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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.
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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.
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Yes
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Lemma 3.1.35 Let \( X \) be a metrizable space and \( B \subseteq X \) Borel. Then \( {\chi }_{B} : X \rightarrow \mathbb{R} \) is Baire.
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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 \( {f}_{n} : X \rightarrow \left\lbrack {0,1}\right\rbrack \) identically equal to 1 on \( {F}_{n} \) and equal to 0 on \( X \smallsetminus U \) . Then the sequence \( \left( {f}_{n}\right) \) converges pointwise to \( {\chi }_{U} \) . Thus, \( U \in \mathcal{B} \) .\n\n(b) Let \( {B}_{0},{B}_{1},{B}_{2},\ldots \) be pairwise disjoint and belong to \( \mathcal{B} \) . Set\n\n\[ {f}_{n} = \mathop{\sum }\limits_{{i \leq n}}{\chi }_{{B}_{i}} \]\n\nBy our hypothesis and 3.1.34(ii), \( {f}_{n} \) is Baire. Since \( \left( {f}_{n}\right) \) converges pointwise to the characteristic function of \( \mathop{\bigcup }\limits_{n}{B}_{n} \), we see that \( \mathop{\bigcup }\limits_{n}{B}_{n} \in \mathcal{B} \) .\n\n(c) Let \( {B}_{0},{B}_{1},{B}_{2},\ldots \) belong to \( \mathcal{B} \) . Put\n\n\[ {f}_{n} = \mathop{\min }\limits_{{i \leq n}}{\chi }_{{B}_{i}} \]\n\nBy our hypothesis and 3.1.34, \( {f}_{n} \) is Baire. As \( \left( {f}_{n}\right) \) converges pointwise to the characteristic function of \( \mathop{\bigcap }\limits_{n}{B}_{n} \), it follows that \( \mathop{\bigcap }\limits_{n}{B}_{n} \in \mathcal{B} \) .\n\nThe result now follows from 3.1.11.
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Yes
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Theorem 3.1.36 (Lebesgue - Hausdorff theorem) Every real-valued Borel function defined on a metrizable space is Baire.
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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.
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Yes
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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 \( \\mathcal{T} \\subseteq {\\mathcal{T}}^{\\prime } \\subseteq {\\mathcal{B}}_{X} \) and each \( {B}_{n} \\in {\\mathcal{T}}^{\\prime } \) .
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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\[ {\\mathcal{T}}^{\\prime } = \\left\{ {{f}^{-1}\\left( U\\right) : U\\text{ open in }X \\times \\mathcal{C}}\\right\}\n\]\n\nAs \( \\left( {X,{\\mathcal{T}}^{\\prime }}\\right) \) is homeomorphic to a subset of \( X \\times \\mathcal{C} \), it is metrizable. Further, if \( X \) is zero-dimensional (separable), so is \( \\left( {X,{\\mathcal{T}}^{\\prime }}\\right) \) .\n\nLet \( U \\subseteq X \) be open with respect to the original topology \( \\mathcal{T} \) . Then\n\n\[ U = {f}^{-1}\\left( {\\{ \\left( {x,\\alpha }\\right) \\in X \\times \\mathcal{C} : x \\in U\\} }\\right)\n\]\n\nand hence belongs to \( {\\mathcal{T}}^{\\prime } \) . Thus, \( {\\mathcal{T}}^{\\prime } \) is finer that \( \\mathcal{T} \) . By 3.1.29, \( f \) is Borel measurable. Therefore, \( {\\mathcal{T}}^{\\prime } \\subseteq {\\mathcal{B}}_{X} \) . It remains to show that each \( {B}_{n} \\in {\\mathcal{T}}^{\\prime } \) . Let\n\n\[ {V}_{n} = \\{ \\left( {x,\\alpha }\\right) \\in X \\times \\mathcal{C} : \\alpha \\left( n\\right) = 1\\} .\n\]\n\nThen \( {V}_{n} \) is open in \( X \\times \\mathcal{C} \) . Since \( {B}_{n} = {f}^{-1}\\left( {V}_{n}\\right) ,{B}_{n} \\in {\\mathcal{T}}^{\\prime } \) .
