}^{\mathbb{N}} \) . Let\n\n\[ {Z}_{n} = \left\{ {x \in X : {H}_{x}\bigcap {V}_{n}\text{ is a singleton }}\right\} . \]\n\nBy 4.12.2, \( {Z}_{n} \) is coanalytic. Each \( {H}_{x} \) is countable and closed, and so if nonempty must have an isolated point. Therefore,\n\n\[ \mathop{\bigcup }\limits_{n}{Z}_{n} = {\pi }_{X}\left( H\right) = {\pi }_{X}\left( B\right) \]\n\nHence, \( {\pi }_{X}\left( B\right) \) is both coanalytic and analytic, and so by Souslin’s theorem, Borel. By the weak reduction principle for coanalytic sets (4.6.5), there exist pairwise disjoint Borel sets \( {B}_{n} \subseteq {Z}_{n} \) such that \( \mathop{\bigcup }\limits_{n}{B}_{n} = \mathop{\bigcup }\limits_{n}{Z}_{n} \) . Let\n\n\[ D = \mathop{\bigcup }\limits_{n}\left\lbrack {\left( {{B}_{n} \times {V}_{n}}\right) \bigcap H}\right\rbrack \]\n\nThen \( D \) is a Borel uniformization of \( H \) . Let \( g : D \rightarrow X \times X \) be the map defined by \( g\left( {x,\alpha }\right) = f\left( \alpha \right) \) . Since \( g \) is one-to-one, the set\n\n\[ C = \{ f\left( \alpha \right) : \left( {x,\alpha }\right) \in D\} \]\n\n is Borel (4.5.4). It clearly uniformizes \( B \) .
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Yes
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Proposition 5.7.3 Let \( X \) be a Polish space and \( \Pi \) a countably separated partition of \( X \) with all equivalence classes countable. Then \( \mathbf{\Pi } \) admits a Borel cross section.
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Proof. Let \( Y \) be a Polish space and \( f : X \rightarrow Y \) a Borel map such that\n\n\[ \n{x\Pi }{x}^{\prime } \Leftrightarrow f\left( x\right) = f\left( {x}^{\prime }\right) .\n\]\n\nDefine\n\n\[ \nB = \{ \left( {y, x}\right) \in Y \times X : f\left( x\right) = y\} .\n\]\n\nThen \( B \) is a Borel set with sections \( {B}_{y} \) countable. By 5.7.2, \( {\pi }_{Y}\left( B\right) \) is Borel and there is a Borel section \( g : {\pi }_{Y}\left( B\right) \rightarrow X \) of \( B \) . Note that \( g \) is one-toone. Take \( S \) to be the range of \( g \) . Then \( S \) is Borel by 4.5.4. Evidently, it is a cross section of \( \mathbf{\Pi } \) .
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Yes
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Theorem 5.8.4 (Kechris [52]) Let \( X, Y \) be Polish spaces. Assume that \( x \rightarrow {\mathcal{I}}_{x} \) is a Borel on Borel map assigning to each \( x \in X \) a \( \sigma \) -ideal \( {\mathcal{I}}_{x} \) of subsets of \( Y \) . Suppose \( B \subseteq X \times Y \) is a Borel set such that for every \( x \in {\pi }_{X}\left( B\right) ,{B}_{x} \notin {\mathcal{I}}_{x} \) . Then \( {\pi }_{X}\left( B\right) \) is Borel, and \( B \) admits a Borel section.
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Proof. Since \( x \rightarrow {\mathcal{I}}_{x} \) is Borel on Borel,\n\n\[{\pi }_{X}\left( B\right) = {\left\{ x : {B}_{x} \in {\mathcal{I}}_{x}\right\} }^{c}\]\n\nis Borel.\n\nIt remains to prove that \( B \) admits a Borel section. Fix a closed subset \( F \) of \( {\mathbb{N}}^{\mathbb{N}} \) and a continuous bijection \( f : F \rightarrow B \) . For each \( s \in {\mathbb{N}}^{ < \mathbb{N}} \) we define a Borel subset \( {B}_{s} \) of \( X \) such that for every \( s, t \in {\mathbb{N}}^{ < \mathbb{N}} \),\n\n(i) \( {B}_{e} = {\pi }_{X}\left( B\right) \) ;\n\n(ii) \( \left| s\right| = \left| t\right| \& s \neq t \Rightarrow {B}_{s} \cap {B}_{t} = \varnothing \) ;\n\n(iii) \( {B}_{s} = \mathop{\bigcup }\limits_{n}{B}_{{s}^{ \frown }n} \) ; and\n\n(iv) \( {B}_{s} \subseteq \left\{ {x \in X : {\left( f\left( \sum \left( s\right) \cap F\right) \right) }_{x} \notin {\mathcal{I}}_{x}}\right\} \) .\n\nWe define such a system of sets by induction on \( \left| s\right| \) . Suppose \( {B}_{t} \) have been defined for every \( t \in {\mathbb{N}}^{ < \mathbb{N}} \) of length \( < n \), and \( s \in {\mathbb{N}}^{ < \mathbb{N}} \) is of length \( n - 1 \) . For any \( k \in \mathbb{N} \), let\n\n\[{D}_{k} = \left\{ {x \in {B}_{s} : {\left( f\left( \sum \left( s \hat{} k\right) \bigcap F\right) \right) }_{x} \notin {\mathcal{I}}_{x}}\right\} .\n\nSince \( f \) is one-to-one and continuous, \( f\left( {\sum \left( {s\widehat{}k}\right) \bigcap F}\right) \) is Borel (4.5.4). Hence, as \( x \rightarrow {\mathcal{I}}_{x} \) is Borel on Borel, each \( {D}_{k} \) is Borel. By (iv), \( {B}_{s} = \mathop{\bigcup }\limits_{k}{D}_{k} \) . Take\n\n\[{B}_{s \land k} = {D}_{k} \smallsetminus \mathop{\bigcup }\limits_{{l < k}}{D}_{l}\]\n\nWe define \( u : {\pi }_{X}\left( B\right) \rightarrow Y \) as follows. Given any \( x \in {\pi }_{X}\left( B\right) \) there is a unique \( \alpha \in F \) (call it \( p\left( x\right) \) ) such that \( x \in {B}_{\alpha \mid k} \) for every \( k \) . Define \( u \) by\n\n\[u\left( x\right) = {\pi }_{Y}\left( {f\left( {p\left( x\right) }\right) }
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Yes
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Example 5.8.3 Let \( X, Y \) be Polish spaces and \( G : X \rightarrow Y \) a closed-valued Borel measurable multifunction. Define \( \mathcal{I} : X \rightarrow \mathcal{P}\left( {\mathcal{P}\left( Y\right) }\right) \) by \[ \mathcal{I}\left( x\right) = \{ I \subseteq Y : I\text{ is meager in }G\left( x\right) \} .
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By imitating the proof of 3.5.18 we can show the following: For every open set \( U \) in \( Y \) and every Borel set \( B \) in \( X \times Y \), the sets \[ {B}^{*U} = \left\{ {x \in X : G\left( x\right) \bigcap U \neq \varnothing }\right. \] \[ \text{&}{B}_{x}\bigcap G\left( x\right) \bigcap U\text{is comeager in}G\left( x\right) \bigcap U\} \] and \[ {B}^{\Delta U} = \left\{ {x \in X : G\left( x\right) \bigcap U \neq \varnothing }\right. \] \[ \text{&}{B}_{x}\bigcap G\left( x\right) \bigcap U\text{is nonmeager in}G\left( x\right) \bigcap U\} \] are Borel. It follows that \( \mathcal{I} \) is Borel on Borel.
