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Proposition 4.3.14 No uncountable analytic \( A \subseteq \mathbb{R} \) has strong measure zero.
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Null
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No
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Theorem 4.3.17 Every coanalytic set is a union of \( {\aleph }_{1} \) Borel sets.
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Proof. Let \( X \) be Polish and \( C \subseteq X \) coanalytic. By the Borel isomorphism theorem (3.3.13), without any loss of generality we may assume that \( X = {\mathbb{N}}^{\mathbb{N}} \) . By 4.1.20, there is a tree \( T \) on \( \mathbb{N} \times \mathbb{N} \) such that\n\n\[ \alpha \in C \Leftrightarrow T\left\lbrack \alpha \right\rbrack \text{is well-founded.} \]\n\nSo,\n\n\[ \alpha \in C \Leftrightarrow {\rho }_{T\left\lbrack \alpha \right\rbrack }\left( e\right) < {\omega }_{1} \]\n\nTherefore,\n\n\[ C = \mathop{\bigcup }\limits_{{\xi < {\omega }_{1}}}{C}_{e}^{\xi } \]\n\nwhere the \( {C}_{e}^{\xi } \) are as in 4.3.16.\n\nThe sets \( {C}_{e}^{\xi },\xi < {\omega }_{1} \), defined in the above proof are called the constituents of \( C \) . Since \( \mathbf{{CH}} \) holds for Borel sets, we now have the following result.\n\nTheorem 4.3.18 A coanalytic set is either countable or of cardinality \( {\aleph }_{1} \) or \( \mathfrak{c} \) .
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Yes
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Theorem 4.3.18 A coanalytic set is either countable or of cardinality \( {\aleph }_{1} \) or \( \mathfrak{c} \) .
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Null
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No
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Lemma 4.4.2 Suppose \( E = \mathop{\bigcup }\limits_{n}{E}_{n} \) cannot be separated from \( F = \mathop{\bigcup }\limits_{m}{F}_{m} \) by a Borel set. Then there exist \( m, n \) such that \( {E}_{n} \) cannot be separated from \( {F}_{m} \) by a Borel set.
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Proof. Suppose for every \( m, n \) there is a Borel set \( {C}_{mn} \) such that\n\n\[ \n{E}_{n} \subseteq {C}_{mn}\text{ and }{F}_{m}\bigcap {C}_{mn} = \varnothing .\n\]\n\nIt is fairly easy to check that the Borel set\n\n\[ \nC = \mathop{\bigcup }\limits_{n}\mathop{\bigcap }\limits_{m}{C}_{mn}\n\]\nseparates \( E \) from \( F \) .
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Yes
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Theorem 4.4.3 (Souslin) A subset A of a Polish space \( X \) is Borel if and only if it is both analytic and coanalytic; i.e., \( {\mathbf{\Delta }}_{1}^{1}\left( X\right) = {\mathcal{B}}_{X} \) .
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Proof. The \
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No
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Proposition 4.4.4 Suppose \( {A}_{0},{A}_{1},\ldots \) are pairwise disjoint analytic subsets of a Polish space \( X \) . Then there exist pairwise disjoint Borel sets \( {B}_{0},{B}_{1},\ldots \) such that \( {B}_{n} \supseteq {A}_{n} \) for all \( n \) .
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Proof. By 4.4.1, for each \( n \) there is a Borel set \( {C}_{n} \) such that\n\n\[ \n{A}_{n} \subseteq {C}_{n}\text{ and }{C}_{n} \cap \mathop{\bigcup }\limits_{{m \neq n}}{A}_{m} = \varnothing .\n\]\n\nTake\n\n\[ \n{B}_{n} = {C}_{n} \cap \mathop{\bigcap }\limits_{{m \neq n}}\left( {X \smallsetminus {C}_{m}}\right)\n\]
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Yes
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Theorem 4.4.5 Let \( E \subseteq X \times X \) be an analytic equivalence relation on a Polish space \( X \) . Suppose \( A \) and \( B \) are disjoint analytic subsets of \( X \) . Assume that \( B \) is invariant with respect to \( E \) (i.e., \( B \) is a union of \( E \) - equivalence classes). Then there is an \( E \) -invariant Borel set \( C \) separating \( A \) from \( B \) .
