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Proposition 3.1.32 Let \( X \) and \( Y \) be metrizable spaces. Then every Baire function \( f : X \rightarrow Y \) is Borel. | Proof. Since every continuous function is Borel and since the limit of a pointwise convergent sequence of Borel functions is Borel (3.1.27), Baire functions are Borel. | Yes |
Lemma 3.1.35 Let \( X \) be a metrizable space and \( B \subseteq X \) Borel. Then \( {\chi }_{B} : X \rightarrow \mathbb{R} \) is Baire. | Proof. Let\n\n\[ \mathcal{B} = \left\{ {B \subseteq X : {\chi }_{B}\text{ is Baire }}\right\} .\n\]\n\n(a) Let \( U \) be open in \( X \) . Write \( U = \mathop{\bigcup }\limits_{n}{F}_{n} \), where the \( {F}_{n} \) ’s are closed and \( {F}_{n} \subseteq {F}_{n + 1} \) . By 2.1.18, there is a continuous function \( {f... | Yes |
Theorem 3.1.36 (Lebesgue - Hausdorff theorem) Every real-valued Borel function defined on a metrizable space is Baire. | Proof. By 3.1.35 the characteristic function of every Borel set is Baire. Hence, by 3.1.34(ii), every simple Borel function is Baire. Now the result follows from 3.1.28. | Yes |
Lemma 3.2.1 Let \( \\left( {X,\\mathcal{T}}\\right) \) be a (zero-dimensional, second countable) metrizable space and \( \\left( {B}_{n}\\right) \) a sequence of Borel subsets of \( X \) . Then there is a (respectively zero-dimensional, second countable) metrizable topology \( {\\mathcal{T}}^{\\prime } \) such that \( ... | Proof. Define \( f : X \\rightarrow X \\times \\mathcal{C} \) by\n\n\[ f\\left( x\\right) = \\left( {x,{\\chi }_{{B}_{0}}\\left( x\\right) ,{\\chi }_{{B}_{1}}\\left( x\\right) ,{\\chi }_{{B}_{2}}\\left( x\\right) ,\\ldots }\\right) .\n\]\n\nThis map is clearly one-to-one. Let\n\n\[ {\\mathcal{T}}^{\\prime } = \\left\{ ... | Yes |
Proposition 3.2.3 Let \( \left( {X,\mathcal{T}}\right) \) be a metrizable space, \( A \subseteq X, Y \) Polish, and \( f : A \rightarrow Y \) any Borel map. Then\n\n(i) there is a finer metrizable topology \( {\mathcal{T}}^{\prime } \) on \( X \) generating the same Borel \( \sigma \) -algebra such that \( f : A \right... | Proof. Fix a countable base \( \left( {U}_{n}\right) \) for \( Y \) . Let \( n \in \mathbb{N} \) . As \( {f}^{-1}\left( {U}_{n}\right) \) is Borel in \( A \), there is a Borel set \( {B}_{n} \) in \( X \) such that \( {f}^{-1}\left( {U}_{n}\right) = A \cap {B}_{n} \) . Take \( {\mathcal{T}}^{\prime } \) as in 3.2.1. Th... | Yes |
Theorem 3.2.4 Suppose \( \left( {X,\mathcal{T}}\right) \) is a Polish space. Then for every Borel set \( B \) in \( X \) there is a finer Polish topology \( {\mathcal{T}}_{B} \) on \( X \) such that \( B \) is clopen with respect to \( {\mathcal{T}}_{B} \) and \( \sigma \left( \mathcal{T}\right) = \sigma \left( {\mathc... | Proof of 3.2.4. Let \( \mathcal{B} \) be the class of all Borel subsets \( B \) of \( X \) such that there is a finer Polish topology \( {\mathcal{T}}_{B} \) generating \( {\mathcal{B}}_{X} \) and making \( B \) clopen.\n\nBy Observation 1, \( \mathcal{B} \) contains all closed sets, and it is clearly closed under comp... | Yes |
Every uncountable Borel subset of a Polish space contains a homeomorph of the Cantor set. In particular, it is of cardinality \( \mathfrak{c} \) . | Proof. Let \( \left( {X,\mathcal{T}}\right) \) be Polish and \( B \) an uncountable Borel subset of \( X \) . By 3.2.4, let \( {\mathcal{T}}^{\prime } \) be a finer Polish topology on \( X \) making \( B \) closed. By 2.6.3, \( \left( {B,{\mathcal{T}}^{\prime } \mid B}\right) \) contains a homeomorph of the Cantor set,... | Yes |
Corollary 3.2.6 Suppose \( \left( {X,\mathcal{T}}\right) \) is a Polish space, \( Y \) a separable metric space, and \( f : X \rightarrow Y \) a Borel map. Then there is a finer Polish topology \( {\mathcal{T}}^{\prime } \) on \( X \) generating the same Borel \( \sigma \) -algebra such that \( f : \left( {X,{\mathcal{... | Null | No |
Theorem 3.2.7 Every uncountable Borel subset of a Polish space contains a homeomorph of the Cantor set. In particular, it is of cardinality \( \mathfrak{c} \) . | Proof. Let \( \left( {X,\mathcal{T}}\right) \) be Polish and \( B \) an uncountable Borel subset of \( X \) . By 3.2.4, let \( {\mathcal{T}}^{\prime } \) be a finer Polish topology on \( X \) making \( B \) closed. By 2.6.3, \( \left( {B,{\mathcal{T}}^{\prime } \mid B}\right) \) contains a homeomorph of the Cantor set,... | Yes |
Example 3.2.8 By 2.6.4, there are exactly \( \mathfrak{c} \) uncountable closed subsets of \( \mathbb{R} \) . Let \( \left\{ {{C}_{\alpha } : \alpha < \mathfrak{c}}\right\} \) be an enumeration of these. We shall get distinct points \( {x}_{\alpha },{y}_{\alpha },\alpha < \mathfrak{c} \), such that \( {x}_{\alpha },{y}... | To define the \( {x}_{\alpha } \) ’s and \( {y}_{\alpha } \) ’s, we proceed by transfinite induction. Choose \( {x}_{0},{y}_{0} \in {C}_{0} \) with \( {x}_{0} \neq {y}_{0} \) . Let \( \alpha < \mathfrak{c} \) . Suppose \( {x}_{\beta },{y}_{\beta } \) has been chosen for all \( \beta < \alpha \) . Let \( D = \left\{ {{x... | Yes |
Example 3.3.1 The closed unit interval \( I = \left\lbrack {0,1}\right\rbrack \) and the Cantor set \( \mathcal{C} \) are Borel isomorphic. | Proof. Let \( D \) be the set of all dyadic rationals in \( I \) and \( E \subset \mathcal{C} \) the set of all sequences of 0 's and 1's that are eventually constant. Define \( f : \mathcal{C} \smallsetminus E \rightarrow I \smallsetminus D \) by\n\n\[ f\left( {{\epsilon }_{0},{\epsilon }_{1},{\epsilon }_{2},\ldots }\... | Yes |
Proposition 3.3.2 Suppose \( \\left( {X,\\mathcal{A}}\\right) \) is a measurable space with \( \\mathcal{A} \) countably generated. Then there is a subset \( Z \) of \( \\mathcal{C} \) and a bimeasurable map \( g : X \\rightarrow \) \( Z \) such that for any \( x, y \) in \( X, g\\left( x\\right) = g\\left( y\\right) \... | Proof. Let \( \\mathcal{G} = \\left\\{ {{A}_{n} : n \\in \\mathbb{N}}\\right\\} \) be a countable generator of \( \\mathcal{A} \) . Define \( g : X \\rightarrow \\mathcal{C} \) by\n\n\[ g\\left( x\\right) = \\left( {{\\chi }_{{A}_{0}}\\left( x\\right) ,{\\chi }_{{A}_{1}}\\left( x\\right) ,{\\chi }_{{A}_{2}}\\left( x\\r... | Yes |