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No
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Theorem 5.8.5 (Kechris [52] and Sarbadhikari [100]) If B is a Borel subset of the product of two Polish spaces \( X \) and \( Y \) such that \( {B}_{x} \) is nonmeager in \( Y \) for every \( x \in {\pi }_{X}\left( B\right) \), then \( B \) admits a Borel uniformization.
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Proof. Apply 5.8.4 with \( {\mathcal{I}}_{x} \) as in example 5.8.2.
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No
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Example 5.8.6 As a special case of 5.8.5 we see that every Borel set \( B \subseteq X \times Y \) with \( {B}_{x} \) a dense \( {G}_{\delta } \) set admits a Borel uniformization. However, there is an \( {F}_{\sigma } \) subset \( E \) of \( \left\lbrack {0,1}\right\rbrack \times {\mathbb{N}}^{\mathbb{N}} \) with sections \( {E}_{x} \) dense and that does not admit a Borel uniformization. Here is an example.
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Let \( C \subseteq \left\lbrack {0,1}\right\rbrack \times {\mathbb{N}}^{\mathbb{N}} \) be a closed set with projection to the first coordinate space \( \left\lbrack {0,1}\right\rbrack \), that does not admit a Borel uniformization. Such a set exists by 5.1.7. For each \( s \in {\mathbb{N}}^{ < \mathbb{N}} \), fix a homeomorphism \( {f}_{s} : \sum \rightarrow \sum \left( s\right) \) . Take\n\n\[ E = \mathop{\bigcup }\limits_{{s \in {\mathbb{N}}^{ < \mathbb{N}}}}\left\{ {\left( {x,{f}_{s}\left( \alpha \right) }\right) : \left( {x,\alpha }\right) \in B}\right\} .\n\]\n\nThis \( E \) works.
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No
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Theorem 5.8.7 (Blackwell and Ryll-Nardzewski [17]) Let \( X, Y \) be Polish spaces, \( P \) a transition probability on \( X \times Y \), and \( B \) a Borel subset of \( X \times Y \) such that \( P\left( {x,{B}_{x}}\right) > 0 \) for all \( x \in {\pi }_{X}\left( B\right) \) . Then \( {\pi }_{X}\left( B\right) \) is Borel, and \( B \) admits a Borel uniformization.
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Proof. Apply 5.8.4 with \( {\mathcal{I}}_{x} \) as in Example 5.8.1.
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No
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Theorem 5.8.8 (Blackwell and Ryll-Nardzewski) Let \( X, Y \) be Polish spaces, \( \mathcal{A} \) a countably generated sub \( \sigma \) algebra of \( {\mathcal{B}}_{X} \), and \( P \) a transition probability on \( X \times Y \) such that for every \( B \in {\mathcal{B}}_{Y}, x \rightarrow P\left( {x, B}\right) \) is \( \mathcal{A} \) -measurable. Suppose \( B \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) is such that \( P\left( {x,{B}_{x}}\right) > 0 \) for all \( x \in {\pi }_{X}\left( B\right) \) . Then \( {\pi }_{X}\left( B\right) \in \mathcal{A} \), and \( B \) admits an \( \mathcal{A} \) -measurable section.
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Proof of 5.8.8. By a slight modification of the argument contained in the proof of 3.4.24 we see that for every \( E \in \mathcal{A} \otimes {\mathcal{B}}_{Y}, x \rightarrow P\left( {x,{E}_{x}}\right) \) is \( \mathcal{A} \) -measurable. As \( {\pi }_{X}\left( B\right) = \left\{ {x \in X : P\left( {x,{B}_{x}}\right) > 0}\right\} \), it follows that \( {\pi }_{X}\left( B\right) \in \mathcal{A} \) . By 5.8.9, there is a \( C \subseteq B \) in \( \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) with compact \( x \) -sections such that \( P\left( {x,{C}_{x}}\right) > 0 \) for every \( x \in {\pi }_{X}\left( B\right) \) . In particular, \( {\pi }_{X}\left( B\right) = {\pi }_{X}\left( C\right) \) . The result follows from Novikov's uniformization theorem (5.7.1).
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Yes
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Lemma 5.8.9 Let \( X, Y,\mathcal{A} \), and \( P \) be as above. For every \( E \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) and every \( \epsilon > 0 \), there is an \( F \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) contained in \( E \) such that \( {F}_{x} \) is compact and \( P\left( {x,{F}_{x}}\right) \geq \epsilon \cdot P\left( {x,{E}_{x}}\right) \) .
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Proof. Let \( \mathcal{M} \) be the class of all sets in \( \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) such that the conclusion of the lemma holds for every \( P \) and every \( \epsilon > 0 \) . By 3.4.20, \( \mathcal{M} \) contains all rectangles \( A \times B \), where \( A \in \mathcal{A} \) and \( B \) Borel in \( Y \) . So, \( \mathcal{M} \) contains all finite disjoint unions of such rectangles. It is fairly routine to check that \( \mathcal{M} \) is a monotone class. Therefore, the result follows from the monotone class theorem.
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Yes
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Proposition 5.8.10 Let \( X, f \), and \( \mathcal{A} \) be as above. An everywhere proper conditional distribution given \( f \) exists if and only if there is an \( \mathcal{A} \) - measurable \( g : X \rightarrow X \) such that \( f\left( {g\left( x\right) }\right) = f\left( x\right) \) for all \( x \) .
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Proof. Suppose an \( \mathcal{A} \) -measurable \( g : X \rightarrow X \) such that \( f \circ g \) is the identity exists. Define\n\n\[ Q\left( {x, B}\right) = \left\{ \begin{array}{ll} 1 & \text{ if }g\left( x\right) \in B \\ 0 & \text{ otherwise. } \end{array}\right. \]\n\nIt is easy to verify that \( Q \) has the desired properties.\n\nConversely, let an everywhere proper conditional distribution \( Q \) given \( f \) exist. Let\n\n\[ S = \{ \left( {x, y}\right) \in X \times X : f\left( x\right) = f\left( y\right) \} .\n\nThen \( S \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} \) and \( Q\left( {x,{S}_{x}}\right) = 1 \) . By 5.8.8, there is an \( \mathcal{A} \) -measurable section \( g \) of \( S \), which is what we are looking for.\n\nSince \( g \) is \( \mathcal{A} \) -measurable, \( g\left( x\right) = g\left( y\right) \) whenever \( f\left( x\right) = f\left( y\right) \) . It follows that there is a Borel function \( h : \mathbb{R} \rightarrow X \) such that \( g\left( x\right) = h\left( {f\left( x\right) }\right) \) for all \( x \) . Then the range of \( f \) equals \( \{ y \in \mathbb{R} : f\left( {h\left( y\right) }\right) = y\} \), which is a Borel set. It follows from the above proposition that whenever the range of \( f \) is not a Borel set, everywhere proper conditional distributions given \( f \) cannot exist.
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Yes
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Proposition 5.8.13 (Feldman and Moore [41]) Every Borel equivalence relation on a Polish space \( X \) with equivalence classes countable is induced by a countable group of Borel automorphisms.