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Yes
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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 \( f : A \rightarrow Y \) is continuous with respect to the new topology \( {\mathcal{T}}^{\prime } \), and\n\n(ii) the map \( f : A \rightarrow Y \) admits a Borel extension \( g : X \rightarrow Y \) .
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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. This answers (i).\n\nTo prove (ii), take \( {\mathcal{T}}^{\prime } \) as above. By 2.2.3, there is a \( {G}_{\delta } \) set \( C \supseteq A \) and a continuous extension \( {g}^{\prime } : C \rightarrow Y \) of \( f \) . Here we are assuming that \( X \) is equipped with the finer topology \( {\mathcal{T}}^{\prime } \) . As \( \mathcal{T} \) and \( {\mathcal{T}}^{\prime } \) generate the same \( \sigma \) -algebra \( {\mathcal{B}}_{X}, C \) is Borel in \( X \) and \( {g}^{\prime } \) measurable relative to the original topology \( \mathcal{T} \) . Extend \( {g}^{\prime } \) to the whole space \( X \) by defining it to be a constant on \( X \smallsetminus C \) .
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Yes
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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 \( \sigma \left( \mathcal{T}\right) = \sigma \left( {\mathcal{T}}_{B}\right) \) .
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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 complementation.\n\nTo show \( \mathcal{B} = {\mathcal{B}}_{X} \), we need to prove only that \( \mathcal{B} \) is closed under countable unions. Let \( {B}_{n} \) belong to \( \mathcal{B} \) and \( B = \bigcup {B}_{n} \) . Let \( {\mathcal{T}}_{n} \) be a finer Polish topology on \( X \) making \( {B}_{n} \) clopen and generating the same Borel \( \sigma \) -algebra. Then \( B \in {\mathcal{T}}_{\infty } \), where \( {\mathcal{T}}_{\infty } \) is the topology generated by \( \bigcup {\mathcal{T}}_{n} \) . By Observation \( 2,{\mathcal{T}}_{\infty } \) is Polish. Take \( {\mathcal{T}}_{B} \) to be the topology generated by \( {\mathcal{T}}_{\infty }\bigcup \left\{ {B}^{c}\right\} \) . By Observation \( 1,{\mathcal{T}}_{B} \) is Polish.
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Yes
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Every uncountable Borel subset of a Polish space contains a homeomorph of the Cantor set. In particular, it is of cardinality \( \mathfrak{c} \) .
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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, say \( K \) . By 2.3.9, \( {\mathcal{T}}^{\prime }\left| {K = \mathcal{T}}\right| K \), and the result follows.
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Yes
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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 \( f : \left( {X,{\mathcal{T}}^{\prime }}\right) \rightarrow Y \) is continuous.
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Null
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No
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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 \( \mathfrak{c} \) .
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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, say \( K \) . By 2.3.9, \( {\mathcal{T}}^{\prime }\left| {K = \mathcal{T}}\right| K \), and the result follows.
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Yes
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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 \( {x}_{\alpha },{y}_{\alpha } \in {C}_{\alpha } \) . Then the set \( A = \left\{ {{x}_{\alpha } : \alpha < \mathfrak{c}}\right\} \) is easily seen to be a Bernstein set.
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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 \( D = \left\{ {{x}_{\beta } : \beta < \alpha }\right\} \bigcup \left\{ {{y}_{\beta } : \beta < \alpha }\right\} \) . Note that \( \left| D\right| = \left| \alpha \right| + \left| \alpha \right| < \mathfrak{c} \) . As \( \left| {C}_{\alpha }\right| = \mathfrak{c} \), we choose distinct points \( {x}_{\alpha },{y}_{\alpha } \) in \( {C}_{\alpha } \smallsetminus D \) .
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Yes
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Example 3.3.1 The closed unit interval \( I = \left\lbrack {0,1}\right\rbrack \) and the Cantor set \( \mathcal{C} \) are Borel isomorphic.