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Proof. First we note the following. Let \( D \) be an analytic subset of \( X \) and \( {D}^{ * } \) the smallest invariant set containing \( D \) . Since\n\n\[ \n{D}^{ * } = {\pi }_{X}\left( {E\bigcap \left( {D \times X}\right) }\right) \n\]\n\nwhere \( {\pi }_{X} : X \times X \rightarrow X \) is the projection to the second coordinate space, \( {D}^{ * } \) is analytic.\n\nWe show that there is a sequence \( \left( {A}_{n}\right) \) of invariant analytic sets and a sequence \( \left( {B}_{n}\right) \) of Borel sets such that\n\n(i) \( A \subseteq {A}_{0} \),\n\n(ii) \( {A}_{n} \subseteq {B}_{n} \subseteq {A}_{n + 1} \), and\n\n(iii) \( B \cap {B}_{n} = \varnothing \).\n\nTake \( {A}_{0} = {A}^{ * } \) . Since \( B \) is invariant, \( {A}_{0} \cap B = \varnothing \) . By 4.4.1, let \( {B}_{0} \) be a Borel set containing \( {A}_{0} \) and disjoint from \( B \) . Suppose \( {A}_{i},{B}_{i},0 \leq i \leq n \) , satisfying (i),(ii), and iii) have been defined. Put \( {A}_{n + 1} = {B}_{n}^{ * } \) . Since \( B \) is invariant, \( {A}_{n + 1} \cap B = \varnothing \) . By 4.4.1, let \( {B}_{n + 1} \) be a Borel set containing \( {A}_{n + 1} \) and disjoint from \( B \) .\n\nHaving defined \( \left( {A}_{n}\right) ,\left( {B}_{n}\right) \), let \( C = \mathop{\bigcup }\limits_{n}{B}_{n} \) . Clearly, \( C \) is a Borel set containing \( A \) and disjoint from \( B \) . Since \( C = \mathop{\bigcup }\limits_{n}{A}_{n} \), it is also invariant.
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Yes
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Proposition 4.5.1 Let \( A \) be an analytic subset of a Polish space, \( Y \) a Polish space, and \( f : A \rightarrow Y \) a one-to-one Borel map. Then \( f : A \rightarrow \) \( f\left( A\right) \) is a Borel isomorphism.
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Proof. Let \( B \subseteq A \) be Borel in \( A \) . We need to show that \( f\left( B\right) \) is Borel in \( f\left( A\right) \) . As both \( B \) and \( C = A \smallsetminus B \) are analytic and \( f \) Borel, \( f\left( B\right) \) and \( f\left( C\right) \) are analytic. Since \( f \) is one-to-one, these two sets are disjoint. So, by 4.4.1, there is a Borel set \( D \subseteq Y \) such that \( f\left( B\right) \subseteq D \) and \( f\left( C\right) \cap D = \varnothing \) . Since \( f\left( B\right) = D \cap f\left( A\right) \), the result follows.
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Yes
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Theorem 4.5.2 Let \( X, Y \) be Polish spaces, \( A \subseteq X \) analytic, and \( f : A \rightarrow Y \) any map. The following statements are equivalent\n\n(i) \( f \) is Borel measurable.\n\n(ii) \( \operatorname{graph}\left( f\right) \) is Borel in \( A \times Y \). \n\n(iii) \( \operatorname{graph}\left( f\right) \) is analytic.
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Proof. We only need to show that (iii) implies (i). The other implications are quite easy to see. Let \( U \) be an open set in \( Y \). As\n\n\[ \n{f}^{-1}\left( U\right) = {\pi }_{X}\left( {\operatorname{graph}\left( f\right) \bigcap \left( {X \times U}\right) }\right) , \n\]\n\nwhere \( {\pi }_{X} : X \times Y \rightarrow X \) is the projection map, it is analytic. Similarly, \( {f}^{-1}\left( {U}^{c}\right) \) is analytic. By 4.4.1, there is a Borel set \( B \subseteq X \) such that\n\n\[ \n{f}^{-1}\left( U\right) \subseteq B\text{ and }B\bigcap {f}^{-1}\left( {U}^{c}\right) = \varnothing . \n\]\n\nSince \( {f}^{-1}\left( U\right) = B \cap A \), it is Borel in \( A \), and the result follows.
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Yes
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Corollary 4.5.5 Let \( X \) be a standard Borel space and \( Y \) a metrizable space. Suppose there is a one-to-one Borel map \( f \) from \( X \) onto \( Y \) . Then \( Y \) is standard Borel and \( f \) a Borel isomorphism.
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Proof. By 4.3.8, \( Y \) is separable. The result follows from 4.5.4.
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Yes
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Theorem 4.5.7 (Blackwell - Mackey theorem, [13]) Let \( X \) be an analytic subset of a Polish space and \( \mathcal{A} \) a countably generated sub \( \sigma \) -algebra of the Borel \( \sigma \) -algebra \( {\mathcal{B}}_{X} \) . Let \( B \subseteq X \) be a Borel set that is a union of atoms of \( \mathcal{A} \) . Then \( B \in \mathcal{A} \) .