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Proof. Let \( \Pi \) be a Borel equivalence relation on \( X \) with equivalence classes countable. By 5.8.11, write\n\n\[ \Pi = \mathop{\bigcup }\limits_{n}{G}_{n} \]\n\nwhere \( {\pi }_{1} \mid {G}_{n} \) is one-to-one, \( {\pi }_{1}\left( {x, y}\right) = x \) ; i.e., the \( {G}_{n} \) ’s are graphs of Borel functions. Let\n\n\[ {H}_{n} = \varphi \left( {G}_{n}\right) \]\n\nwhere \( \varphi \left( {x, y}\right) = \left( {y, x}\right) \) . Then \( {\pi }_{2} \mid {H}_{n} \) is one-to-one, where \( {\pi }_{2}\left( {x, y}\right) = y \) . Let\n\n\[ X \times X \smallsetminus \Delta = \mathop{\bigcup }\limits_{k}\left( {{U}_{k} \times {V}_{k}}\right) \]\n\n\( {U}_{k},{V}_{k} \) open, where \( \Delta = \{ \left( {x, x}\right) : x \in X\} \) . Note that \( {U}_{k} \cap {V}_{k} = \varnothing \) . Put\n\n\[ {D}_{nmk} = \left( {{G}_{n}\bigcap {H}_{m}}\right) \bigcap \left( {{U}_{k} \times {V}_{k}}\right) . \]\n\nNote that \( {\pi }_{1}\left| {{D}_{nmk}\text{and}{\pi }_{2}}\right| {D}_{nmk} \) are one-to-one, and\n\n\[ {\pi }_{1}\left( {D}_{nmk}\right) \bigcap {\pi }_{2}\left( {D}_{nmk}\right) = \varnothing . \]\n\nSo, there is a Borel automorphism \( {g}_{nmk} \) of \( X \) given by\n\n\[ {g}_{nmk}\left( x\right) = \left\{ \begin{array}{ll} y & \text{ if }\left( {x, y}\right) \in {D}_{nmk}\text{ or }\left( {y, x}\right) \in {D}_{nmk}, \\ x & \text{ otherwise. } \end{array}\right. \]\n\nClearly,\n\n\[ \mathbf{\Pi } = \Delta \bigcup \mathop{\bigcup }\limits_{{nmk}}\operatorname{graph}\left( {g}_{nmk}\right) \]\n\nNow take \( G \) to be the group of automorphisms generated by \( \left\{ {g}_{nmk}\right. \) : \( n, m, k \in \mathbb{N}\} \) .
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Yes
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Theorem 5.9.1 (Miller [85]) Every partition \( \Pi \) of a Polish space \( X \) into \( {G}_{\delta } \) sets such that the saturation of every basic open set is simultaneously \( {F}_{\sigma } \) and \( {G}_{\delta } \) admits a section \( s : X \rightarrow X \) that is Borel measurable of class 2. In particular, such partitions admit a \( {G}_{\delta } \) cross section.
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Proof. Let \( \left( {U}_{n}\right) \) be a countable base for the topology of \( X \) . Let \( \left( {V}_{n}\right) \) be an enumeration of \( \left\{ {{U}_{n}^{ * } : n \in \mathbb{N}}\right\} \bigcup \left\{ {{\left( {U}_{n}^{ * }\right) }^{c} : n \in \mathbb{N}}\right\} \) . Let \( {\mathcal{T}}^{\prime } \) be the topology on \( X \) generated by \( \left\{ {{U}_{n} : n \in \mathbb{N}}\right\} \bigcup \left\{ {{V}_{n} : n \in \mathbb{N}}\right\} \) . Note that every \( {\mathcal{T}}^{\prime } \) open set is an \( {F}_{\sigma } \) set in \( X \) relative to the original topology of \( X \) . Consider the map \( f : X \rightarrow X \times {2}^{\mathbb{N}} \) defined by\n\n\[ f\left( x\right) = \left( {x,{\chi }_{{V}_{0}}\left( x\right) ,{\chi }_{{V}_{1}}\left( x\right) ,{\chi }_{{V}_{2}}\left( x\right) ,\ldots }\right) ,\;x \in X.\]\n\nThe map \( f \) is one-to-one and of class 2 . Let \( G \) be the range of \( f \) . It is quite easy to see that\n\n\[ {\mathcal{T}}^{\prime } = \left\{ {{f}^{-1}\left( W\right) : W\text{ open in }G}\right\}\]\n\nArguing as in the proof of 3.2.5, it is easily seen that \( G \) is a \( {G}_{\delta } \) set in \( X \times {2}^{\mathbb{N}} \) . Therefore, by 2.2.1, \( \left( {X,{\mathcal{T}}^{\prime }}\right) \) is Polish. As argued in 5.1.13,\n\n\[ \left\lbrack x\right\rbrack = \bigcap \left\{ {{U}_{n}^{ * } : {U}_{n}^{ * } \supseteq \left\lbrack x\right\rbrack }\right\}\]\n\nSo, each \( \Pi \) -equivalence class is closed relative to \( {\mathcal{T}}^{\prime } \) .\n\nLet \( \mathcal{L} \) be the set of all invariant subsets of \( X \) that are clopen relative to \( {\mathcal{T}}^{\prime } \) . We claim that the multifunction \( x \rightarrow \left\lbrack x\right\rbrack \) is \( {\mathcal{L}}_{\sigma } \) -measurable. Let \( \mathcal{S} = {\left\{ {V}_{n} : n \in \mathbb{N}\right\} }_{d} \), the set of all finite intersections of sets in \( \left\{ {{V}_{n} : n \in \mathbb{N}}\right\} \) . Any \( {\mathcal{T}}^{\prime } \) -open \( W \) is of the form \( U\bigcup V, U \) open relative to the original toplogy of \( X \) and \( V \) a union of sets in \( \mathcal{S} \) . Then \( {W}^{ * } = {U}^{ * }\bigcup V \), which proves our claim.\n\nBy the selection theorem of Kuratowski and Ryll-Nardzewski, there exists an \( {\mathcal{L}}_{\sigma } \) -measurable selection \( s \) for \( x \rightarrow \left\lbrack x\right\rbrack \) . In particular, \( s \) is continuous with respect to \( {\mathcal{T}}^{\prime } \) . The associated cross section \( S = \{ x \in s\left( x\right) = x\} \) is \( {\mathcal{T}}^{\prime } \) - closed and so is a \( {G}_{\delta } \) set relative to the original toplogy of \( X \) .
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Yes
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Theorem 5.9.2 (Srivastava [114]) Every Borel measurable partition \( \mathbf{\Pi } \) of a Polish space \( X \) into \( {G}_{\delta } \) sets admits a Borel cross section.
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Proof. (Kechris) For \( x \in X \) let \( \left\lbrack x\right\rbrack \) denote the member of \( \mathbf{\Pi } \) containing \( x \) . Consider the multifunction \( p : X \rightarrow X \) defined by\n\n\[ p\left( x\right) = \operatorname{cl}\left( \left\lbrack x\right\rbrack \right) \]\n\nThen \( p : X \rightarrow X \) is a closed-valued measurable multifunction. Further, for every \( x, y \in X, x \equiv y \Leftrightarrow p\left( x\right) = p\left( y\right) \) (5.9.1).\n\nNow consider \( F\left( X\right) \), the set of nonempty closed subsets of \( X \) with Effros Borel structure. By 3.3.10, it is standard Borel. Note that \( p \) considered as a map from \( X \) to \( F\left( X\right) \) is measurable. Let\n\n\[ P = \{ \left( {F, x}\right) \in F\left( X\right) \times X : p\left( x\right) = F\} . \]\n\nThe set \( P \) is Borel. For \( F \in F\left( X\right) \), let \( {\mathcal{I}}_{F} \) be the \( \sigma \) -ideal of subsets of \( X \) that are meager in \( F \) . As the multifunction \( F \rightarrow F \) from \( F\left( x\right) \) to \( X \) is measurable, by \( {5.8.3}, F \rightarrow {\mathcal{I}}_{F} \) is Borel on Borel. By the Baire category theorem, \( {P}_{F} \notin {\mathcal{I}}_{F} \) for each \( F \) . Therefore, by 5.8.4, \( D = {\pi }_{F\left( X\right) }\left( P\right) \) is Borel, and there is a Borel section \( q : D \rightarrow X \) of \( P \) . Let\n\n\[ S = \{ x \in X : x = q\left( {p\left( x\right) }\right) \} . \]\n\nClearly \( S \) is a Borel cross section of \( \mathbf{\Pi } \) .
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Yes
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Theorem 5.9.5 \( \operatorname{irr}\left( A\right) / \sim \) is standard Borel if and only if \( A \) is GCR.