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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\[ f\left( {{\epsilon }_{0},{\epsilon }_{1},{\epsilon }_{2},\ldots }\right) = \mathop{\sum }\limits_{{n \in \mathbb{N}}}{\epsilon }_{n}/{2}^{n + 1}. \]\n\nIt is easy to check that \( f \) is a homeomorphism from \( \mathcal{C} \smallsetminus E \) onto \( I \smallsetminus D \) . Since both \( D \) and \( E \) are countably infinite, there is a bijection \( g : E \rightarrow D \) . The function \( h : I \rightarrow \mathcal{C} \) obtained by piecing \( f \) and \( g \) together is clearly a Borel isomorphism from \( I \) onto \( \mathcal{C} \) .
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Yes
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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 \( X, g\\left( x\\right) = g\\left( y\\right) \) if and only if \( x \) and \( y \) belong to the same atom of \( \\mathcal{A} \) .
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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\[ g\\left( x\\right) = \\left( {{\\chi }_{{A}_{0}}\\left( x\\right) ,{\\chi }_{{A}_{1}}\\left( x\\right) ,{\\chi }_{{A}_{2}}\\left( x\\right) ,\\ldots }\\right) .\n\]\n\nTake \( Z = g\\left( X\\right) \) . By 3.1.29, \( g \) is measurable. Also note that for any two \( x, y \) in \( X, g\\left( x\\right) = g\\left( y\\right) \) if and only if \( x \) and \( y \) belong to the same \( {A}_{i} \) ’s. Recall that the atoms of \( \\mathcal{A} \) are precisely the sets of the form \( \\mathop{\\bigcap }\\limits_{n}{A}^{\\epsilon \\left( n\\right) } \) , \( \\left( {\\epsilon \\left( 0\\right) ,\\epsilon \\left( 1\\right) ,\\ldots }\\right) \\in \\mathcal{C} \) . (See the proof of 3.1.15.) It follows that \( g\\left( x\\right) = g\\left( y\\right) \) if and only if \( x \) and \( y \) belong to the same atom of \( \\mathcal{A} \) . As\n\n\[ g\\left( {A}_{n}\\right) = Z\\bigcap \\{ \\alpha \\in \\mathcal{C} : \\alpha \\left( n\\right) = 1\\}\n\]\n\n it is Borel in \( Z \) . Now observe that\n\n\[ \\mathcal{B} = \\{ B \\in \\mathcal{A} : g\\left( B\\right) \\text{ is Borel in }Z\\}\n\]\n\n is a \( \\sigma \) -algebra containing \( {A}_{n} \) for all \( n \) . So, \( {g}^{-1} \) is also measurable.
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Yes
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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 \( X \) .
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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 3.3.2, get a metrizable space \( Z \) and a bimeasurable map \( g : X \\rightarrow Z \) such that for any \( x,{x}^{\\prime } \\in X, g\\left( x\\right) = g\\left( {x}^{\\prime }\\right) \) if and only if \( x \) and \( {x}^{\\prime } \) belong to the same atom of \( \\sigma \\left( \\left( {B}_{n}\\right) \\right) \) . Hence, for \( x,{x}^{\\prime } \\in A, f\\left( x\\right) = f\\left( {x}^{\\prime }\\right) \) if and only if \( g\\left( x\\right) = g\\left( {x}^{\\prime }\\right) \) . Set \( B = g\\left( A\\right) \) and define \( h : B \\rightarrow Y \) by\n\n\[ h\\left( z\\right) = f\\left( x\\right) \]\n\nwhere \( x \\in A \) is such that \( g\\left( x\\right) = z \) . It is easy to see that \( h \) is well-defined and \( {h}^{-1}\\left( {U}_{n}\\right) = g\\left( {B}_{n}\\right) \\cap B \) . Hence, \( h \) is Borel. By 3.2.3, there is a Borel extension \( {h}^{\\prime } : Z \\rightarrow Y \) of \( h \) . The composition \( {h}^{\\prime } \\circ g \) is clearly a measurable extension of \( f \) to \( X \) .
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Yes
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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.
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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.
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Yes
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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.