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Proof. Let \( \left\{ {{B}_{n} : n \in \mathbb{N}}\right\} \) be a countable generator of \( \mathcal{A} \) . Consider the map \( f : X \rightarrow {2}^{\mathbb{N}} \) defined by\n\n\[ f\left( x\right) = \left( {{\chi }_{{B}_{0}}\left( x\right) ,{\chi }_{{B}_{1}}\left( x\right) ,\ldots }\right) ,\;x \in X. \]\n\nThen \( \mathcal{A} = {f}^{-1}\left( {\mathcal{B}}_{{2}^{\mathbb{N}}}\right) \) . In particular, \( f : X \rightarrow {2}^{\mathbb{N}} \) is Borel measurable. So, \( f\left( B\right) \) and \( f\left( {B}^{c}\right) \) are disjoint analytic subsets of \( {2}^{\mathbb{N}} \) . By 4.4.1, there is a Borel set \( C \) containing \( f\left( B\right) \) and disjoint from \( f\left( {B}^{c}\right) \) . Clearly, \( B = {f}^{-1}\left( C\right) \), and so it belongs to \( \mathcal{A} \) .
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Yes
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Corollary 4.5.10 Let \( X \) be an analytic subset of a Polish space and \( {\mathcal{A}}_{1} \) , \( {\mathcal{A}}_{2} \) two countably generated sub \( \sigma \) -algebras of \( {\mathcal{B}}_{X} \) with the same set of atoms. Then \( {\mathcal{A}}_{1} = {\mathcal{A}}_{2} \) . In particular, if \( \mathcal{A} \) is a countably generated sub \( \sigma \) -algebra containing all the singletons, then \( \mathcal{A} = {\mathcal{B}}_{X} \) .
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Null
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No
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Theorem 4.6.1 (The generalized first separation theorem, Novikov[90]) Let \( \\left( {A}_{n}\\right) \) be a sequence of analytic subsets of a Polish space \( X \) such that \( \\bigcap {A}_{n} = \\varnothing \) . Then there exist Borel sets \( {B}_{n} \\supseteq {A}_{n} \) such that \( \\bigcap {B}_{n} = \\varnothing \) .
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Proof of 4.6.1. (Mokobodzki [86]) Let \( \\left( {A}_{n}\\right) \) be a sequence of analytic sets that is not Borel separated and such that \( \\mathop{\\bigcap }\\limits_{n}{A}_{n} = \\varnothing \) . For each \( n \), fix a continuous surjection \( {f}_{n} : {\\mathbb{N}}^{\\mathbb{N}} \\rightarrow {A}_{n} \) . We get a sequence \( {\\alpha }_{0},{\\alpha }_{1},\\ldots \) in \( {\\mathbb{N}}^{\\mathbb{N}} \) such that for every \( k > 0 \) the sequence\n\n\[ \n{f}_{0}\\left( {\\sum \\left( {{\\alpha }_{0} \\mid k}\\right) }\\right) ,{f}_{1}\\left( {\\sum \\left( {{\\alpha }_{1} \\mid \\left( {k - 1}\\right) }\\right) }\\right) ,\\ldots ,{f}_{k - 1}\\left( {\\sum \\left( {{\\alpha }_{k - 1} \\mid 1}\\right) }\\right) ,{A}_{k},{A}_{k + 1},\\ldots \n\]\n\nis not Borel separated.\n\nTo see that such a sequence exists we proceed by induction. Write \( {A}_{0} = \\mathop{\\bigcup }\\limits_{n}{f}_{0}\\left( {\\sum \\left( n\\right) }\\right) \) . By 4.6.2, there exists \( {\\alpha }_{0}\\left( 0\\right) \\in \\mathbb{N} \) such that the sequence \( {f}_{0}\\left( {\\sum \\left( {{\\alpha }_{0}\\left( 0\\right) }\\right) }\\right) ,{A}_{1},{A}_{2},\\ldots \) is not Borel separated. Write \( {f}_{0}\\left( {\\sum \\left( {{\\alpha }_{0}\\left( 0\\right) }\\right) }\\right) = \\mathop{\\bigcup }\\limits_{m}{f}_{0}\\left( {\\sum \\left( {{\\alpha }_{0}\\left( 0\\right) m}\\right) }\\right) \) and \( {A}_{1} = \\mathop{\\bigcup }\\limits_{n}{f}_{1}\\left( {\\sum \\left( n\\right) }\\right) \) . Apply 4.6.2 again to get \( {\\alpha }_{0}\\left( 1\\right) ,{\\alpha }_{1}\\left( 0\\right) \\in \\mathbb{N} \) such that the sequence \( {f}_{0}\\left( {\\sum \\left( {{\\alpha }_{0}\\left( 0\\right) {\\alpha }_{0}\\left( 1\\right) }\\right) }\\right) ,{f}_{1}\\left( {\\sum \\left( {{\\alpha }_{1}\\left( 0\\right) }\\right) }\\right) ,{A}_{2},{A}_{2},\\ldots \n\n
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No
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Lemma 4.6.2 Let \( \left( {E}_{n}\right) \) be a sequence of subsets of \( X, k \in \mathbb{N} \), and \( {E}_{i} = \) \( \mathop{\bigcup }\limits_{n}{E}_{in} \) for \( i \leq k \) . Suppose \( \left( {E}_{n}\right) \) is not Borel separated. Then there exist \( {n}_{0},{n}_{1},\ldots ,{n}_{k} \) such that the sequence \( {E}_{0{n}_{0}},{E}_{1{n}_{1}},\ldots ,{E}_{k{n}_{k}},{E}_{k + 1},{E}_{k + 2},\ldots \) is not Borel separated.