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Its proof makes crucial uses of 5.4.3 and 4.5.4. We refer the interested reader to [4] and [43] for a proof.
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No
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Theorem 5.10.1 (The reflection theorem) Let \( X \) be a Polish space and \( \Phi \subseteq \mathcal{P}\left( X\right) {\mathbf{\Pi }}_{1}^{1} \) on \( {\mathbf{\Pi }}_{1}^{1} \) . For every \( {\mathbf{\Pi }}_{1}^{1} \) set \( A \in \Phi \) there is a Borel \( B \subseteq A \) in \( \Phi \) .
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Proof. Suppose there is a \( {\mathbf{\Pi }}_{1}^{1} \) set \( A \subseteq X \) in \( \Phi \) that does not contain a Borel set belonging to \( \Phi \) . We shall get a contradiction. Let \( \varphi \) be a \( {\mathbf{\Pi }}_{1}^{1} \) -norm on \( A \) and\n\n\[ C = \left\{ {\left( {x, y}\right) : y{ < }_{\varphi }^{ * }x}\right\} \]\n\nWe claim that\n\n\[ x \notin A \Leftrightarrow {C}_{x} \in \Phi . \]\n\n\( \left( *\right) \)\n\nSuppose \( x \notin A \) . Then \( {C}_{x} = A \in \Phi \) . Conversely, if \( x \in A \), then \( {C}_{x} \) is a Borel subset of \( A \) . So by our assumptions, \( {C}_{x} \notin \Phi \) .\n\nSince \( \Phi \) is \( {\mathbf{\Pi }}_{1}^{1} \) on \( {\mathbf{\Pi }}_{1}^{1},{A}^{c} \) is \( {\mathbf{\Pi }}_{1}^{1} \) by \( \left( \star \right) \) . Hence, by Souslin’s theorem, it is Borel, contradicting our assumption again.
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Yes
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Theorem 5.10.2 Let \( X, Y \) be Polish spaces and \( A \subseteq X \times Y \) analytic with sections \( {A}_{x} \) countable. Then every coanalytic set \( B \) containing \( A \) contains a Borel set \( E \supseteq A \) with all sections countable.
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Proof. Let \( C = {B}^{c} \) . Define \( \Phi \subseteq \mathcal{P}\left( {X \times Y}\right) \) by\n\n\[ D \in \Phi \Leftrightarrow {D}^{c} \subseteq B\& \forall x\left( {\left( {D}^{c}\right) }_{x}\right. \text{is countable})\text{.}\]\n\nUsing 4.3.7 we can easily check that \( \Phi \) is \( {\mathbf{\Pi }}_{1}^{1} \) on \( {\mathbf{\Pi }}_{1}^{1} \) . Since \( {A}^{c} \in \Phi \), by 5.10.1 there is a Borel set \( D \) in \( \Phi \) contained in \( {A}^{c} \) . Take \( E = {D}^{c} \) .
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Yes
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Theorem 5.10.3 (Lusin) Every analytic set with countable sections, in the product of two Polish spaces, can be covered by countably many Borel graphs.
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Proof. The result immediately follows from 5.10.2 and 5.8.11.
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No
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Proposition 5.10.4 (Burgess) Let \( X \) be Polish, \( E \) an analytic equivalence relation on \( X \), and \( C \subseteq X \times X \) a coanalytic set containing \( E \) . Then there is a Borel equivalence relation \( B \) such that \( E \subseteq B \subseteq C \) .
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Proof of 5.10.4. Applying 5.10.5 repeatedly, by induction on \( n \) we can define a sequence of Borel sets \( \left( {B}_{n}\right) \) such that\n\n\[ E \subseteq {B}_{n} \subseteq \mathcal{E}\left( {B}_{n}\right) \subseteq {B}_{n + 1} \subseteq C \]\n\nfor all \( n \) . Take \( B = \mathop{\bigcup }\limits_{n}{B}_{n} \) .
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No
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Lemma 5.10.5 Let \( X \) be a Polish space, \( P \) analytic, \( C \) coanalytic, and \( \mathcal{E}\left( P\right) \subseteq C \) . Then there is a Borel set \( B \) containing \( P \) such that\n\n\[ \mathcal{E}\left( B\right) \subseteq C\text{.} \]
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Proof. Define \( \Phi \subseteq \mathcal{P}\left( {X \times X}\right) \) by\n\n\[ D \in \Phi \Leftrightarrow \mathcal{E}\left( {D}^{c}\right) \subseteq C. \]\n\n\( \Phi \) is \( {\mathbf{\Pi }}_{1}^{1} \) on \( {\mathbf{\Pi }}_{1}^{1} \) . Further, \( {P}^{c} \in \Phi \) . By the reflection theorem (5.10.1), there is a Borel set \( D \) in \( \Phi \) that is contained in \( {P}^{c} \) . Take \( B = {D}^{c} \) .
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Yes
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Corollary 5.10.6 For every analytic equivalence relation \( E \) on a Polish space \( X \) there exist Borel equivalence relations \( {B}_{\alpha },\alpha < {\omega }_{1} \), such that \( E = \) \( \mathop{\bigcap }\limits_{{\alpha < {\omega }_{1}}}{B}_{\alpha }. \)
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Proof. By 4.3.17, write \( E = \mathop{\bigcap }\limits_{{\alpha < {\omega }_{1}}}{C}_{\alpha },{C}_{\alpha } \) coanalytic. By 5.10.4, for each \( \alpha \) there exists a Borel equivalence relation \( {B}_{\alpha } \) such that \( E \subseteq {B}_{\alpha } \subseteq {C}_{\alpha } \) .
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Yes
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Theorem 5.11.4 Every countably generated sub \( \sigma \) -algebra of the Borel \( \sigma \) - algebra of a Polish space has a minimal complement.
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Proof of 5.11.4. Let \( X \) be Polish and \( \mathcal{C} \) a countably generated sub \( \sigma \) -algebra of \( {\mathcal{B}}_{X} \) . \n\nCase 1. There is a cocountable atom \( A \) of \( \mathcal{C} \) .\n\nLet \( f : X \smallsetminus A \rightarrow A \) be a one-to-one map. Take\n\n\[ \mathcal{D} = \sigma \left( {\{ \{ x, f\left( x\right) \} : x \in X \smallsetminus A\} \bigcup {\mathcal{B}}_{A \smallsetminus f\left( {A}^{c}\right) }}\right) . \]\n\nBy 5.11.5, \( \mathcal{D} \) is a minimal complement of \( \mathcal{C} \) .\n\nCase 2. There is an uncountable atom \( A \) of \( \mathcal{C} \) such that \( X \smallsetminus A \) is also uncountable.\n\nLet \( f : A \rightarrow {A}^{c} \) be a Borel isomorphism and \( g : X \rightarrow X \) the map that equals \( f \) on \( A \) and the identity on \( {A}^{c} \) . Take\n\n\[ \mathcal{D} = {g}^{-1}\left( {\mathcal{B}}_{X}\right) \]\n\nBy 5.11.5, \( \mathcal{D} \) is a minimal complement of \( \mathcal{C} \) .\n\nCase 3. All atoms of \( \mathcal{C} \) are countable. Since \( \mathcal{C} \) is countably generated with all atoms countable, by 5.8.12 there exists a countable partition \( {G}_{n} \) of \( X \) such that each \( {G}_{n} \) is a partial cross section of the set of atoms of \( \mathcal{C} \) . It is easy to choose the \( {G}_{n} \) ’s in such a way that for distinct \( {G}_{n} \) and \( {G}_{m} \) , \( {G}_{n}\bigcup {G}_{m} \) is not a partial cross section of the set of atoms of \( \mathcal{C} \) . The result follows by 5.11.5 by taking\n\n\[ \mathcal{D} = \sigma \left( \left\{ {{G}_{n} : n \in \mathbb{N}}\right\} \right) . \]
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Yes
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Lemma 5.11.5 Let \( X \) be Polish and \( \mathcal{C} \) a countably generated sub \( \sigma \) -algebra of \( {\mathcal{B}}_{X} \) . Suppose \( \mathcal{D} \) is a countably generated sub \( \sigma \) -algebra of \( {\mathcal{B}}_{X} \) such that every atom \( A \) of \( \mathcal{D} \) is a partial cross section of the atoms of \( \mathcal{C} \) . Further, assume that for any two distinct atoms \( {C}_{1},{C}_{2} \) of \( \mathcal{D},{C}_{1}\bigcup {C}_{2} \) is not a partial cross section of the set of atoms of \( \mathcal{C} \) . Then \( \mathcal{D} \) is a minimal complement of \( \mathcal{C} \).