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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 \( g : {X}^{\prime } \rightarrow {Y}^{\prime } \) extending \( f \) between Borel subsets \( {X}^{\prime } \) and \( {Y}^{\prime } \) of \( \widehat{X} \) (the completion of \( X \) ) and \( Z \) respectively. Since \( X = {g}^{-1}\left( Y\right) \), it is Borel in \( {X}^{\prime } \) and hence in \( \widehat{X} \) .
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Yes
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Theorem 3.3.10 The Effros Borel space of a Polish space is standard Borel.
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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\[ \mathcal{Z} = \{ \operatorname{cl}\left( F\right) \in F\left( Y\right) : F \in F\left( X\right) \}\]\n\nwhere closure is relative to \( Y \) .\n\nNote that \( \mathcal{Z} \subseteq K\left( Y\right) \) and\n\n\[ K \in \mathcal{Z} \Leftrightarrow K\bigcap X\text{ is dense in }K \]\n\nThe result will be proved if we show the following.\n\n(i) The map \( F \rightarrow \operatorname{cl}\left( F\right) \) from \( \left( {F\left( X\right) ,\mathcal{E}\left( X\right) }\right) \) onto \( \mathcal{Z} \) is an isomorphism, and\n\n(ii) \( \mathcal{Z} \) is a \( {G}_{\delta } \) set in \( K\left( Y\right) \) .\n\nClearly, \( F \rightarrow \operatorname{cl}\left( F\right) \) is one-to-one on \( F\left( X\right) \) . Further, for any \( F \in F\left( X\right) \) and any \( U \) open in \( Y \) ,\n\n\[ \operatorname{cl}\left( F\right) \bigcap U \neq \varnothing \Leftrightarrow F\bigcap \left( {U\bigcap X}\right) \neq \varnothing .\n\nHence, (i) follows.\n\nWe now prove (ii). We have\n\n\[ K \in \mathcal{Z} \Leftrightarrow K \cap \mathop{\bigcap }\limits_{n}{U}_{n}\text{ is dense in }K \]\n\nTherefore, by the Baire category theorem,\n\n\[ K \in \mathcal{Z} \Leftrightarrow \forall n\left( {K\bigcap {U}_{n}\text{ is dense in }K}\right) \]\n\n\[ \Leftrightarrow \;\forall n\forall m\left( {K\bigcap {V}_{m} \neq \varnothing \Rightarrow K\bigcap {V}_{m}\bigcap {U}_{n} \neq \varnothing }\right) .\n\nThus,\n\n\[ Z = \mathop{\bigcap }\limits_{n}\mathop{\bigcap }\limits_{m}\left\{ {K \in F\left( Y\right) : K\bigcap {V}_{m} = \varnothing \text{ or }K\bigcap {V}_{m}\bigcap {U}_{n} \neq \varnothing }\right\} \]\n\nand the result follows.
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Yes
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Theorem 3.3.13 (The Borel isomorphism theorem) Any two uncountable standard Borel spaces are Borel isomorphic.
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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, \( B \) is Borel isomorphic to \( \mathcal{C} \), and the proof is complete.
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Yes
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Lemma 3.3.14 Every standard Borel space \( B \) is Borel isomorphic to a Borel subset of \( \mathcal{C} \) .
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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 2.1.32, every standard Borel space is isomorphic to a Borel subset of the Hilbert cube, and hence of \( \mathcal{C} \) .
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Yes
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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 \( B \) .