Proposition 3.3.4 Let \( \\left( {X,\\mathcal{A}}\\right) \) be a measurable space, \( Y \) a Polish space, \( A \\subseteq X \), and \( f : A \\rightarrow Y \) a measurable map. Then \( f \) admits a measurable extension to \( X \) . | Proof. Fix a countable base \( \\left( {U}_{n}\\right) \) for \( Y \) . For every \( n \), choose \( {B}_{n} \\in \\mathcal{A} \) such that \( {f}^{-1}\\left( {U}_{n}\\right) = {B}_{n}\\bigcap A \) . Without loss of generality, we assume that \( \\mathcal{A} = \\sigma \\left( \\left( {B}_{n}\\right) \\right) \) . By 3.... | Yes |
Proposition 3.3.6 Let \( X \) and \( Y \) be measurable spaces and \( f : X \rightarrow Y \) , \( g : Y \rightarrow X \) one-to-one, bimeasurable maps. Then \( X \) and \( Y \) are isomorphic. | Proof. As \( f \) and \( g \) are bimeasurable, the set \( E \) described in the proof of the Schröder - Bernstein theorem (1.2.3) is measurable. So the bijection \( h : X \rightarrow Y \) obtained there is bimeasurable. | Yes |
Proposition 3.3.7 Let \( X \) be a second countable metrizable space. Then the following statements are equivalent.\n\n(i) \( X \) is standard Borel.\n\n(ii) \( X \) is Borel in its completion \( \widehat{X} \) .\n\n(iii) \( X \) is homeomorphic to a Borel subset of a Polish space. | Proof. Clearly, (ii) implies (iii), and (i) follows from (iii). We show that (i) implies (ii).\n\nLet \( X \) be standard Borel. Then, there is a Polish space \( Z \), a Borel subset \( Y \) of \( Z \), and a Borel isomorphism \( f : X \rightarrow Y \) . By 3.3.5, there is a Borel isomorphism \( g : {X}^{\prime } \righ... | Yes |
Theorem 3.3.10 The Effros Borel space of a Polish space is standard Borel. | Proof. Let \( Y \) be a compact metric space containing \( X \) as a dense subspace. By 2.2.7, \( X \) is a \( {G}_{\delta } \) set in \( Y \) . Write \( X = \bigcap {U}_{n},{U}_{n} \) open in \( Y \) . Let \( \left( {V}_{n}\right) \) be a countable base for \( Y \) . Now consider\n\n\[ \mathcal{Z} = \{ \operatorname{c... | Yes |
Theorem 3.3.13 (The Borel isomorphism theorem) Any two uncountable standard Borel spaces are Borel isomorphic. | Proof of 3.3.13. Let \( B \) be an uncountable standard Borel space. Without loss of generality, we assume that \( B \) is a Borel subset of some Polish space. By 3.3.14, there is a bimeasurable bijection from \( B \) into \( \mathcal{C} \). By 3.2.7, \( B \) contains a homeomorph of the Cantor set. By 3.3.6, \( B \) i... | Yes |
Lemma 3.3.14 Every standard Borel space \( B \) is Borel isomorphic to a Borel subset of \( \mathcal{C} \) . | Proof. By 3.3.1, \( I \) and \( \mathcal{C} \) are Borel isomorphic. Therefore, the Hilbert cube \( {I}^{\mathbb{N}} \) and \( {\mathcal{C}}^{\mathbb{N}} \) are isomorphic. But \( {\mathcal{C}}^{\mathbb{N}} \) is homeomorphic to \( \mathcal{C} \) . Thus, the Hilbert cube and the Cantor set are Borel isomorphic. By 2.1.... | Yes |
Proposition 3.3.15 For every Borel subset \( B \) of a Polish space \( X \), there is a Polish space \( Z \) and a continuous bijection from \( Z \) onto \( B \) . | Proof. Let \( \mathcal{B} \) be the set of all \( B \subseteq X \) such that there is a continuous bijection from a Polish space \( Z \) onto \( B \) . We show that \( \mathcal{B} = {\mathcal{B}}_{X} \) . Since every open subset of \( X \) is Polish,(2.2.1), open sets belong to \( \mathcal{B} \) . By 3.1.11, it is suff... | Yes |
Corollary 3.3.16 Two standard Borel spaces are Borel isomorphic if and only if they are of the same cardinality. | Null | No |
Theorem 3.3.17 Every Borel subset of a Polish space is a continuous image of \( {\mathbb{N}}^{\mathbb{N}} \) and a one-to-one, continuous image of a closed subset of \( {\mathbb{N}}^{\mathbb{N}} \) . | Proof. The result follows directly from 3.3.15, 2.6.9, and 2.6.13. | No |
Theorem 3.3.18 For every infinite Borel subset \( X \) of a Polish space, \( \left| {\mathcal{B}}_{X}\right| = \mathfrak{c} \) | Proof. Without loss of generality, we assume that \( X \) is uncountable. Since \( X \) contains a countable infinite set, \( \left| {\mathcal{B}}_{X}\right| \geq \mathfrak{c} \) . By 2.6.6, the cardinality of the set of continuous maps from \( {\mathbb{N}}^{\mathbb{N}} \) to \( X \) is \( \mathfrak{c} \) . Therefore, ... | Yes |
Theorem 3.3.22 (Ramsey - Mackey theorem) Suppose \( \left( {X,\mathcal{B}}\right) \) is a standard Borel space and \( f : X \rightarrow X \) a Borel isomorphism. Then there is a Polish topology \( \mathcal{T} \) on \( X \) generating \( \mathcal{B} \) and making \( f \) a homeomorphism. | Proof. If \( X \) is countable, we equip \( X \) with the disrete topology, and the result follows. So, we assume that \( X \) is uncountable. By the Borel isomorphism theorem, there is a Polish topology \( {\mathcal{T}}_{0} \) generating \( \mathcal{B} \) . Suppose for some \( n \in \mathbb{N} \), a Polish topology \(... | Yes |
Example 3.4.1 Let \( X \) be uncountable and \( \mathcal{A} \) the countable-cocountable \( \sigma \) -algebra. For \( A \in \mathcal{A} \), let\n\n\[ \mu \left( A\right) = \left\{ \begin{array}{ll} 1 & \text{ if }A\text{ is uncountable,} \\ 0 & \text{ otherwise. } \end{array}\right. \]\n\nThen \( \mu \) is a measure o... | Null | No |
Example 3.4.2 Let \( \\left( {X,\\mathcal{A}}\\right) \) be a measurable space and \( x \\in X \) . For \( A \\in \\mathcal{A} \) , let\n\n\[ \n{\\delta }_{x}\\left( A\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ if }x \\in A \\\\ 0 & \\text{ otherwise. } \\end{array}\\right. \n\]\n\nThen \( {\\delta }_{x} \) is ... | Null | No |
Lemma 3.4.5 Let \( \\left( {X,\\mathcal{B}}\\right) \) be a measurable space and \( \\mathcal{A} \) an algebra such that \( \\sigma \\left( \\mathcal{A}\\right) = \\mathcal{B} \) . Suppose \( {\\mu }_{1} \) and \( {\\mu }_{2} \) are finite measures on \( \\left( {X,\\mathcal{B}}\\right) \) such that \( {\\mu }_{1}\\lef... | Proof. Let\n\n\[ \n\\mathcal{M} = \\left\\{ {A \\in \\mathcal{B} : {\\mu }_{1}\\left( A\\right) = {\\mu }_{2}\\left( A\\right) }\\right\\} \n\]\n\nBy our hypothesis \( \\mathcal{A} \\subseteq \\mathcal{M} \) . By (iii) and (iv) above, \( \\mathcal{M} \) is a monotone class. The result follows from 3.1.14. | Yes |