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Proof. Under the hypothesis, \( \mathcal{C} \vee \mathcal{D} \) is a countably generated sub \( \sigma \) - algebra of \( {\mathcal{B}}_{X} \) with atoms singletons. Hence, by 4.5.7, \( \mathcal{C} \vee \mathcal{D} = {\mathcal{B}}_{X} \) . Let \( {\mathcal{D}}^{ * } \) be a proper countably generated sub \( \sigma \) -algebra of \( \mathcal{D} \) . By the corollary to 4.5.7, there is an atom \( A \) of \( {\mathcal{D}}^{ * } \) that is not an atom of \( \mathcal{D} \) . Hence, it is a union of more than one atom of \( \mathcal{D} \) . Hence, there exist two distinct points \( x, y \) of \( A \) that belong to the same atom of \( \mathcal{C} \) . This implies that there is no \( E \in \mathcal{C}\bigvee {\mathcal{D}}^{ * } \) containing exactly one of \( x, y \) . So, \( \mathcal{C}\bigvee {\mathcal{D}}^{ * } \neq {\mathcal{B}}_{X} \) . The result now follows from 5.11.1 and 5.11.2.
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Yes
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Theorem 5.12.1 (Arsenin, Kunugui [60]) Let \( B \subseteq X \times Y \) be a Borel set, \( X, Y \) Polish, such that \( {B}_{x} \) is \( \sigma \) -compact for every \( x \) . Then \( {\pi }_{X}\left( B\right) \) is Borel, and \( B \) admits a Borel uniformization.
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Proof of 5.12.1. Write \( B = \mathop{\bigcup }\limits_{n}{B}_{n} \), the \( {B}_{n} \) ’s Borel with compact sections. That this can be done follows from 5.12.3. Then\n\n\[{\pi }_{X}\left( B\right) = \mathop{\bigcup }\limits_{n}{\pi }_{X}\left( {B}_{n}\right)\]\n\nSince the projection of a Borel set with compact sections is Borel (4.7.11), each \( {\pi }_{X}\left( {B}_{n}\right) \), and hence \( {\pi }_{X}\left( B\right) \), is Borel. Let\n\n\[{D}_{n} = {\pi }_{X}\left( {B}_{n}\right) \smallsetminus \mathop{\bigcup }\limits_{{m < n}}{\pi }_{X}\left( {B}_{m}\right)\]\n\nThen the \( {D}_{n} \) ’s are Borel and pairwise disjoint. Further, the set\n\n\[C = \mathop{\bigcup }\limits_{n}\left( {{B}_{n}\bigcap \left( {{D}_{n} \times Y}\right) }\right)\]\n\n is a Borel subset of \( B \) with compact sections such that \( {\pi }_{X}\left( C\right) = {\pi }_{X}\left( B\right) \) . By Novikov’s uniformization theorem (5.7.1), \( C \) admits a Borel uniformization, and our result follows.
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Yes
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Theorem 5.12.3 Let \( X, Y \) be Polish spaces and \( A \subseteq X \times Y \) a Borel set with sections \( {A}_{x} \) \(\sigma\) -compact. Then \( A = \mathop{\bigcup }\limits_{n}{B}_{n} \), where each \( {B}_{n} \) is Borel with \( {\left( {B}_{n}\right) }_{x} \) compact for all \( x \) and all \( n \) .
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Proof. The result trivially follows from 5.12.2 by taking \( B = {A}^{c} \) .
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No
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Proposition 5.12.4 Let \( B \subseteq X \times Y \) be a Borel set with sections \( {B}_{x} \) that are \( {G}_{\delta } \) sets in \( Y \) . Then there exist Borel sets \( {B}_{n} \) with open sections such that \( B = \mathop{\bigcap }\limits_{n}{B}_{n} \) .
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Proof. Let \( Z \) be a compact metric space containing (a homeomorph of) \( Y \) . Then \( B \) is Borel in \( X \times Z \) with sections \( {G}_{\delta } \) sets (2.2.7). By 5.12.3, there exist Borel sets \( {C}_{n} \) in \( X \times Z \) with sections compact such that \( \left( {X \times Z}\right) \smallsetminus B = \) \( \mathop{\bigcup }\limits_{n}{C}_{n} \) . Take \( {B}_{n} = \left( {X \times Y}\right) \smallsetminus {C}_{n} \) .
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Yes
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Proposition 5.12.6 Let \( X \) be a Polish space and \( \mathcal{B} \subseteq F\left( X\right) \) hereditary. Then \( {\Omega }_{{D}_{\mathcal{B}}} = {\mathcal{B}}_{\sigma } \cap F\left( X\right) \) .
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Proof. Fix a closed set \( A \subseteq X \) and a countable base \( \left( {U}_{n}\right) \) for \( X \) .\n\nLet \( {D}^{\infty }\left( A\right) = \varnothing \) . Then\n\n\[ A = \mathop{\bigcup }\limits_{{\alpha < {\left| A\right| }_{D}}}\left( {{D}^{\alpha }\left( A\right) \smallsetminus {D}^{\alpha + 1}\left( A\right) }\right) \]\n\n\[ = \mathop{\bigcup }\limits_{{\alpha < {\left| A\right| }_{D}}}\mathop{\bigcup }\limits_{n}\left\{ {{U}_{n} \cap {D}^{\alpha }\left( A\right) : \operatorname{cl}\left( {{U}_{n} \cap {D}^{\alpha }\left( A\right) }\right) \in \mathcal{B}}\right\} \]\n\n\[ = \;\mathop{\bigcup }\limits_{{\alpha < {\left| A\right| }_{D}}}\mathop{\bigcup }\limits_{n}\{ \operatorname{cl}\left( {{U}_{n}\bigcap {D}^{\alpha }\left( A\right) }\right) : \operatorname{cl}\left( {{U}_{n}\bigcap {D}^{\alpha }\left( A\right) }\right) \in \mathcal{B}\} . \]\n\nThe last equality holds because \( A \) is closed. Thus, \( A \in {\mathcal{B}}_{\sigma } \).\n\nTo prove the converse, take an \( A \in {\mathcal{B}}_{\sigma } \cap F\left( X\right) \) . Suppose \( {D}^{\infty }\left( A\right) \neq \varnothing \) . We shall get a contradiction. Write \( A = \mathop{\bigcup }\limits_{m}{B}_{m},{B}_{m} \in \mathcal{B} \) . By the Baire category theorem, there exist \( n \) and \( m \) such that\n\n\[ \varnothing \neq {D}^{\infty }\left( A\right) \bigcap {U}_{n} \subseteq {D}^{\infty }\left( A\right) \bigcap {B}_{m} \]\n\nThis implies that\n\n\[ {D}^{{\left| A\right| }_{D} + 1}\left( A\right) \neq {D}^{{\left| A\right| }_{D}}\left( A\right) \]\n\nWe have arrived at a contradiction.