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Proof. We prove the result by induction on \( k \) .\n\nInitial step: \( k = 0 \) . Suppose the result is not true. Hence, for every \( n \) , there is a sequence \( {\left( {B}_{in}\right) }_{i \in \mathbb{N}} \) of Borel sets such that\n\n(i) \( \mathop{\bigcap }\limits_{i}{B}_{in} = \varnothing \) ,\n\n(ii) \( {B}_{0n} \supseteq {E}_{0n} \), and\n\n(iii) \( {B}_{in} \supseteq {E}_{i} \) for all \( i \) .\n\nLet\n\n\[ {B}_{i} = \mathop{\bigcup }\limits_{n}{B}_{in}\;\text{ if }\;i = 0, \]\n\n\[ = \mathop{\bigcap }\limits_{n}{B}_{in}\;\text{if}\;i > 0\text{.} \]\n\nThen \( {B}_{i} \supseteq {E}_{i} \), the \( {B}_{i} \) ’s are Borel and \( \bigcap {B}_{i} = \varnothing \) . This contradicts the hypothesis that \( \left( {E}_{n}\right) \) is not Borel separated, and we have proved the result for \( k = 0 \) .\n\nInductive step. Suppose \( k > 0 \) and the result is true for all integers less than \( k \) . By the induction hypothesis, there are integers \( {n}_{0},{n}_{1},\ldots ,{n}_{k - 1} \) such that \( {E}_{0{n}_{0}},{E}_{1{n}_{1}},\ldots ,{E}_{k - 1{n}_{k - 1}},{E}_{k},{E}_{k + 1},\ldots \) is not Borel separated. By the initial step, there is an \( {n}_{k} \) such that \( {E}_{0{n}_{0}},{E}_{1{n}_{1}},\ldots ,{E}_{k{n}_{k}},{E}_{k + 1},{E}_{k + 2},\ldots \) is not Borel separated.
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Yes
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Corollary 4.6.3 Let \( \\left( {A}_{n}\\right) \) be a sequence of analytic subsets of a Polish space \( X \) such that \( \\lim \\sup {A}_{n} = \\varnothing \) . Then there exist Borel sets \( {B}_{n} \\supseteq {A}_{n} \) such that \( \\lim \\sup {B}_{n} = \\varnothing \) .
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Null
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No
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Theorem 4.6.5 (Weak reduction principle for coanalytic sets) Let \( {C}_{0},{C}_{1},{C}_{2},\ldots \) be a sequence of coanalytic subsets of a Polish space such that \( \bigcup {C}_{n} \) is Borel. Then there exist pairwise disjoint Borel sets \( {B}_{n} \subseteq {C}_{n} \) such that \( \bigcup {B}_{n} = \bigcup {C}_{n} \) .
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Proof. Let \( {A}_{n} = X \smallsetminus {C}_{n} \), where \( X = \mathop{\bigcup }\limits_{n}{C}_{n} \) . Then \( \left( {A}_{n}\right) \) is a sequence of analytic sets such that \( \mathop{\bigcap }\limits_{n}{A}_{n} = \varnothing \) . By 4.6.1, there exist Borel sets \( {D}_{n} \supseteq {A}_{n} \) such that \( \mathop{\bigcap }\limits_{n}{D}_{n} = \varnothing \) . Take\n\n\[ \n{B}_{n} = {B}_{n}^{\prime } \smallsetminus \mathop{\bigcup }\limits_{{m < n}}{B}_{m}^{\prime }\n\]\n\nwhere \( {B}_{n}^{\prime } = X \smallsetminus {D}_{n} \) .