Example 3.4.4 Let \( X \) be a nonempty set. For \( A \subseteq X \), let \( \mu \left( A\right) \) denote the number of elements in \( A \) . ( \( \mu \left( A\right) \) is \( \infty \) if \( A \) is infinite.) Then \( \mu \) is a measure on \( \mathcal{P}\left( X\right) \), called the counting measure. | Let \( \left( {X,\mathcal{A},\mu }\right) \) be a measure space. The following are easy to check.\n\n(i) \( \mu \) is monotone: If \( A \) and \( B \) are measurable sets with \( A \subseteq B \), then \( \mu \left( A\right) \leq \mu \left( B\right) . \)\n\n(ii) \( \mu \) is countably subadditive: For any sequence \( \... | Yes |
Lemma 3.4.5 Let \( \left( {X,\mathcal{B}}\right) \) be a measurable space and \( \mathcal{A} \) an algebra such that \( \sigma \left( \mathcal{A}\right) = \mathcal{B} \) . Suppose \( {\mu }_{1} \) and \( {\mu }_{2} \) are finite measures on \( \left( {X,\mathcal{B}}\right) \) such that \( {\mu }_{1}\left( A\right) = {\... | Proof. Let\n\n\[ \mathcal{M} = \left\{ {A \in \mathcal{B} : {\mu }_{1}\left( A\right) = {\mu }_{2}\left( A\right) }\right\} \]\n\nBy our hypothesis \( \mathcal{A} \subseteq \mathcal{M} \) . By (iii) and (iv) above, \( \mathcal{M} \) is a monotone class. The result follows from 3.1.14. | Yes |
Theorem 3.4.6 Let \( \mathcal{A} \) be an algebra on \( X \) and \( \mu \) a \( \sigma \) -finite measure on \( \mathcal{A} \) . Then there is a unique measure \( \nu \) on \( \sigma \left( \mathcal{A}\right) \) that extends \( \mu \) . | Null | No |
Example 3.4.7 Let \( \mathcal{A} \) be the algebra on \( \mathbb{R} \) consisting of finite disjoint unions of nondegenerate intervals (3.1.4). For any interval \( I \), let \( \left| I\right| \) denote the length of \( I \) . Let \( {I}_{0},{I}_{1},\ldots ,{I}_{n} \) be pairwise disjoint intervals and \( A = \) \( \ma... | By 3.4.6, there is a unique measure on \( \sigma \left( \mathcal{A}\right) = {\mathcal{B}}_{\mathbb{R}} \) extending \( \lambda \) . We call this measure the Lebesgue measure on \( \mathbb{R} \) and denote it by \( \lambda \) itself. | No |
Example 3.4.8 Let \( \left( {X,\mathcal{A},\mu }\right) \) and \( \left( {Y,\mathcal{B},\nu }\right) \) be \( \sigma \) -finite measure spaces. Let \( Z = X \times Y \) and let \( \mathcal{D} \) be the algebra of finite disjoint unions of measurable rectangles (3.1.5). Let \( \mu \times \nu \) be the finitely additive ... | Null | No |
Example 3.4.9 Let \( \left( {{X}_{n},{\mathcal{A}}_{n},{\mu }_{n}}\right), n \in \mathbb{N} \), be a sequence of probability spaces and \( X = \mathop{\prod }\limits_{n}{X}_{n} \) . For any nonempty, finite \( F \subseteq \mathbb{N} \), let \( {\pi }_{F} : X \rightarrow \) \( \mathop{\prod }\limits_{{n \in F}}{X}_{n} \... | Null | No |
Exercise 3.4.11 Show that \( {\overline{\mathcal{A}}}^{\mu } \) consists of all sets of the form \( {A\Delta N} \) where \( A \in \mathcal{A} \) and \( N \) is null. Further, \( \bar{\mu }\left( {A\Delta N}\right) = \mu \left( A\right) \) defines a measure on the completion. | Null | No |
Example 3.4.13 Let \( \\left( {X,\\mathcal{A},\\mu }\\right) \) be a \( \\sigma \) -finite measure space. Define \( {\\mu }^{ * } \) : \( \\mathcal{P}\\left( X\\right) \\rightarrow \\left\\lbrack {0,\\infty }\\right\\rbrack \) by\n\n\[ \n{\\mu }^{ * }\\left( A\\right) = \\inf \\{ \\mu \\left( B\\right) : B \\in \\mathc... | Null | No |
Lemma 3.4.14 Let \( X \) be a metrizable space and \( \mu \) a finite measure on \( X \) . Then \( \mu \) is regular; i.e., for every Borel set \( B \) , | Proof. Consider the class \( \mathcal{D} \) of all sets \( B \) satisfying the above conditions. We show that \( \mathcal{D} = {\mathcal{B}}_{X} \) . Let \( B \) be closed. Therefore, it is a \( {G}_{\delta } \) set. Write \( B = \mathop{\bigcap }\limits_{n}{U}_{n} \), the \( {U}_{n} \) ’s open and nonincreasing. Since... | Yes |
Theorem 3.4.17 If \( E \subseteq \mathbb{R} \) is a Lebesgue measurable set of positive Lebesgue measure, then the set\n\n\[ E - E = \{ x - y : x, y \in E\} \]\n\n is a neighborhood of 0 . | Proof. By 3.4.16 (ii), the function \( f\left( x\right) = \lambda \left( {E\bigcap \left( {E + x}\right) }\right), x \in \mathbb{R} \), is continuous. Since \( f\left( 0\right) = \lambda \left( E\right) > 0 \), there is a nonempty open interval \( \left( {-a, a}\right) \) such that \( f\left( x\right) > 0 \) for every ... | Yes |
Example 3.4.18 Let \( G \) be the additive group \( \mathbb{R} \) of real numbers, \( \mathbb{Q} \) the subgroup of rationals, and \( \mathbf{\Pi } \) the partition of \( \mathbb{R} \) consisting of all the cosets of \( \mathbb{Q} \) . The partition \( \mathbf{\Pi } \) is known as the Vitali partition. By \( \mathbf{{A... | Suppose not. Two cases arise. Either \( \lambda \left( S\right) = 0 \) or \( \lambda \left( S\right) > 0 \) . Assume first that \( \lambda \left( S\right) = 0 \) . Then, as \( \mathbb{R} = \mathop{\bigcup }\limits_{{r \in \mathbb{O}}}\left( {r + S}\right) \) , \( \lambda \left( \mathbb{R}\right) = 0 \), which is a cont... | Yes |
Theorem 3.4.19 Let \( X \) be a Polish space, \( \mu \) a finite Borel measure on \( X \) , and \( \epsilon > 0 \) . Then there is a compact set \( K \) such that \( \mu \left( {X \smallsetminus K}\right) < \epsilon \) . | Proof. Fix a compatible complete metric \( d \leq 1 \) on \( X \) . Take a regular system \( \left\{ {{F}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) of nonempty closed sets such that\n\n(i) \( {F}_{e} = X \) ,\n\n(ii) \( {F}_{s} = \mathop{\bigcup }\limits_{n}{F}_{s \cap n} \), and\n\n(iii) diameter \( \left( ... | Yes |
Theorem 3.4.20 Let \( \left( {X,\mathcal{T}}\right) \) be a Polish space and \( \mu \) a finite Borel measure on \( X \) . Then for every Borel subset \( B \) of \( X \) and every \( \epsilon > 0 \), there is a compact \( K \subseteq B \) such that \( \mu \left( {B \smallsetminus K}\right) < \epsilon \) . | Proof. By 3.2.4, there is a Polish topology \( {\mathcal{T}}_{B} \) on \( X \) finer than \( \mathcal{T} \) generating the same Borel \( \sigma \) -algebra such that \( B \) is clopen with respect to \( {\mathcal{T}}_{B} \) . By 3.4.19, there is a compact set \( K \) relative to \( {\mathcal{T}}_{B} \) contained in \( ... | Yes |