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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 sufficient to show that \( \mathcal{B} \) is closed under countable intersections and countable disjoint unions. Let \( {B}_{0},{B}_{1},{B}_{2},\ldots \) belong to \( \mathcal{B} \) . Fix Polish spaces \( {Z}_{0},{Z}_{1},\ldots \) and continuous bijections \( {g}_{i} : {Z}_{i} \rightarrow {B}_{i} \) . Let\n\n\[ Z = \left\{ {z \in \prod {Z}_{i} : {g}_{0}\left( {z}_{0}\right) = {g}_{1}\left( {z}_{1}\right) = \cdots }\right\} .\n\]\n\nThen \( Z \) is closed in \( \mathop{\prod }\limits_{i}{Z}_{i} \) . Therefore, \( Z \) is Polish. Define \( g : Z \rightarrow X \) by\n\n\[ g\left( z\right) = {g}_{0}\left( {z}_{0}\right) \]\n\nThen \( g \) is a one-to-one, continuous map from \( Z \) onto \( \bigcap {B}_{i} \) . Thus, \( \mathcal{B} \) is closed under countable intersections.\n\nLet us next assume that \( {B}_{0},{B}_{1},\ldots \) are pairwise disjoint. Choose \( {g}_{i},{Z}_{i} \) as before. Take \( Z = \bigoplus {Z}_{i} \), the direct sum of the \( {Z}_{i} \) ’s. Define \( g : Z \rightarrow X \) by\n\n\[ g\left( z\right) = {g}_{i}\left( z\right) \text{ if }z \in {Z}_{i} \]\n\nThen \( Z \) is a Polish space, and \( g \) is a one-to-one, continuous map from \( Z \) onto \( \bigcup {B}_{i} \) . So, \( \mathcal{B} \) is also closed under countable disjoint unions.
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Yes
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Corollary 3.3.16 Two standard Borel spaces are Borel isomorphic if and only if they are of the same cardinality.
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Null
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No
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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 \( {\mathbb{N}}^{\mathbb{N}} \) .
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Proof. The result follows directly from 3.3.15, 2.6.9, and 2.6.13.
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No
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Theorem 3.3.18 For every infinite Borel subset \( X \) of a Polish space, \( \left| {\mathcal{B}}_{X}\right| = \mathfrak{c} \)
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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, by 3.3.17, \( \left| {\mathcal{B}}_{X}\right| \leq \mathfrak{c} \) . The result follows from the Schröder - Bernstein Theorem.
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Yes
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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.
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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 \( {\mathcal{T}}_{n} \) generating \( \mathcal{B} \) has been defined. Let \( \left\{ {{B}_{i}^{n} : i \in \mathbb{N}}\right\} \) be a countable base for \( \left( {X,{\mathcal{T}}_{n}}\right) \) . Consider\n\n\[ \mathcal{D} = \left\{ {f\left( {B}_{i}^{n}\right) : i \in \mathbb{N}}\right\} \bigcup \left\{ {{f}^{-1}\left( {B}_{i}^{n}\right) : i \in \mathbb{N}}\right\} .\n\]\n\nBy 3.2.5, there is a Polish topology \( {\mathcal{T}}_{n + 1} \) finer than \( {\mathcal{T}}_{n} \) making each element of \( \mathcal{D} \) open. Now take \( \mathcal{T} \) to be the topology generated by \( \bigcup {\mathcal{T}}_{n} \) . By Observation 2 following 3.2.4, \( \mathcal{T} \) is Polish. A routine argument now completes the proof.
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Yes
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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 on \( \mathcal{A} \) .
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Null
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No
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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 a measure on \( \\mathcal{A} \), called the Dirac measure at \( x \) .
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Null
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No
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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 \( {\\mu }_{1}\\left( A\\right) = {\\mu }_{2}\\left( A\\right) \) for every \( A \\in \\mathcal{A} \) . Then \( {\\mu }_{1}\\left( A\\right) = {\\mu }_{2}\\left( A\\right) \) for every \( A \\in \\mathcal{B} \) .
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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.
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Yes
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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.
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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 \( \left( {A}_{n}\right) \) of measurable sets,\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{n}{A}_{n}}\right) \leq \mathop{\sum }\limits_{0}^{\infty }\mu \left( {A}_{n}\right) \]\n\n(iii) If the \( {A}_{n} \) ’s are measurable and nondecreasing, then\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{n}{A}_{n}}\right) = \lim \mu \left( {A}_{n}\right) \]\n\n(iv) If \( \mu \) is finite and \( \left( {A}_{n}\right) \) a nonincreasing sequence of measurable sets, then\n\n\[ \mu \left( {\mathop{\bigcap }\limits_{n}{A}_{n}}\right) = \lim \mu \left( {A}_{n}\right) \]
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Yes
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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 \( {\mu }_{1}\left( A\right) = {\mu }_{2}\left( A\right) \) for every \( A \in \mathcal{A} \) . Then \( {\mu }_{1}\left( A\right) = {\mu }_{2}\left( A\right) \) for every \( A \in \mathcal{B} \) .