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Yes
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Theorem 4.7.1 (Saint Raymond[97]) Let \( {A}_{0} \) and \( {A}_{1} \) be disjoint analytic subsets of \( X \times Y \) with the sections \( {\left( {A}_{0}\right) }_{x}, x \in X \), closed in \( Y \). Then there is a sequence \( \left( {B}_{n}\right) \) of Borel subsets of \( X \) such that \[ {A}_{1} \subseteq \mathop{\bigcup }\limits_{n}\left( {{B}_{n} \times {V}_{n}}\right) \text{ and }{A}_{0}\bigcap \mathop{\bigcup }\limits_{n}\left( {{B}_{n} \times {V}_{n}}\right) = \varnothing . \]
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Proof. By 4.4.1, there is a Borel set containing \( {A}_{1} \) and disjoint from \( {A}_{0} \). So, without any loss of generality, we assume that \( {A}_{1} \) is Borel. For each \( n \), let \[ {C}_{n} = \left\{ {x \in X : {V}_{n} \subseteq {\left( {A}_{0}\right) }_{x}^{c}}\right\} \] Then \( {C}_{n} \) is coanalytic and \[ {\left( {A}_{0}\right) }^{c} = \mathop{\bigcup }\limits_{n}\left( {{C}_{n} \times {V}_{n}}\right) \] Note that \( \left( {\left( {{C}_{n} \times {V}_{n}}\right) \bigcap {A}_{1}}\right) \) is a sequence of coanalytic sets whose union is Borel. Hence, by 4.6.5, there exist Borel sets \( {D}_{n} \subseteq \left( {{C}_{n} \times {V}_{n}}\right) \bigcap {A}_{1} \) such that \[ \bigcup {D}_{n} = \mathop{\bigcup }\limits_{n}\left( {{A}_{1}\bigcap \left( {{C}_{n} \times {V}_{n}}\right) }\right) = {A}_{1}. \] By 4.4.1, there exist Borel sets \( {B}_{n} \) such that \[ {\pi }_{X}\left( {D}_{n}\right) \subseteq {B}_{n} \subseteq {C}_{n} \] where \( {\pi }_{X} : X \times Y \rightarrow X \) is the projection map. It is now fairly easy to see that \( \left( {B}_{n}\right) \) satisfies \( \left( \star \right) \).
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Yes
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Theorem 4.7.2 (Kunugui, Novikov) Suppose \( B \subseteq X \times Y \) is any Borel set with sections \( {B}_{x} \) open, \( x \in X \) . Then there is a sequence \( \left( {B}_{n}\right) \) of Borel subsets of \( X \) such that\n\n\[ B = \bigcup \left( {{B}_{n} \times {V}_{n}}\right) \]
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Proof. Apply 4.7.1 to \( {A}_{0} = {B}^{c} \) and \( {A}_{1} = B \) .
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No
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Corollary 4.7.3 Let \( {A}_{0} \) and \( {A}_{1} \) be disjoint analytic subsets of \( X \times Y \) with sections \( {\left( {A}_{0}\right) }_{x} \) and \( {\left( {A}_{1}\right) }_{x} \) closed for all \( x \in X \) . Then there exist disjoint Borel sets \( {B}_{0} \) and \( {B}_{1} \) with closed sections such that \( {A}_{0} \subseteq {B}_{0} \) and \( {A}_{1} \subseteq {B}_{1} \) .
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Null
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No
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Corollary 4.7.4 Suppose \( B \subseteq X \times Y \) is a Borel set with the sections \( {B}_{x} \) closed. Then there is a Polish topology \( \mathcal{T} \) finer than the given topology on \( X \) generating the same Borel \( \sigma \) -algebra such that \( B \) is closed relative to the product topology on \( X \times Y, X \) being equipped with the new topology \( \mathcal{T} \) .
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Proof. By 4.7.2, write\n\n\[ \n{B}^{c} = \mathop{\bigcup }\limits_{n}\left( {{B}_{n} \times {V}_{n}}\right) \n\]\n\nthe \( {B}_{n} \) ’s Borel. By 3.2.5, take a finer Polish topology \( \mathcal{T} \) on \( X \) generating the same Borel \( \sigma \) -algebra such that \( {B}_{n} \) is \( \mathcal{T} \) -open.