Theorem 3.4.23 (The isomorphism theorem for measure spaces) If \( \mu \) is a continuous probability on a standard Borel space \( X \), then there is a Borel isomorphism \( h : X \rightarrow I \) such that for every Borel subset \( B \) of \( I,\lambda \left( B\right) = \) \( \mu \left( {{h}^{-1}\left( B\right) }\right... | Proof. By the Borel isomorphism theorem (3.3.13), we can assume that \( X = I \) . Let \( F : I \rightarrow I \) be the distribution function of \( \mu \) . So, \( F \) is a continuous, nondecreasing map with \( F\left( 0\right) = 0 \) and \( F\left( 1\right) = 1 \) . Let\n\n\[ N = \left\{ {y \in I : {F}^{-1}\left( {\{... | Yes |
Proposition 3.4.24 Let \( X, Y \), and \( P \) be as above. Then for every \( A \in \) \( \mathcal{A} \otimes {\mathcal{B}}_{Y} \), the map \( x \rightarrow P\left( {x,{A}_{x}}\right) \) is measurable. | Proof. Let\n\n\[ \mathcal{B} = \left\{ {A \in \mathcal{A}\bigotimes {\mathcal{B}}_{Y} : \text{ the map }x \rightarrow P\left( {x,{A}_{x}}\right) \text{ is measurable }}\right\} .\n\]\n\nIt is obvious that \( \mathcal{B} \) contains all the measurable rectangles and is closed under finite disjoint unions. Clearly, \( \m... | Yes |
Proposition 3.5.1 The collection \( \mathcal{D} \) of all subsets of a topological space \( X \) having the Baire property forms a \( \sigma \) -algebra. | Proof. Closure under countable unions: Let \( {A}_{0},{A}_{1},{A}_{2},\ldots \) belong to \( \mathcal{D} \) . Take open sets \( {U}_{0},{U}_{1},{U}_{2},\ldots \) such that \( {A}_{n}\Delta {U}_{n} \) is meager for each \( n \) . Since\n\n\[ \left( {\mathop{\bigcup }\limits_{n}{A}_{n}}\right) \Delta \left( {\mathop{\big... | Yes |
Corollary 3.5.2 Every Borel subset of a metrizable space has the Baire property. | Null | No |
Proposition 3.5.4 The following statements are equivalent.\n\n(i) \( X \) is a Baire space.\n\n(ii) Every comeager set in \( X \) is dense in \( X \) .\n\n(iii) The intersection of countably many dense open sets in \( X \) is dense in \( X \) . | Null | No |
Proposition 3.5.6 Let \( X \) be a second countable Baire space and \( \left( {U}_{n}\right) \) a countable base for \( X \) . Let \( U \) be an open set in \( X \) .\n\n(i) For every sequence \( \left( {A}_{n}\right) \) of subsets of \( X,\bigcap {A}_{n} \) is comeager in \( U \) if and only if \( {A}_{n} \) is comeag... | Proof. Suppose \( \bigcap {A}_{n} \) is comeager in \( U \) . Then clearly each of \( {A}_{n} \) is comeager in \( U \) . Conversely, if each of \( {A}_{n} \) is comeager in \( U \), then \( U \smallsetminus {A}_{n} \) is meager in \( U \) for all \( n \) . So, \( \mathop{\bigcup }\limits_{n}\left( {U \smallsetminus {A... | Yes |
Proposition 3.5.7 A topological group is Baire if and only if it is of second category in itself. | Proof. The \ | No |
Proposition 3.5.8 Let \( Y \) be a second countable topological space and \( f \) : \( X \rightarrow Y \) Baire measurable. Then there is a comeager set \( A \) in \( X \) such that \( f \mid A \) is continuous. | Proof. Take a countable base \( \left( {V}_{n}\right) \) for \( Y \) . Since \( f \) is Baire measurable, for each \( n \) there is a meager set \( {I}_{n} \) in \( X \) such that \( {f}^{-1}\left( {V}_{n}\right) \Delta {I}_{n} \) is open. Let \( I = \mathop{\bigcup }\limits_{n}{I}_{n} \) . Plainly, \( f \mid \left( {X... | Yes |
Proposition 3.5.9 Let \( G \) be a completely metrizable group and \( H \) a second countable group. Then every Baire measurable homomorphism \( \varphi : G \rightarrow H \) is continuous. In particular, every Borel homomorphism \( \varphi : G \rightarrow H \) is continuous. | Proof. By 3.5.8, there is a meager set \( I \) in \( G \) such that \( \varphi \mid \left( {G \smallsetminus I}\right) \) is continuous. Now take any sequence \( \left( {g}_{k}\right) \) in \( G \) converging to an element \( g \) . Let\n\n\[ J = \left( {{g}^{-1} \cdot I}\right) \bigcup \mathop{\bigcup }\limits_{k}\lef... | Yes |
Example 3.5.10 Let \( {\mathbb{Q}}^{ + } \) be the multiplicative group of positive rational numbers and \( \varphi : {\mathbb{Q}}^{ + } \rightarrow \mathbb{Z} \) the homomorphism satisfying \( \varphi \left( p\right) = 0 \) for primes \( p > 2 \) and \( \varphi \left( 2\right) = 1 \) . Since \( {\mathbb{Q}}^{ + } \) i... | Null | No |
Theorem 3.5.12 (Pettis theorem) Let \( G \) be a Baire topological group and \( H \) a nonmeager subset with \( {BP} \) . Then there is a neighborhood \( V \) of the identity contained in \( {\mathrm{H}}^{-1}\mathrm{H} \) . | Proof. Since \( H \) is nonmeager with BP, there is a nonempty open set \( U \) such that \( {H\Delta U} \) is meager. Let \( g \in U \) . Choose a neighborhood \( V \) of the identity such that \( {gV}{V}^{-1} \subseteq U \) . We show that for every \( h \in V, H \cap {Hh} \) is nonmeager, in particular, nonempty. It ... | Yes |
Corollary 3.5.13 Every nonmeager Borel subgroup \( H \) of a Polish group \( G \) is clopen. | Proof. Let \( H \) be a Borel subgroup of \( G \) that is not meager. By 3.5.12, \( H \) contains a neighborhood of the identity. Hence, \( H \) is open. Since \( {H}^{c} \) is the union of the remaining cosets of \( H \), which are all open, it is open too. | Yes |
Lemma 3.5.14 Let \( X \) be a Baire space and \( Y \) second countable. Suppose \( A \subseteq X \times Y \) is a closed, nowhere dense set. Then\n\n\[ \left\{ {x \in X : {A}_{x}}\right. \text{is nowhere dense}\} \]\n\nis a dense \( {G}_{\delta } \) set. | Proof. Take any \( A \subseteq X \times Y \), closed and nowhere dense. Fix a countable base \( \left( {V}_{n}\right) \) for \( Y \) . Let \( U = {A}^{c} \) . Then \( U \) is dense and open. Let\n\n\[ {W}_{n} = \left\{ {x \in X : {U}_{x}\bigcap {V}_{n} \neq \varnothing }\right\} .\n\nAs\n\n\[ {W}_{n} = {\pi }_{X}\left(... | Yes |
Lemma 3.5.15 Let \( X \) be a Baire space, \( Y \) second countable, and suppose \( A \subseteq X \times Y \) has BP. The following statements are equivalent.\n\n(i) \( A \) is meager.\n\n(ii) \( \left\{ {x \in X : {A}_{x}}\right. \) is meager \( \} \) is comeager. | Proof. (ii) follows from (i) by 3.5.14. Now assume that \( A \) is nonmeager. Since \( A \) has BP, there exist nonempty open sets \( U \) and \( V \) in \( X \) and \( Y \) respectively such that \( A \) is comeager in \( U \times V \) . Therefore, from what we have just proved, \( {A}^{*V} \) is comeager in \( U \) .... | Yes |