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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.
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Yes
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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 \( \mu \) .
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Null
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No
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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 \( A = \) \( \mathop{\bigcup }\limits_{{k = 0}}^{n}{I}_{k} \) . Set\n\n\[ \lambda \left( A\right) = \mathop{\sum }\limits_{{k = 0}}^{n}\left| {I}_{k}\right| \]\n\nThen \( \lambda \) is a \( \sigma \) -finite measure on \( \mathcal{A} \) .
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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.
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No
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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 measure on \( \mathcal{D} \) satisfying\n\n\[ \n\mu \times \nu \left( {A \times B}\right) = \mu \left( A\right) \cdot \nu \left( B\right) .\n\]\n\nThen \( \mu \times \nu \) is countably additive. (Show this.)
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Null
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No
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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 \( {\pi }_{F} : X \rightarrow \) \( \mathop{\prod }\limits_{{n \in F}}{X}_{n} \) be the canonical projection map. Define\n\n\[ \n\mathcal{A} = \left\{ {{\pi }_{F}^{-1}\left( R\right) : R \in {\bigotimes }_{n \in F}{\mathcal{A}}_{n}, F\text{ finite }}\right\} .\n\]\n\nThen \( \mathcal{A} \) is an algebra that generates the product \( \sigma \) -algebra \( {\bigotimes }_{n}{\mathcal{A}}_{n} \) . Define \( \mathop{\prod }\limits_{n}{\mu }_{n} \) on \( \mathcal{A} \) by\n\n\[ \n\mathop{\prod }\limits_{n}{\mu }_{n}\left( {{\pi }_{F}^{-1}\left( R\right) }\right) = \left( {{ \times }_{i \in F}{\mu }_{i}}\right) \left( R\right)\n\]\n\nThen \( \mathop{\prod }\limits_{n}{\mu }_{n} \) defines a probability on \( \mathcal{A} \) . By 3.4.6, there is a unique probability on \( {\bigotimes }_{n}{\mathcal{A}}_{n} \) that extends \( \mathop{\prod }\limits_{n}{\mu }_{n} \) . We call it the product of the \( {\mu }_{n} \) ’s and denote it by \( \mathop{\prod }\limits_{n}{\mu }_{n} \) . If \( \left( {{X}_{n},{\mathcal{A}}_{n},{\mu }_{n}}\right) \) are the same, say \( {\mu }_{n} = \mu \) for all \( n \), then we shall denote the product measure simply by \( {\mu }^{\mathbb{N}} \) .
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Null
|
No
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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.
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Null
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No
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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\[ \n{\\mu }^{ * }\\left( A\\right) = \\inf \\{ \\mu \\left( B\\right) : B \\in \\mathcal{A}\\& A \\subseteq B\\} .\n\]\n\nIt is routine to check that \( {\\mu }^{ * } \) is an outer measure on \( X \) . The set function \( {\\mu }^{ * } \) is called the outer measure induced by \( \\mu \) . Clearly, for every set \( A \) there is a set \( B \\in \\mathcal{A} \) such that \( A \\subseteq B \) and \( \\mu \\left( B\\right) = {\\mu }^{ * }\\left( A\\right) \) . Note that if \( {B}^{\\prime } \) is another measurable set containing \( A \) then \( B \\smallsetminus {B}^{\\prime } \) is null.