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Yes
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Example 4.7.9 (H. Sarbadhikari) Let \( A \subseteq \left\lbrack {0,1}\right\rbrack \) be an analytic non-Borel set and \( E \subseteq \left\lbrack {0,1}\right\rbrack \times {\mathbb{N}}^{\mathbb{N}} \) a closed set whose projection is \( A \) . Set \( B = \) \( E\bigcup \left( {\left( {\left\lbrack {0,1}\right\rbrack \smallsetminus A}\right) \times {\mathbb{N}}^{\mathbb{N}}}\right) \) and \( f : B \rightarrow \left\lbrack {0,1}\right\rbrack \) the characteristic function of \( E \) . We claim that there is no Borel extension \( F : \left\lbrack {0,1}\right\rbrack \times {\mathbb{N}}^{\mathbb{N}} \rightarrow \left\lbrack {0,1}\right\rbrack \) of \( f \) such that \( y \rightarrow F\left( {x, y}\right) \) is continuous.
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Suppose not. Consider \( C = {F}^{-1}(\left( {0,1\rbrack }\right) \) . Then \( C \) is a Borel set with sections \( {C}_{x} \) open and whose projection is \( A \) . Hence \( A \) is Borel. (See the paragraph below.) We have arrived at a contradiction.
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Yes
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Theorem 4.7.11 (Novikov) Let \( X \) and \( Y \) be Polish spaces and \( B \) a Borel subset of \( X \times Y \) with sections \( {B}_{x} \) compact. Then \( {\pi }_{X}\left( B\right) \) is Borel in \( X \) .
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Proof. (Srivastava) Since every Polish space is homeomorphic to a \( {G}_{\delta } \) subset of the Hilbert cube \( \mathbb{H} \), without any loss of generality, we assume that \( Y \) is a compact metric space. Note that the sections \( {B}_{x} \) are closed in \( Y \) . By 4.7.4, there is a finer Polish topology on \( X \) generating the same Borel \( \sigma \) -algebra and making \( B \) closed in \( X \times Y \) . Hence, by 2.3.24, \( {\pi }_{X}\left( B\right) \) is closed in \( X, X \) being equipped with the new topology. But the Borel structure of \( X \) is the same with respect to both the topologies. The result follows.
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Yes
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Corollary 4.7.12 Let \( X, Y \) be Polish spaces with \( {Y\sigma } \) -compact (equivalently, locally compact). Then the projection of every Borel set \( B \) in \( X \times Y \) with \( x \) -sections closed in \( Y \) is Borel.
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Proof. Write \( Y = \mathop{\bigcup }\limits_{n}{Y}_{n},{Y}_{n} \) compact. Then\n\n\[ \n{\pi }_{X}\left( B\right) = \mathop{\bigcup }\limits_{n}{\pi }_{X}\left( {B\bigcap \left( {X \times {Y}_{n}}\right) }\right) \n\] \n\nNow apply 4.7.11.
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Yes
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Theorem 4.8.1 Let \( \left( {G, \cdot }\right) \) be a Polish group and \( H \) a closed subgroup. Suppose \( E = \left\{ {\left( {x, y}\right) : x \cdot {y}^{-1} \in H}\right\} \) ; i.e., \( E \) is the equivalence relation induced by the right cosets. Then the \( \sigma \) -algebra of invariant Borel sets is countably generated.
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Proof. Let \( \left\{ {{U}_{n} : n \in \mathbb{N}}\right\} \) be a countable base for the topology of \( G \) . Put\n\n\[ \n{B}_{n} = \mathop{\bigcup }\limits_{{y \in H}}y \cdot {U}_{n} \n\]\n\nSo, the \( {B}_{n} \) ’s are Borel (in fact, open). We show that \( \left\{ {{B}_{n} : n \in \mathbb{N}}\right\} \) generates \( \mathcal{B} \) .\n\nLet \( {H}_{1} \) and \( {H}_{2} \) be two distinct cosets. Since \( H \) is closed, \( {H}_{1} \) and \( {H}_{2} \) are closed. Since they are disjoint, there is a basic open set \( {U}_{n} \) such that \( {U}_{n}\bigcap {H}_{1} \neq \varnothing \) and \( {U}_{n}\bigcap {H}_{2} = \varnothing \) . Then \( {H}_{1} \subset {B}_{n} \) and \( {B}_{n}\bigcap {H}_{2} = \varnothing \) . It follows that the right cosets are precisely the atoms of \( \sigma \left( \left\{ {{B}_{n} : n \in \mathbb{N}}\right\} \right) \) . The result now follows from 4.5.7.
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Yes
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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
|
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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Yes
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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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Yes
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Example 4.8.13 Consider the additive group \( \left( {\mathbb{R}, + }\right) \) of real numbers. Let \( \left( {\mathbb{R},\mathcal{T}}\right) \) be the topological sum \( \left( {\mathbb{R}\smallsetminus \{ 0\} \text{, usual topology}}\right) \oplus \{ 0\} \) So, \( \mathcal{T} \) is generated by the usual open sets and \( \{ 0\} \) . Clearly, \( \mathcal{T} \) is a Polish topology on \( \mathbb{R} \) inducing the usual Borel \( \sigma \) -algebra. In particular, the addition \( \left( {x, y}\right) \rightarrow \) \( x + y \) is Borel. If \( \left( {\mathbb{R},\mathcal{T}}\right) \) were a topological group it would be discrete, which is not the case.