Theorem 3.5.16 (Kuratowski - Ulam theorem) Let \( X, Y \) be second countable Baire spaces and suppose \( A \subseteq X \times Y \) has the Baire property. The following are equivalent .\n\n(i) \( A \) is meager (comeager).\n\n(ii) \( \left\{ {x \in X : {A}_{x}}\right. \) is meager (comeager) \( \} \) is comeager.\n\n(... | Null | No |
Proposition 3.5.18 Let \( \\left( {X,\\mathcal{A}}\\right) \) be a measurable space and \( Y \) a Polish space. For every \( A \\in \\mathcal{A}\\bigotimes {\\mathcal{B}}_{Y} \) and \( U \) open in \( Y \), the sets \( {A}^{\\Delta U},{A}^{*U} \), and \( \\left\\{ {x \\in X : {A}_{x}}\\right. \) is meager in \( \\left.... | Proof. Fix a countable base \( \\left( {U}_{n}\\right) \) for \( Y \). Step 1. Let \[ \\mathcal{B} = \\left\\{ {A \\subseteq X \\times Y : {A}^{\\Delta U} \\in \\mathcal{A}\\text{ for all open }U}\\right\\} . \] We show that \( \\mathcal{A} \\otimes {\\mathcal{B}}_{Y} \\subseteq \\mathcal{B} \). Let \( A = B \\times V,... | Yes |
Every \( \sigma \) -finite complete measure space is Marczewski complete. | We prove this now. Let \( \left( {X,\mathcal{B},\mu }\right) \) be a \( \sigma \) -finite complete measure space. First assume that \( {\mu }^{ * }\left( A\right) < \infty \) . Take \( \widehat{A} \) to be a measurable set containing \( A \) with \( {\mu }^{ * }\left( A\right) = \mu \left( \widehat{A}\right) \) . In th... | No |
Example 3.5.21 Let \( X \) be a topological space and \( A \subseteq X \) . Take \( {A}^{ * } \) to be the union of all open sets \( U \) such that \( A \) is comeager in \( U \) . We first show that \( {A}^{ * } \smallsetminus A \) is meager. | Let \( \mathcal{U} \) be a maximal family of pairwise disjoint open sets \( U \) such that \( A \) is comeager in \( U \) . Let \( W = \bigcup \mathcal{U} \) . By the maximality of \( \mathcal{U},{A}^{ * } \subseteq \operatorname{cl}\left( W\right) \) . By the Banach category theorem, \( A \) is comeager in \( W \) . N... | Yes |
Theorem 3.5.22 (Marczewski) If \( \left( {X,\mathcal{B}}\right) \) is a measurable space with \( \mathcal{B} \) Marczewski complete, then \( \mathcal{B} \) is closed under the Souslin operation. | Proof. Let \( \left\{ {{B}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) be a system of sets in \( \mathcal{B} \) . We have to show that \( B = \mathcal{A}\left( \left\{ {B}_{s}\right\} \right) \in \mathcal{B} \) . Without loss of generality we assume that the system \( \left\{ {B}_{s}\right\} \) is regular. For... | Yes |
Proposition 3.6.1 (i) For every \( 1 \leq \alpha < {\omega }_{1} \), \[ {\mathbf{\sum }}_{\alpha }^{0},{\mathbf{\Pi }}_{\alpha }^{0} \subseteq {\mathbf{\Delta }}_{\alpha + 1}^{0} \] | Proof. Since every closed (open) set in a metrizable space is a \( {G}_{\delta } \) set (respectively an \( {F}_{\sigma } \) set),(i) is true for \( \alpha = 1 \) . A simple transfinite induction argument completes the proof of (i) for all \( \alpha \) . | Yes |
Proposition 3.6.3 Every set of additive class \( \alpha > 2 \) is a countable disjoint union of multiplicative class \( < \alpha \) sets. | Proof. Let \( A \) be a set of additive class \( \alpha > 2 \) . Write \( A = \bigcup {A}_{n} \), where \( {A}_{n} \) is of multiplicative class less than \( \alpha \) . Let \( {B}_{n} = {\left( \mathop{\bigcup }\limits_{{i < n}}{A}_{i}\right) }^{c} \) . Then \( {B}_{n} \) is of additive class \( < \alpha \) . Write \(... | Yes |
Proposition 3.6.3 Every set of additive class \( \alpha > 2 \) is a countable disjoint union of multiplicative class \( < \alpha \) sets. | Proof. Let \( A \) be a set of additive class \( \alpha > 2 \) . Write \( A = \bigcup {A}_{n} \), where \( {A}_{n} \) is of multiplicative class less than \( \alpha \) . Let \( {B}_{n} = {\left( \mathop{\bigcup }\limits_{{i < n}}{A}_{i}\right) }^{c} \) . Then \( {B}_{n} \) is of additive class \( < \alpha \) . Write \(... | Yes |
Theorem 3.6.6 Let \( 1 \leq \alpha < {\omega }_{1} \) and \( {\mathbf{\Gamma }}_{\alpha } \) the pointclass of \( {\mathbf{\Pi }}_{\alpha }^{0} \) or \( {\mathbf{\sum }}_{\alpha }^{0} \) sets. For every second countable metrizable space \( Y \), there exists a \( U \in {\mathbf{\Gamma }}_{\alpha }\left( {{\mathbb{N}}^{... | Proof. We proceed by induction on \( \alpha \) . Let \( \left( {V}_{n}\right) \) be a countable base for the topology of \( Y \) with at least one \( {V}_{n} \) empty. Define \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times Y \) by \[ \left( {x, y}\right) \in U \Leftrightarrow y \in \mathop{\bigcup }\limits_{n}{V}_{x\le... | Yes |
Theorem 3.6.7 Let \( 1 \leq \alpha < {\omega }_{1} \) and \( {\mathbf{\Gamma }}_{\alpha } \) the pointclass of additive or multiplicative class \( \alpha \) sets. Then for every uncountable Polish space \( X \) , there is a \( U \in {\mathbf{\Gamma }}_{\alpha }\left( {X \times X}\right) \) universal for \( {\mathbf{\Ga... | Proof. Since \( X \) is uncountable Polish, it has a subset, say \( Y \), homeomorphic to \( {\mathbb{N}}^{\mathbb{N}} \) . By 3.6.6, there is \( U \subseteq Y \times X \) universal for \( {\mathbf{\Gamma }}_{\alpha }\left( X\right) \) . By 3.6.4(iii), \( V \cap \left( {Y \times X}\right) = U \) for some \( V \in {\mat... | Yes |
Corollary 3.6.8 Let \( X \) be any uncountable Polish space and \( 1 \leq \alpha < {\omega }_{1} \) . Then there exists an additive class \( \alpha \) set that is not of multiplicative class \( \alpha \) . | Proof. Let \( U \subseteq X \times X \) be universal for \( {\mathbf{\sum }}_{\alpha }^{0}\left( X\right) \) . Take\n\n\[ A = \{ x \in X : \left( {x, x}\right) \in U\} . \]\n\nSince \( {\mathbf{\sum }}_{\alpha }^{0} \) is closed under continuous preimages, \( A \) is of additive class \( \alpha \) . We claim that \( A ... | Yes |
Proposition 3.6.9 Let a pointclass \( \mathbf{\Delta } \) be closed under complementation and continuous preimages. Then for no Polish space \( X \) is there a set in \( \mathbf{\Delta }\left( {X \times X}\right) \) universal for \( \mathbf{\Delta }\left( X\right) \) . | Proof. Suppose there is a Polish space \( X \) and a \( U \in \mathbf{\Delta }\left( {X \times X}\right) \) universal for \( \mathbf{\Delta }\left( X\right) \) . Take\n\n\[ A = \{ x \in X : \left( {x, x}\right) \in U\} . \]\n\nSince \( \mathbf{\Delta } \) is closed under continuous preimages, \( A \in \mathbf{\Delta } ... | Yes |