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Yes
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Proposition 5.12.7 Let \( X \) be Polish and \( D \) a derivative on \( X \) such that\n\n\[ \n\\{ \\left( {A, B}\\right) \\in F\\left( X\\right) \\times F\\left( X\\right) : A \\subseteq D\\left( B\\right) \\} \n\]\n\nis analytic. Then\n\n(i) \( {\\Omega }_{D} \) is coanalytic, and\n\n(ii) for every analytic \( \\mathcal{A} \\subseteq {\\Omega }_{D} \), \n\n\[ \n\\sup \\left\\{ {{\\left| A\\right| }_{D} : A \\in \\mathcal{A}}\\right\\} < {\\omega }_{1} \n\]
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Proof. Assertion (i) follows from the following equivalence:\n\n\[ \nA \\notin {\\Omega }_{D} \\Leftrightarrow \\exists B\\left( {B \\neq \\varnothing \\& B \\subseteq A\\& B \\subseteq D\\left( B\\right) }\\right) .\n\]\n\n(The sets \( A \) and \( B \) are closed in \( X \).)\n\nSuppose (ii) is false for some analytic \( \\mathcal{A} \\subseteq {\\Omega }_{D} \). Then,\n\n\[ \n\\sup \\left\\{ {{\\left| A\\right| }_{D} : A \\in \\mathcal{A}}\\right\\} = {\\omega }_{1} \n\]\n\nDefine \( R \\subseteq {2}^{\\mathbb{N} \\times \\mathbb{N}} \\times F\\left( X\\right) \) as follows:\n\n\[ \n\\left( {x, A}\\right) \\in R \\Leftrightarrow x \\in L{O}^{ * }\\& \n\]\n\n\[ \n\\exists f \\in F{\\left( X\\right) }^{\\mathbb{N}}\\lbrack f\\left( 0\\right) = A\\& \n\]\n\n\[ \n\\forall m \\in D\\left( x\\right) \\{ f\\left( m\\right) \\neq \\varnothing \\& \n\]\n\n\[ \n\\left( {m \\neq 0 \\Rightarrow \\forall n{ < }_{x}^{ * }m\\left( {f\\left( m\\right) \\subseteq D\\left( {f\\left( n\\right) }\\right) }\\right) }\\right) \\} \\rbrack .\n\]\n\nIt is fairly easy to check that \( R \) is analytic and that for \( \\varnothing \\neq A \\in {\\Omega }_{D} \),\n\n\[ \nR\\left( {x, A}\\right) \\Leftrightarrow x \\in W{O}^{ * }\\& \\left| x\\right| \\leq {\\left| A\\right| }_{D}.\n\]\n\nBy our assumptions,\n\n\[ \nx \\in W{O}^{ * } \\Leftrightarrow \\exists A \\in \\mathcal{A}\\left( {R\\left( {x, A}\\right) }\\right) .\n\]\n\nThis implies that \( W{O}^{ * } \) is analytic, which is not the case, and our result is proved.
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Yes
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Theorem 5.13.1 (Lopez-Escobar) A subset \( A \) of \( {X}_{L} \) is invariant (with respect to the logic action) and Borel, if and only if there is a sentence \( \sigma \) of \( {L}_{{\omega }_{1}\omega } \) such that \( A = {A}_{\sigma } \) .
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Proof. The sufficient part of this result is proved by induction on formulae of \( {L}_{{\omega }_{1}\omega } \) as follows:\n\nFor every formula \( \phi \left\lbrack {{v}_{0},{v}_{1},\ldots ,{v}_{k - 1}}\right\rbrack \), the set\n\n\[ \n{A}_{\phi, k} = \left\{ {\left( {x,{n}_{0},{n}_{1},\ldots ,{n}_{k - 1}}\right) : {\mathcal{A}}_{x} \vDash \phi \left\lbrack {{n}_{0},{n}_{1},\ldots ,{n}_{k - 1}}\right\rbrack }\right\} \n\]\n\nis Borel.\n\nThe necessary part is also proved by induction, but the induction in this case is a bit subtle. We proceed as follows. Let \( {\left( \mathbb{N}\right) }^{k} \) denote the set of all one-to-one finite sequences in \( \mathbb{N} \) of length \( k \) and for any \( s \in {\left( \mathbb{N}\right) }^{k} \) ,\n\n\[ \n\left\lbrack s\right\rbrack = \left\{ {g \in {S}_{\infty } : s \prec {g}^{-1}}\right\} \n\]\n\nClearly, \( \left\{ {\left\lbrack s\right\rbrack : s \in {\left( \mathbb{N}\right) }^{k}}\right\} \) form a base for the topology of \( {S}_{\infty } \) .\n\nSuppose \( A \) is a Borel set in \( {X}_{L} \) . Then, for every \( k \) there is a formula \( \phi \left\lbrack {{v}_{0},{v}_{1},\ldots ,{v}_{k - 1}}\right\rbrack \) of \( {L}_{{\omega }_{1}\omega } \) such that\n\n\[ \n{A}_{\phi, k} = \left\{ {\left( {x, s}\right) : s \in {\left( \mathbb{N}\right) }^{k}\& x \in {A}^{*\left\lbrack s\right\rbrack }}\right\} .\n\]\n\nThis is proved by induction on \( A \) using basic identities on Vaught transforms. We invite readers to complete the proof themselves. (Otherwise consult \( \left\lbrack {\left\lbrack {53}\right\rbrack \text{, p.97}}\right\rbrack \) .)\n\nNow, if \( A \subseteq {X}_{L} \) is an invariant Borel set, then \( {A}^{ * } = A \) and the result follows from the above assertion by taking \( k = 0 \) .
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No
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Theorem 5.13.8 Topological Vaught conjecture holds if \( G \) is a locally compact Polish group.
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Assuming 5.13.9, we prove 5.13.8 as follows: Let \( G \) be a locally compact Polish group acting continuously on a Polish space \( X \) . Write \( G = \mathop{\bigcup }\limits_{n}{K}_{n} \) , \( {K}_{n} \) compact. Then, for \( x, y \in X \) ,\n\n\[ \exists g \in G\left( {y = g \cdot x}\right) \Leftrightarrow \exists n\exists g \in {K}_{n}\left( {y = g \cdot x}\right) . \]\n\nSince \( {K}_{n} \) is compact and the set \( \left\{ {\left( {x, y, g}\right) \in X \times X \times {K}_{n} : y = g \cdot x}\right\} \) is closed, the equivalence relation induced by the group action is an \( {F}_{\sigma } \) set. Our result now follows from 5.13.9.
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Yes
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Theorem 5.13.9 Let \( E \) be an analytic equivalence relation on a Polish space \( X \) with all equivalence classes \( {F}_{\sigma } \) . Then the number of equivalence classes is \( \leq {\aleph }_{0} \) or perfectly many.
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Proof of 5.13.9. Let \( X \) be a Polish space and \( E \) an analytic equivalence relation on \( X \) with all its equivalence classes \( {F}_{\sigma } \) sets. Further assume that there are uncountably many \( E \) -equivalence classes. Fix a countable base \( \left( {V}_{n}\right) \) for the topology of \( X \) . Let \( P \) be the union of all basic open sets which is contained in countably many equivalence classes and \( Q \) its saturation; i.e., \( Q = \operatorname{proj}\left( {E\bigcap \left( {P \times P}\right) }\right) \) . Thus
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No
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Proposition 5.13.10 Suppose \( X \) is a Polish space and \( E \) an equivalence relation on \( X \) which is meager in \( {X}^{2} \) . Then \( E \) has perfectly many equivalence classes.