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Null
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No
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Example 4.8.14 (G. Hjorth) Under AC, there is a discontinuous group isomorphism \( \varphi : \mathbb{R} \rightarrow \mathbb{R} \) . Take \( G \) to be \( \mathbb{R} \times \mathbb{R} \) with the product topology and the group operation defined by\n\n\[ \left( {r, s}\right) \cdot \left( {p, q}\right) = \left( {r + {2}^{\varphi \left( s\right) }p, s + q}\right) ,\]\n\ni.e., the group is a semidirect product of two copies of \( \mathbb{R} \) with respect to the homomorphism \( \bar{\varphi } : \mathbb{R} \rightarrow \operatorname{Aut}\left( \mathbb{R}\right) \) naturally induced by \( \varphi ,\bar{\varphi }\left( s\right) \left( p\right) = \) \( {2}^{\varphi \left( s\right) }p \) .
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Null
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No
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Theorem 4.9.1 (Moschovakis) Every \( {\mathbf{\Pi }}_{1}^{1} \) set \( A \) in a Polish space \( X \) admits a \( {\mathbf{\Pi }}_{1}^{1} \) -norm \( \varphi : A \rightarrow {\omega }_{1} \) .
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Null
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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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No
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Example 4.9.17 Let \( {U}_{0},{U}_{1} \) be a universal pair of analytic subsets of \( {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \) (4.1.17). Suppose there exist pairwise disjoint analytic sets \( {V}_{0} \subseteq \) \( {U}_{0},{V}_{1} \subseteq {U}_{1} \) such that \( {V}_{0}\bigcup {V}_{1} = {U}_{0}\bigcup {U}_{1} \) . By the first separation theorem for analytic sets,(4.4.1), there is a Borel set \( B \) containing \( {V}_{0} \) and disjoint from \( {V}_{1} \) . We claim that \( B \) is universal Borel, which contradicts 3.6.9. To prove our claim, take any Borel \( E \subseteq {\mathbb{N}}^{\mathbb{N}} \) . Since \( {U}_{0},{U}_{1} \) is a universal pair of analytic sets, there is an \( \alpha \) such that \( E = {\left( {U}_{0}\right) }_{\alpha } \) and \( {E}^{c} = {\left( {U}_{1}\right) }_{\alpha } \) . Plainly, \( E = {B}_{\alpha } \) .
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Null
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No
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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 4.9.21 Let \( {C}_{0} \) and \( {C}_{1} \) be disjoint coanalytic subsets of \( I = \left\lbrack {0,1}\right\rbrack \) that are not Borel separated; i.e., there is no Borel set containing \( {C}_{0} \) and disjoint from \( {C}_{1} \) . Let\n\n\[ \n{A}_{0} = \left( {I\times \{ 0\} }\right) \bigcup \left( {{C}_{0} \times \left\lbrack {0,3/4}\right\rbrack }\right)\n\]\n\nand\n\n\[ \n{A}_{1} = \left( {I\times \{ 1\} }\right) \bigcup \left( {{C}_{1} \times \left\lbrack {1/4,1}\right\rbrack }\right) .\n\]\n\nClearly, \( {A}_{0},{A}_{1} \) are disjoint coanalytic sets with sections closed. Suppose there is a Borel map \( u : I \times I \rightarrow I \) such that \( u \mid B \) is the characteristic\n\nfunction of \( {A}_{1} \), where \( B = {A}_{0}\bigcup {A}_{1} \) . Then, the set\n\n\[ \nE = \{ x \in I : u\left( {x,1/2}\right) = 0\}\n\]\n\nis Borel and separates \( {C}_{0} \) from \( {C}_{1} \).