Theorem 3.6.10 (Reduction theorem for additive classes) Let \( X \) be a metrizable space and \( 1 < \alpha < {\omega }_{1} \) . Suppose \( \left( {A}_{n}\right) \) is a sequence of additive class \( \alpha \) sets in \( X \) . Then there exist \( {B}_{n} \subseteq {A}_{n} \) such that\n\n(a) The \( {B}_{n} \) ’s are p... | Proof. Write\n\n\[ \n{A}_{n} = \mathop{\bigcup }\limits_{m}{C}_{nm} \n\]\n\n\( \left( *\right) \)\n\nwhere the \( {C}_{nm} \) ’s are of ambiguous class \( \alpha \) . If \( \alpha > 1 \), this is always possible. If \( \alpha = 1 \), it is possible if \( X \) is zero-dimensional and second countable (3.6.1). Enumerate ... | Yes |
Theorem 3.6.11 (Separation theorem for multiplicative classes) Let \( X \) be metrizable and \( 1 < \alpha < {\omega }_{1} \) . Then for every sequence \( \left( {A}_{n}\right) \) of multiplicative class \( \alpha \) sets with \( \bigcap {A}_{n} = \varnothing \), there exist ambiguous class \( \alpha \) sets \( {B}_{n}... | Proof. By 3.6.10, there exist pairwise disjoint additive class \( \alpha \) sets \( {C}_{n} \subseteq \) \( {A}_{n}^{c} \) such that \( \mathop{\bigcup }\limits_{n}{C}_{n} = \mathop{\bigcup }\limits_{n}{A}_{n}^{c} = X \) . Obviously, the \( {C}_{n} \) ’s are of ambiguous class \( \alpha \) . Take \( {B}_{n} = {C}_{n}^{... | Yes |
Example 3.6.12 (a) Fix a homeomorphism \( \alpha \rightarrow \left( {{\alpha }_{0},{\alpha }_{1}}\right) \) from \( {\mathbb{N}}^{\mathbb{N}} \) onto \( {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \) . Let \( \gamma \) be any countable ordinal and \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{... | Null | No |
Proposition 3.6.13 Let \( X \) be metrizable and \( 2 < \alpha < {\omega }_{1} \) . Suppose \( A \in \) \( {\mathbf{\Delta }}_{\alpha }^{0}\left( X\right) \) . Then there is a sequence \( \left( {A}_{n}\right) \) of ambiguous class \( < \alpha \) sets such that \( A = \lim {A}_{n} \) . | Proof. We write\n\n\[ A = \mathop{\bigcup }\limits_{n}{C}_{n} = \mathop{\bigcap }\limits_{n}{D}_{n} \]\n\nwhere the \( {C}_{n} \) ’s are multiplicative class \( < \alpha \) sets, the \( {D}_{n} \) ’s are additive class \( < \alpha \) sets, \( {C}_{n} \subseteq {C}_{n + 1} \), and \( {D}_{n + 1} \subseteq {D}_{n} \) . B... | Yes |
Proposition 3.6.14 Let \( 2 < \alpha < {\omega }_{1} \) and \( X \) an uncountable Polish space. There exists a sequence \( {A}_{n} \) in \( {\mathbf{\Pi }}_{\alpha }^{0}\left( X\right) \) with \( \limsup {A}_{n} = \varnothing \) such that there does not exist \( {B}_{n} \supseteq {A}_{n} \) in \( {\mathbf{\sum }}_{\al... | Proof. Take \( A \in {\mathbf{\sum }}_{\alpha + 1}^{0}\left( X\right) \smallsetminus {\mathbf{\Pi }}_{\alpha + 1}^{0}\left( X\right) \) . Such a set exists by 3.6.8. By 3.6.3, we can find disjoint sets \( {A}_{n} \in {\mathbf{\Pi }}_{\alpha }^{0}\left( X\right) \) with union \( A \) . Quite trivially, \( \lim \sup {A}_... | Yes |
Theorem 3.6.15 Suppose \( X, Y \) are metrizable spaces with \( Y \) second countable and \( 2 < \alpha < {\omega }_{1} \) . Then for every Borel function \( f : X \rightarrow Y \) of class \( \alpha \), there is a sequence \( \left( {f}_{n}\right) \) of Borel maps from \( X \) to \( Y \) of class \( < \alpha \) such t... | Proof of 3.6.15. Let \( d \) be a totally bounded compatible metric on \( Y \) . Such a metric exists by 2.1.32 and 2.3.12. By 3.6.16, there is a sequence \( \left( {g}_{m}\right) \) of class \( \alpha \) functions, with range finite, converging to \( f \) uniformly. Without any loss of generality, we assume that for a... | No |
Lemma 3.6.16 Suppose \( Y \) is totally bounded. Then every \( f : X \rightarrow Y \) of class \( \alpha ,\alpha > 1 \), is the limit of a uniformly convergent sequence of class \( \alpha \) functions \( {f}_{n} : X \rightarrow Y \) of finite range. | Proof. Take any \( \epsilon > 0 \) . We shall obtain a function \( g : X \rightarrow Y \) of class \( \alpha \) such that the range of \( g \) is finite and \( d\left( {g\left( x\right), f\left( x\right) }\right) < \epsilon \) for all \( x \) . Let \( \left\{ {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right\} \) be an \( \epsil... | Yes |
Lemma 3.6.17 Let \( f : X \rightarrow Y \) be of class \( \alpha > 2 \) with range contained in a finite set \( E = \left\{ {{y}_{1},{y}_{2},\ldots ,{y}_{n}}\right\} \) . Then \( f \) is the limit of a sequence of functions of class \( < \alpha \) with values in \( E \) . | Proof. Let \( {A}_{i} = {f}^{-1}\left( {y}_{i}\right), i = 1,2,\ldots, n \) . Then \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) are pairwise disjoint, ambiguous class \( \alpha \) sets with union \( X \) . By 3.6.13, for each \( i \) there is a sequence \( \left( {A}_{im}\right) \) of sets of ambiguous class \( < \alpha \) su... | Yes |
Proposition 4.1.1 Let \( X \) be a Polish space and \( A \subseteq X \) . The following statements are equivalent.\n\n(i) \( A \) is analytic.\n\n(ii) There is a Polish space \( Y \) and a Borel set \( B \subseteq X \times Y \) whose projection is \( A \) .\n\n(iii) There is a continuous map \( f : {\mathbb{N}}^{\mathb... | Proof. (i) trivially implies (ii).\n\nLet \( Y \) be a Polish space and \( B \) a Borel subset of \( X \times Y \) such that \( {\pi }_{X}\left( B\right) = A \), where \( {\pi }_{X} : X \times Y \rightarrow X \) is the projection map. By 3.3.17, there is a continuous map \( g \) from \( {\mathbb{N}}^{\mathbb{N}} \) ont... | Yes |
Proposition 4.1.2 (i) The pointclass \( {\mathbf{\sum }}_{1}^{1} \) is closed under countable unions, countable intersections and Borel preimages. Consequently, \( {\mathbf{\Pi }}_{1}^{1} \) is closed under these operations. | Proof. We first prove (i).\n\nClosure under Borel preimages: Let \( X \) and \( Z \) be Polish spaces, \( A \subseteq X \) analytic, and \( f : Z \rightarrow X \) a Borel map. Choose a Borel subset \( B \) of \( X \times X \) whose projection is \( A \) . Let\n\n\[ C = \{ \left( {z, x}\right) \in Z \times X : \left( {f... | Yes |