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Proof. Let \( E \subseteq \mathop{\bigcup }\limits_{n}{F}_{n},{F}_{n} \) closed and nowhere dense in \( {X}^{2} \) . Without any loss of generality, we further assume that the diagonal \( \left\{ {\left( {x, y}\right) \in {X}^{2} : x = y}\right\} \) is contained in each of \( {F}_{n} \) .\n\nFor each \( s \in {2}^{ < \mathbb{N}} \), we define a nonempty open set \( U\left( s\right) \) in \( X \) satisfying the following properties.\n\n(i) \( \operatorname{diameter}\left( {U\left( s\right) }\right) \leq {2}^{-\left| s\right| } \) .\n\n(ii) \( s \prec t \Rightarrow \operatorname{cl}\left( {U\left( t\right) }\right) \subseteq U\left( s\right) \) .\n\n(iii) If \( s \neq {s}^{\prime } \) and \( \left| s\right| = \left| {s}^{\prime }\right| \), then \( \left( {U\left( s\right) \times U\left( {s}^{\prime }\right) }\right) \cap {F}_{\left| s\right| } = \varnothing \) . In particular, \( {U}_{s} \) and \( {U}_{s} \) ’s are disjoint.\n\nWe define \( \left\{ {U\left( s\right) : s \in {2}^{ < \mathbb{N}}}\right\} \) by induction on \( \left| s\right| \) . Take \( U\left( e\right) \) to be any nonempty open set of diameter less than 1 disjoint from \( {F}_{0} \) . Since \( {F}_{0} \) is closed nowhere dense, such a set exists. Suppose \( n \) is a positive integer and \( U\left( s\right) \) has been defined for every sequence \( s \) of length less than \( n \) . Consider the set \( {F}_{n}^{{2}^{n}} \) . It is closed and nowhere dense in \( {X}^{{2}^{n + 1}} \) . Hence, there is an open set of the form \( \mathop{\prod }\limits_{{s \in {2}^{n - 1}}}\left( {U\left( {s \hat{} 0}\right) \times U\left( {s \hat{} 1}\right) }\right) \) contained in \( \mathop{\prod }\limits_{{s \in {2}^{n - 1}}}\left( {U\left( s\right) \times U\left( s\right) }\right) \) disjoint from \( {F}_{n}^{{2}^{n}} \) . We can further assume that the diameter of \( U\left( {s \hat{} \epsilon }\right) ,\left| s\right| = n - 1 \) and \( \epsilon = 0 \) or 1, is less than \( {2}^{-n} \), and that its closure is contained in \( U\left( s\right) \) .\n\nFor \( \alpha \in {2}^{\omega } \), let \( f\left( \alpha \right) \) be the unique element of \( X \) that belongs to each of \( U\left( {\alpha \mid n}\right) \) . It is easy to see that the range of \( f \) is a perfect set of pairwise \( E \) -inequivalent elements.
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Yes
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Theorem 5.13.12 (Sami) Topological Vaught conjecture holds if \( G \) is abelian.
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Proof. Assume that the number of orbits is uncountable. We shall show that there is a perfect set of inequivalent elements.\n\nLet \( E \) be the equivalence relation on \( X \) defined by\n\n\[ \n{xEy} \Leftrightarrow {G}_{x} = {G}_{y} \n\]\n\nwhere \( {G}_{x} \) is the stabilizer of \( x \) . Let \( y = g \cdot x \) for some \( g \in G \) . Then \( {G}_{x} = {g}^{-1} \cdot {G}_{y} \cdot g = {G}_{y} \), as \( G \) is abelian. Thus,\n\n\[ \nx{E}_{a}y \Rightarrow {xEy} \n\]\n\nwhere \( {E}_{a} \) is the equivalence relation induced by the action. Now note that\n\n\[ \n{xEy} \Leftrightarrow \forall g\left( {g \cdot x = x \Leftrightarrow g \cdot y = y}\right) . \n\]\n\nHence, \( E \) is coanalytic.\n\nSuppose there are uncountably many \( E \) -equivalence classes. Then by Silver’s theorem, there is a perfect set of \( E \) -inequivalent elements. In particular, there is a perfect set of \( {E}_{a} \) -inequivalent elements.\n\nNow assume that the set of \( E \) -equivalence classes is countable. We shall show that \( {E}_{a} \) is Borel. Our proof will then follow from Silver’s theorem.\n\nLet \( Y \subseteq X \) be an \( E \) -equivalence class. It is sufficient to show that \( {E}_{a} \cap \left( {Y \times Y}\right) \) is Borel. Let \( x \in Y \) and \( H = {G}_{x} \) . The partition of \( G \) by the cosets of \( H \) is lower-semicontinuous. Hence, there is a Borel cross-section \( S \) for this partition. For \( x, y \in Y \), we have the following:\n\n\[ \nx{E}_{a}y \Leftrightarrow \left( {\exists \text{ a unique }g \in S}\right) \left( {y = g \cdot x}\right) ; \n\]\n\n i.e., \( {E}_{a} \cap \left( {Y \times Y}\right) \) is a one-to-one projection of the Borel set\n\n\[ \n\{ \left( {x, y, g}\right) : g \in S\text{ and }y = g \cdot x\} . \n\]\n\nHence, \( {E}_{a} \) is Borel.
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Yes
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Lemma 5.13.14 Suppose \( \left\{ {{A}_{\alpha } : \alpha < {\omega }_{1}}\right\} \) is a family of Borel subsets of a Polish space \( X \) and \( E \) the equivalence relation on \( X \) defined by\n\n\[ \n{xEy} \Leftrightarrow \forall \alpha \left( {x \in {A}_{\alpha } \Leftrightarrow y \in {A}_{\alpha }}\right), x, y \in X.\n\]\n\n\( \left( *\right) \)\n\nThen the number of \( E \) -equivalence classes is \( \leq {\aleph }_{1} \) or perfectly many.
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Proof of 5.13.14. Although the proof of the lemma is messy looking, ideawise it is quite simple. Assume that the number of \( E \) -equivalence classes is \( > {\aleph }_{1} \) . We shall then show that there are perfectly many \( E \) -equivalence classes. The following fact will be used repeatedly in the proof of the lemma.\n\nFact. Suppose \( Z \) is a subset of \( X \) of cardinality \( > {\aleph }_{1} \) such that no two distinct elements \( Z \) are \( E \) -equivalent. Then there is an \( \alpha < {\omega }_{1} \) such that both \( Z \cap {A}_{\alpha } \) and \( Z \cap {A}_{\alpha }^{c} \) are of cardinality \( > {\aleph }_{1} \) .\n\nWe prove this fact by contradiction. If possible, let for every \( \alpha < {\omega }_{1} \) at least one of \( Z\bigcap {A}_{\alpha } \) and \( Z\bigcap {A}_{\alpha }^{c} \) be of cardinality \( \leq {\aleph }_{1} \) . Denote one such set by \( {M}_{\alpha } \) . We claim that \( Z \smallsetminus \mathop{\bigcup }\limits_{\alpha }{M}_{\alpha } \) is a singleton. Suppose not. Let \( x \) , \( y \) be two distinct elements of \( Z \smallsetminus \mathop{\bigcup }\limits_{\alpha }{M}_{\alpha } \) . Since \( x, y \) are \( E \) -inequivalent, by \( \left( \star \right) \) there exists an \( \alpha < {\omega }_{1} \) such that exactly one of \( x \) and \( y \) belong to \( {A}_{\alpha } \) . It follows that at least one of \( x, y \) belong to \( {M}_{\alpha } \) . But this is not the case. Hence, \( Z \smallsetminus \mathop{\bigcup }\limits_{\alpha }{M}_{\alpha } \) contains at most one point. It follows that the cardinality of \( Z \) is at most \( {\aleph }_{1} \), and we have arrived at a contradiction.\n\nFix a compatible complete metric on \( X \) . Following our usual notation, for \( \epsilon = 0 \) or 1, we set\n\n\[ \n{A}_{\alpha }^{\epsilon } = \left\{ \begin{array}{ll} {A}_{\alpha } & \text{ if }\epsilon = 0 \\ {A}_{\alpha }^{c} & \text{ if }\epsilon = 1 \end{array}\right.\n\]\n\nSince \( {A}_{\alpha }^{\epsilon } \) analytic, there is a continuous map \( {f}_{\alpha }^{\epsilon } : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) whose range is \( {A}_{\alpha }^{\epsilon } \) . We can arrange matters so that for every \( s \in {\mathbb{N}}^{ < \mathbb{N}} \), the diameter of \( {f}_{\alpha }^{\epsilon }\left( {\sum \left( s\right) }\right) \) is at most \( {2}^{-\left| s\right| } \) .\n\nFix any subset \( Z \) of \( X \) of cardinality \( > {\aleph }_{1} \) consisting of pairwise \( E \) -
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Yes
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Example 5.13.15 Let \( L \) be a first order language whose non-logical symbols consists of exactly one binary relation symbol. So, \( {X}_{L} = {2}^{\omega \times \omega } \) . We claim that in this case the equivalence relation \( {E}_{a} \) induced by the logic action is not Borel. Suppose not. Then \( {E}_{a} \in {\mathbf{\sum }}_{\beta }^{0} \) for some \( \beta < {\omega }_{1} \) .