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Null
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No
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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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Example 4.10.2 (Separation capacity) Let \( X \) be a polish space. Define \( I : \mathcal{P}\left( {X \times X}\right) \rightarrow \{ 0,1\} \) by\n\n\[ I\left( A\right) = \left\{ \begin{array}{ll} 0 & \text{ if }{\pi }_{1}\left( A\right) \cap {\pi }_{2}\left( A\right) = \varnothing , \\ 1 & \text{ otherwise,} \end{array}\right. \]\n\nwhere \( {\pi }_{1} \) and \( {\pi }_{2} \) are the two projection maps on \( X \times X \) . For \( A \subseteq X \times X \) , set\n\n\[ R\left\lbrack A\right\rbrack = {\pi }_{1}\left( A\right) \times {\pi }_{2}\left( A\right) \]\n\ni. e., \( R\left\lbrack A\right\rbrack \) is the smallest rectangle containing \( A \) . Clearly, \( I\left( A\right) = 0 \) if and only if \( R\left\lbrack A\right\rbrack \) is disjoint from the diagonal in \( X \times X \) . It is easy to verify that \( I \) is a capacity on \( X \times X \) . Later in this section, using this capacity we shall give a rather beautiful proof of the first separation theorem for analytic sets. Because of this, \( I \) is called the separation capacity on \( X \times X \) .
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Null
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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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Example 4.10.6 Consider \( I : \mathcal{P}\left( {\mathbb{N}}^{\mathbb{N}}\right) \rightarrow \{ 0,1\} \) defined by \[ I\left( A\right) = \left\{ \begin{array}{ll} 0 & \text{ if }A\text{ is contained in a }{K}_{\sigma }\text{ set,} \\ 1 & \text{ otherwise. } \end{array}\right. \] Then \( I \) satisfies the conditions (i),(ii), and (iii) of the definition of a capacity. Further, if \( \left( {K}_{n}\right) \) is a nonincreasing sequence of compact sets with intersection, say, \( K \), then \( I\left( {K}_{n}\right) \rightarrow I\left( K\right) \) . But since no open set in \( {\mathbb{N}}^{\mathbb{N}} \) is contained in a \( {K}_{\sigma }, I \) is not right-continuous over compacta.
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Null
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No
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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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Yes
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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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Corollary 4.12.2 Let \( X, Y \) be Polish spaces and \( B \subseteq X \times Y \) a Borel set. Then the set\n\n\[ Z = \\left\\{ {x \\in X : {B}_{x}\\text{ is a singleton }}\\right\\} \]\n\nis coanalytic.
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Null
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No
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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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Example 5.1.7 (Blackwell[16]) Let \( {C}_{1},{C}_{2} \) be two disjoint coanalytic subsets of \( \left\lbrack {0,1}\right\rbrack \) that cannot be separated by Borel sets. The existence of such sets has been shown in (4.9.17). Let \( {B}_{i} \) be a closed subset of \( \left\lbrack {0,1}\right\rbrack \times \sum \left( i\right) \) whose projection is \( \left\lbrack {0,1}\right\rbrack \smallsetminus {C}_{i}, i = 1 \) or 2 . Take \( B = {B}_{1} \cup {B}_{2} \) . Then \( B \) is a closed subset of \( \left\lbrack {0,1}\right\rbrack \times {\mathbb{N}}^{\mathbb{N}} \) whose projection is \( \left\lbrack {0,1}\right\rbrack \) . Suppose there exists a Borel section \( f : \left\lbrack {0,1}\right\rbrack \rightarrow {\mathbb{N}}^{\mathbb{N}} \) of \( B \) . Then \( {f}^{-1}\left( {\sum \left( 2\right) }\right) \) is a Borel set containing \( {C}_{1} \) and disjoint from \( {C}_{2} \) . But no such Borel set exists. Thus \( B \) does not admit a Borel uniformization.
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Null
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No
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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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No
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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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Yes
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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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Yes
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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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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\[ \chi \left( x\right) = \left( {{\chi }_{{B}_{0}}\left( x\right) ,{\chi }_{{B}_{1}}\left( x\right) ,{\chi }_{{B}_{2}}\left( x\right) ,\ldots }\right) ,\;x \in X. \]\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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Yes
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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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Yes
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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\n\n\[ \mathcal{I}\left( x\right) = \{ I \subseteq Y : I\text{ is meager in }G\left( x\right) \} .\n\nBy imitating the proof of 3.5.18 we can show the following:\n\nFor every open set \( U \) in \( Y \) and every Borel set \( B \) in \( X \times Y \), the sets\n\n\[ {B}^{*U} = \left\{ {x \in X : G\left( x\right) \bigcap U \neq \varnothing }\right.\n\n\[ \text{&}{B}_{x}\bigcap G\left( x\right) \bigcap U\text{is comeager in}G\left( x\right) \bigcap U\}\n\nand\n\n\[ {B}^{\Delta U} = \left\{ {x \in X : G\left( x\right) \bigcap U \neq \varnothing }\right.\n\n\[ \text{&}{B}_{x}\bigcap G\left( x\right) \bigcap U\text{is nonmeager in}G\left( x\right) \bigcap U\}\n\nare Borel.\n\nIt follows that \( \mathcal{I} \) is Borel on Borel.
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Null
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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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