Theorem 4.1.4 For every Polish space \( X \), there is an analytic set \( U \subseteq \) \( {\mathbb{N}}^{\mathbb{N}} \times X \) such that \( A \subseteq X \) is analytic if and only if \( A = {U}_{\alpha } \) for some \( \alpha \) ; i.e., \( U \) is universal for \( {\mathbf{\sum }}_{1}^{1}\left( X\right) \) . | Proof. Let \( C \subseteq {\mathbb{N}}^{\mathbb{N}} \times \left( {X \times {\mathbb{N}}^{\mathbb{N}}}\right) \) be a universal closed set. The existence of such a set is shown in 3.6.6. Let\n\n\[ U = \left\{ {\left( {\alpha, x}\right) \in {\mathbb{N}}^{\mathbb{N}} \times X : \left( {\alpha, x,\beta }\right) \in C\text... | Yes |
Theorem 4.1.5 Let \( X \) be an uncountable Polish space.\n\n(i) There is an analytic set \( U \subseteq X \times X \) such that for every analytic set \( A \subseteq X \), there is an \( x \in X \) with \( A = {U}_{x} \).\n\n(ii) There is a subset of \( X \) that is analytic but not Borel. | Proof. (i) Since \( X \) is uncountable Polish, it contains a homeomorph of \( {\mathbb{N}}^{\mathbb{N}} \), say \( Y \) (2.6.5). The set \( Y \) is a \( {G}_{\delta } \) set in \( X \) (2.2.7). Take \( U \subseteq Y \times X \) as in 4.1.4.\n\n(ii) Let\n\n\[ A = \{ x \in X : \left( {x, x}\right) \in U\} .\n\]\n\nSince... | Yes |
Proposition 4.1.7 Let \( n \) be a positive integer.\n\n(i) The pointclasses \( {\mathbf{\sum }}_{n}^{1} \) and \( {\mathbf{\Pi }}_{n}^{1} \) are closed under countable unions, countable intersections and Borel preimages.\n\n(ii) \( {\mathbf{\Delta }}_{n}^{1} \) is a \( \sigma \) -algebra.\n\n(iii) The pointclass \( {\... | Proof. Clearly, (ii) follows from (i). So, we prove (i) and (iii) only. We proceed by induction on \( n \) . Let \( n > 1 \) and \( {\mathbf{\Pi }}_{n - 1}^{1} \) and \( {\mathbf{\sum }}_{n - 1}^{1} \) have all the closure properties stated in (i) and (iii). The arguments contained in the proof of 4.1.2 show that \( {\... | No |
Proposition 4.1.9 For every \( n \geq 1 \) ,\n\n\[ \n{\mathbf{\sum }}_{n}^{1}\bigcup {\mathbf{\Pi }}_{n}^{1} \subseteq {\mathbf{\Delta }}_{n + 1}^{1} \n\] | Proof. We prove the result by induction on \( n \) . Let \( X \) be a Polish space and \( A \subseteq X \) analytic. As \( {\mathbf{\Delta }}_{1}^{1} \subseteq {\mathbf{\Pi }}_{1}^{1} \), it follows that \( {\mathbf{\sum }}_{1}^{1} \subseteq {\mathbf{\sum }}_{2}^{1} \) . Since \( {\mathbf{\sum }}_{1}^{1} \) is closed u... | Yes |
Lemma 4.1.10 Let \( n \geq 1,\mathbf{\Gamma } \) either \( {\mathbf{\sum }}_{n}^{1} \) or \( {\mathbf{\Pi }}_{n}^{1} \), and \( X \) a Polish space. There is a \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times X \) in \( \mathbf{\Gamma } \) such that \( A \subseteq X \) is in \( \mathbf{\Gamma } \) if and only if \( A = ... | Proof. The result is proved by induction. Suppose \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times X \) is universal for \( {\mathbf{\sum }}_{1}^{1}\left( X\right) \) . Then \( {U}^{c} \) is universal for \( {\mathbf{\Pi }}_{1}^{1}\left( X\right) \) . Let \( C \subseteq {\mathbb{N}}^{\mathbb{N}} \times (X \times \) \( {... | No |
Theorem 4.1.11 Let \( X \) be an uncountable Polish space and \( n \geq 1 \) .\n\n(i) There is a set \( U \in {\mathbf{\sum }}_{n}^{1}\left( {X \times X}\right) \) such that for every \( A \in {\mathbf{\sum }}_{n}^{1}\left( X\right) \), there is an \( x \) with \( A = {U}_{x} \) .\n\n(ii) There is a subset of \( X \) t... | Proof. The result is proved in exactly the same way as 4.1.5. | No |
Theorem 4.1.13 Let \( X \) be a Polish space, \( d \) a compatible complete metric on \( X \), and \( A \subseteq X \). The following statements are equivalent.\n\n(i) \( A \) is analytic.\n\n(ii) There is a regular system \( \left\{ {{F}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) of closed subsets of \( X \)... | Proof. (ii) implies (iii) is obvious.\n\n(iii) \( \Rightarrow \) (i): Let \( \left\{ {F}_{s}\right\} \) be a system of closed sets in \( X \) such that\n\n\[ A = \mathcal{A}\left( \left\{ {F}_{s}\right\} \right) \]\n\ni.e.,\n\n\[ x \in A \Leftrightarrow \exists \alpha \forall n\left( {x \in {F}_{\alpha \mid n}}\right) ... | Yes |
Theorem 4.1.14 The pointclass \( {\mathbf{\sum }}_{1}^{1} \) is closed under the Souslin operation. | Proof. By 1.13.1, the Souslin operation is idempotent; i.e., for any family \( \mathcal{F} \) of sets \( \mathcal{A}\left( {\mathcal{A}\left( \mathcal{F}\right) }\right) = \mathcal{A}\left( \mathcal{F}\right) \) . Since \( {\mathbf{\sum }}_{1}^{1} = \mathcal{A}\left( \mathcal{F}\right) \), where \( \mathcal{F} \) is th... | Yes |
Proposition 4.1.20 Let \( A \subseteq {\mathbb{N}}^{\mathbb{N}} \) . The following statements are equivalent.\n\n(i) \( A \) is coanalytic.\n\n(ii) There is a tree \( T \) on \( \mathbb{N} \times \mathbb{N} \) such that\n\n\[ \alpha \in A \Leftrightarrow T\left\lbrack \alpha \right\rbrack \text{ is well-founded } \]\n\... | Proof. Let \( A \subseteq {\mathbb{N}}^{\mathbb{N}} \) be a coanalytic set. Then \( {A}^{c} \) is analytic. Let \( C \) be a closed set in \( {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \) such that \( {\pi }_{1}\left( C\right) = {A}^{c} \), where \( {\pi }_{1} : {\mathbb{N}}^{\mathbb{N}} \times {\mathbb... | Yes |
Example 4.1.21 Let \( g : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \) be a Borel function. Define \[ f\left( x\right) = \mathop{\sup }\limits_{y}g\left( {x, y}\right) ,\;x \in X. \] Assume that \( f\left( x\right) < \infty \) for all \( x \) . The function \( f \) need not be Borel. To see this, take an anal... | Proof. Let \( v : \mathbb{R} \rightarrow \mathbb{R} \) be a Borel function such that \( v\left( x\right) \leq f\left( x\right) \) for all \( x \) . For \( n \in \mathbb{Z} \), let \[ {B}_{n} = \{ x \in \mathbb{R} : n \leq v\left( x\right) < n + 1\} . \] Fix an enumeration \( \left\{ {{r}_{m} : m \in \mathbb{N}}\right\}... | Yes |
Proposition 4.1.22 (H. Sarbadhikari [99]) For every A-function \( f \) : \( \mathbb{R} \rightarrow \mathbb{R} \) dominating a Borel function there is a Borel \( g : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \) such that \( f\left( x\right) = \mathop{\sup }\limits_{y}g\left( {x, y}\right) \) . | Proof. Let \( v : \mathbb{R} \rightarrow \mathbb{R} \) be a Borel function such that \( v\left( x\right) \leq f\left( x\right) \) for all \( x \) . For \( n \in \mathbb{Z} \), let\n\n\[ \n{B}_{n} = \{ x \in \mathbb{R} : n \leq v\left( x\right) < n + 1\} .\n\] \n\nFix an enumeration \( \left\{ {{r}_{m} : m \in \mathbb{N... | Yes |