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It follows that \( W{O}^{\alpha } = \{ x \in {WO} : \left| x\right| \leq \alpha \} \in {\mathbf{\sum }}_{\beta }^{0} \) for every \( \alpha < {\omega }_{1} \) . Now take any Borel set \( A \) in \( {\mathbb{N}}^{\mathbb{N}} \) which is not of additive class \( \beta \) . Since \( {WO} \) is \( {\mathbf{\Pi }}_{1}^{1} \) -complete, there is a continuous function \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow {LO} \) such that \( A = {f}^{-1}\left( {WO}\right) \) . But by the boundedness theorem, \( A = {f}^{-1}\left( {W{O}^{\alpha }}\right) \) for some \( \alpha \) . It follows that \( A \in {\mathbf{\sum }}_{\beta }^{0} \), and we have arrived at a contradiction.
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Yes
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Theorem 5.13.19 (Stern) Assume analytic determinacy. Let \( E \) be an analytic equivalence relation on a Polish space \( X \) such that all but countably many equivalence classes are of bounded Borel rank. Then the number of equivalence classes is \( \leq {\aleph }_{0} \) or perfectly many.
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The proof this result is beyond the scope of this book.
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No
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Theorem 5.14.1 (Kondô’s theorem) Let \( X, Y \) be Polish spaces. Every coanalytic set \( C \subseteq X \times Y \) admits a coanalytic uniformization.
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We shall show that there is a sequence of coanalytic norms on a given co-analytic set with certain \
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No
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Corollary 5.14.5 Let \( X \) be a Polish space and \( A \subseteq X \) coanalytic. Then A admits a very good \( {\mathbf{\Pi }}_{1}^{1} \) -scale.
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Proof. By 2.6.9 there is a closed set \( D \subseteq {\mathbb{N}}^{\mathbb{N}} \) and a continuous bijection \( f : D \rightarrow X \) . Now, \( {f}^{-1}\left( A\right) \cap D \) is a \( {\mathbf{\Pi }}_{1}^{1} \) subset of \( {\mathbb{N}}^{\mathbb{N}} \) and hence admits a very good \( {\mathbf{\Pi }}_{1}^{1} \) -scale by 5.14.4. The scale on \( A \) is now obtained by transfer via the function \( f \) .
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Yes
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We consider the discretization of the boundary value problem for the ordinary differential equation\n\n\[ - {u}^{\prime \prime }\left( x\right) = f\left( {x, u\left( x\right) }\right) ,\;x \in \left\lbrack {0,1}\right\rbrack \]\n\n(2.1)\n\nwith boundary condition\n\n\[ u\left( 0\right) = u\left( 1\right) = 0. \]\n\n(2.2)\n\nHere, \( f : \left\lbrack {0,1}\right\rbrack \times \mathbb{R} \rightarrow \mathbb{R} \) is a given continuous function, and we are looking for a twice continuously differentiable solution \( u : \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \).
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For the approximate solution we choose an equidistant subdivision of the interval \( \left\lbrack {0,1}\right\rbrack \) by setting\n\n\[ {x}_{j} = {jh},\;j = 0,\ldots, n + 1, \]\n\nwhere the step size is given by \( h = 1/\left( {n + 1}\right) \) with \( n \in \mathbb{N} \). At the internal grid points \( {x}_{j}, j = 1,\ldots, n \), we replace the differential quotient in the differential equation (2.1) by the difference quotient\n\n\[ {u}^{\prime \prime }\left( {x}_{j}\right) \approx \frac{1}{{h}^{2}}\left\lbrack {u\left( {x}_{j + 1}\right) - {2u}\left( {x}_{j}\right) + u\left( {x}_{j - 1}\right) }\right\rbrack \]\nto obtain the system of equations\n\n\[ - \frac{1}{{h}^{2}}\left\lbrack {{u}_{j - 1} - 2{u}_{j} + {u}_{j + 1}}\right\rbrack = f\left( {{x}_{j},{u}_{j}}\right) ,\;j = 1,\ldots, n, \]\n\nfor approximate values \( {u}_{j} \) to the exact solution \( u\left( {x}_{j}\right) \). This system has to be complemented by the two boundary conditions \( {u}_{0} = {u}_{n + 1} = 0 \). For an abbreviated notation we introduce the \( n \times n \) matrix\n\n\[ A = \frac{1}{{h}^{2}}\left( \begin{array}{rrrrrr} 2 & - 1 & & & & \\ - 1 & 2 & - 1 & & & \\ & - 1 & 2 & - 1 & & \\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ & & & - 1 & 2 & - 1 \\ & & & & - 1 & 2 \end{array}\right) \]\n\nand the vectors \( U = {\left( {u}_{1},\ldots ,{u}_{n}\right) }^{T} \) and \( F\left( U\right) = {\left( f\left( {x}_{1},{u}_{1}\right) ,\ldots, f\left( {x}_{n},{u}_{n}\right) \right) }^{T} \). Then our system of equations, including the boundary conditions, reads\n\n\[ {AU} = F\left( U\right) \]\n\n(2.3)\n\nFor obvious reasons, the above matrix \( A \) is called a tridiagonal matrix, and the vector \( F \) is diagonal; i.e., the \( j \) th component of \( F \) depends only on the \( j \) th component of \( u \). If (2.1) is a linear differential equation, i.e., if \( f \) depends linearly on the second variable \( u \), then the tridiagonal system of equations (2.3) also is linear.
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Yes
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Consider the linear integral equation\n\n\[ \varphi \left( x\right) - {\int }_{0}^{1}K\left( {x, y}\right) \varphi \left( y\right) {dy} = f\left( x\right) ,\;x \in \left\lbrack {0,1}\right\rbrack ,\]
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For the numerical approximation we replace the integral by the rectangular sum\n\n\[ {\int }_{0}^{1}K\left( {x, y}\right) \varphi \left( y\right) {dy} \approx \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}K\left( {x,{x}_{k}}\right) \varphi \left( {x}_{k}\right) \]\n\nwith equidistant grid points \( {x}_{k} = k/n, k = 1,\ldots, n \) . If we require the approximated equation to be satisfied only at the grid points, we arrive at the system of linear equations\n\n\[ {\varphi }_{j} - \frac{1}{n}\mathop{\sum }\limits_{{k = 1}}^{n}K\left( {{x}_{j},{x}_{k}}\right) {\varphi }_{k} = f\left( {x}_{j}\right) ,\;j = 1,\ldots, n, \]\n\nfor approximate values \( {\varphi }_{j} \) to the exact solution \( \varphi \left( {x}_{j}\right) \) .
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Yes
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