We show that \( {WF} \) is \( {\mathbf{\Pi }}_{1}^{1} \) -complete. | Observe that\n\n\[ T \in {WF} \Leftrightarrow T \in \operatorname{Tr}\& \forall \beta \exists n\left( {T\left( {\beta \mid n}\right) = 0}\right) .\n\]\n\nTherefore, \( {WF} = {\forall }^{{\mathbb{N}}^{\mathbb{N}}}E \), where\n\n\[ E = \left\{ {\left( {T,\beta }\right) \in {2}^{{\mathbb{N}}^{ < \mathbb{N}}} \times {\mat... | Yes |
We now show that \( {WO} \) is \( {\mathbf{\Pi }}_{1}^{1} \) -complete. It is sufficient to show that there is a continuous map \( R : {Tr} \rightarrow {2}^{\mathbb{N} \times \mathbb{N}} \) such that \( {WF} = {R}^{-1}\left( {WO}\right) \) . | Fix a bijection \( u : \mathbb{N} \rightarrow {\mathbb{N}}^{ < \mathbb{N}} \) . To each \( T \in {Tr} \), associate a binary relation \( R\left( T\right) \) on \( \mathbb{N} \) as follows:\n\n\[ {kR}\left( T\right) l \Leftrightarrow \left( {u\left( k\right), u\left( l\right) \notin T\& k \leq l}\right) \]\n\n\[ \vee \l... | Yes |
Proposition 4.2.5 Let \( X \) be an uncountable Polish space. Then\n\n\[ U\left( X\right) = \{ K \in K\left( X\right) : K\text{ is uncountable }\} \]\n\nis \( {\mathbf{\sum }}_{1}^{1} \) -complete. | Proof. We first show that \( U\left( X\right) \in {\mathbf{\sum }}_{1}^{1} \) . Let \( P\left( X\right) \) denote the set of all nonempty perfect subsets of \( X \) . Then \( P\left( X\right) \) is Borel in \( K\left( X\right) \) . To see this, take a countable base \( \left( {V}_{n}\right) \) for \( X \) . We have\n\n... | Yes |
Theorem 4.3.1 Let \( \mu \) be a \( \sigma \) -finite measure on \( \left( {X,{\mathcal{B}}_{X}}\right), X \) Polish. Then every analytic subset of \( X \) is \( \mu \) -measurable. | Null | No |
Theorem 4.3.2 Every analytic subset of a Polish space has the Baire property. | Null | No |
Theorem 4.3.4 (B. V. Rao[95]) Let \( X \) be an uncountable Polish space and \( U \subseteq X \times X \) universal analytic. Then\n\n\[ U \notin \mathcal{P}\left( X\right) \bigotimes \mathcal{B} \]\n\nwhere \( \mathcal{B} \) is as in 4.3.3. | Proof. Suppose \( U \in \mathcal{P}\left( X\right) \otimes \mathcal{B} \) . We shall get a contradiction. From 3.1.7, there are \( {C}_{0},{C}_{1},{C}_{2},\ldots \subseteq X \) and \( {D}_{0},{D}_{1},{D}_{2},\ldots \) in \( \mathcal{B} \) such that\n\n\( U \in \sigma \left( \left\{ {{C}_{i} \times {D}_{i} : i \in \math... | Yes |
Theorem 4.3.5 Every uncountable analytic set contains a homeomorph of the Cantor set and hence is of cardinality \( \mathfrak{c} \) . | Proof. Let \( X \) be a Polish space and \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) a continuous map whose range is uncountable. We first show that there is a Cantor scheme \( \left\{ {{F}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) of closed subsets of \( {\mathbb{N}}^{\mathbb{N}} \) such that whenever \( \left... | Yes |
Proposition 4.3.7 Let \( X \) be a Polish space and \( A \subseteq X \) . The following statements are equivalent.\n\n(i) \( A \) is analytic.\n\n(ii) There is a closed set \( C \subseteq X \times {\mathbb{N}}^{\mathbb{N}} \) such that\n\n\[ A = \left\{ {x \in X : {C}_{x}}\right. \text{is uncountable}\} \text{.} \]\n\n... | Proof. (i) \( \Rightarrow \) (ii): Let \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) be a continuous map with range \( A \) and \( {\pi }_{1} : {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \rightarrow {\mathbb{N}}^{\mathbb{N}} \) the projection map. Note that \( {\pi }_{1} \) is continuous and \( {\pi... | Yes |
Theorem 4.3.8 (S. Simpson [79]) Let \( X \) be an analytic subset of a Polish space, \( Y \) a metrizable space, and \( f : X \rightarrow Y \) a Borel map. Then \( f\left( X\right) \) is separable. | Proof. Without any loss of generality, we assume that \( X \) is Polish and \( Y = f\left( X\right) \) . Suppose \( Y \) is not separable. Then there is an uncountable closed discrete subspace \( Z \) of \( Y \) . As \( \left| X\right| = \mathfrak{c},\left| Y\right| \leq \mathfrak{c} \), and hence \( \left| Z\right| \l... | Yes |
Corollary 4.3.9 Every Borel homomorphism \( \varphi : G \rightarrow H \) from a completely metrizable group \( G \) to a metrizable group \( H \) is continuous. | Proof. Let \( \left( {g}_{n}\right) \) be a sequence in \( G \) converging to \( g \) . Replacing \( G \) by the closed subgroup generated by \( \left\{ {{g}_{n} : n \in \mathbb{N}}\right\} \), we assume that \( G \) is Polish. By 4.3.8, \( \varphi \left( G\right) \) is separable. The result follows from 3.5.9. | No |
Proposition 4.3.10 (i) Every countable set of reals has strong measure zero. | Proof. (i) and (ii) are immediate consequences of the definition. We prove (iii) now. Let \( \left( {A}_{n}\right) \) be a sequence of strong measure zero sets. Take any sequence \( \left( {a}_{n}\right) \) of positive real numbers. Choose pairwise disjoint infinite subsets \( {I}_{0},{I}_{1},{I}_{2},\ldots \) of \( \m... | No |
Proposition 4.3.11 Let \( A \subseteq \left\lbrack {0,1}\right\rbrack \) be a strong measure zero set and \( f \) : \( \left\lbrack {0,1}\right\rbrack \rightarrow \mathbb{R} \) a continuous map. Then the set \( f\left( A\right) \) has strong measure zero. | Proof. Let \( \left( {a}_{n}\right) \) be any sequence of positive real numbers. We have to show that there exist open intervals \( {J}_{n}, n \in \mathbb{N} \), such that \( \left| {J}_{n}\right| \leq {a}_{n} \) and \( f\left( A\right) \subseteq \mathop{\bigcup }\limits_{n}{J}_{n} \) . Since \( f \) is uniformly conti... | Yes |
Example 4.3.12 It is easy to see that there is no sequence \( \left( {I}_{n}\right) \) of open intervals such that the length of \( {I}_{n} \) is at most \( {3}^{-\left( {n + 1}\right) } \) and \( \left( {I}_{n}\right) \) cover the Cantor ternary set \( \mathcal{C} \) . Hence, \( \mathcal{C} \) is not a strong measure ... | Null | No |
Proposition 4.3.13 No set of reals containing a perfect set has strong measure zero. | Null | No |
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