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Null
|
No
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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 \( B \) ,
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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 \( \mu \) is finite, \n\n\[ \mu \left( B\right) = \inf \mu \left( {U}_{n}\right) = \lim \mu \left( {U}_{n}\right) . \]\n\nThus every closed set has the above property. \( \mathcal{D} \) is clearly closed under complementation.\n\nNow let \( {B}_{0},{B}_{1},{B}_{2},\ldots \) belong to \( \mathcal{D} \), and \( B = \mathop{\bigcup }\limits_{n}{B}_{n} \) . Fix \( \epsilon > 0 \) . Choose \( N \) such that \( \mu \left( {B \smallsetminus \mathop{\bigcup }\limits_{{i < N}}{B}_{i}}\right) < \epsilon /2 \) . For each \( 0 \leq i \leq N \), there is a closed set \( {F}_{i} \subseteq {B}_{i} \) such that \( \mu \left( {{B}_{i} \smallsetminus {F}_{i}}\right) < \epsilon /\left( {2\left( {N + 1}\right) }\right) \) . It is easy to check that \( \mu \left( {B \smallsetminus \mathop{\bigcup }\limits_{{i < N}}{F}_{i}}\right) < \epsilon \)\n\nTo show the other equality, choose closed sets \( {F}_{i} \subseteq {B}_{i}^{c} \) such that \( \mu \left( {B}_{i}^{c}\right. \smallsetminus \) \( \left. {F}_{i}\right) < \epsilon /{2}^{i + 1} \) . As \( {B}^{c} \smallsetminus \bigcap {F}_{i} \subseteq \bigcup \left( {{B}_{i}^{c} \smallsetminus {F}_{i}}\right) \), it follows that \( \mu \left( {{B}^{c} \smallsetminus \bigcap {F}_{i}}\right) < \epsilon \) . Take \( U = {\left( \bigcap {F}_{i}\right) }^{c} \) . Then \( U \) is an open set containing \( B \) such that \( \mu \left( {U \smallsetminus B}\right) < \) \( \epsilon \) . It follows that \( \mathcal{D} \) is closed under countable unions too. The result follows.
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Yes
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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 .
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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 \( x \in \left( {-a, a}\right) \) . In particular, \( E \cap (E + \) \( x) \neq \varnothing \) for every \( x \in \left( {-a, a}\right) \) . It follows that \( \left( {-a, a}\right) \subseteq E - E \) .
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Yes
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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 \( \mathbf{{AC}} \), there exists a set \( S \) intersecting each coset in exactly one point. We claim that \( S \) is not Lebesgue measurable.
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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 contradiction. Now, let \( \lambda \left( S\right) > 0 \) . By 3.4.17, \( S - S \) contains a nonempty open interval. Hence, there are distinct points \( x, y \) in \( S \) such that \( x - y \) is rational. We have arrived at a contradiction again.
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Yes
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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 \( \mu \left( {X \smallsetminus K}\right) < \epsilon \) .
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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 \( \left( {F}_{s}\right) \leq 1/{2}^{\left| s\right| } \) .\n\nTo see that such a system exists, we proceed by induction on \( \left| s\right| \) . Suppose \( {F}_{s} \) has been defined. Since \( X \) is second countable, there is a sequence \( \left( {U}_{n}\right) \) of open sets of diameter \( \leq {2}^{-\left| s\right| } \) covering \( {F}_{s} \), and further, \( {F}_{s}\bigcap {U}_{n} \neq \varnothing \) for all \( n \) . Take \( {F}_{s \cap n} = \operatorname{cl}\left( {{F}_{s} \cap {U}_{n}}\right) \) .\n\nBy an easy induction, we now define positive integers \( {n}_{0},{n}_{1},{n}_{2},\ldots \) such that the following conditions hold: for every \( s = \left( {{m}_{0},{m}_{1},\ldots ,{m}_{k - 1}}\right) \) with \( {m}_{i} \leq {n}_{i} \)\n\n\[ \mu \left( {{F}_{s} \smallsetminus \mathop{\bigcup }\limits_{{j \leq {n}_{k}}}{F}_{s \cap j}}\right) < \frac{\epsilon }{{2}^{k + 1} \cdot {n}_{0}\ldots \cdot {n}_{k - 1}}. \]\n\nSet\n\n\[ K = \mathop{\bigcap }\limits_{k}\mathop{\bigcup }\limits_{s}{F}_{s} \]\n\nwhere the union varies over all \( s = \left( {{m}_{0},{m}_{1},\ldots ,{m}_{k - 1}}\right) \) with \( {m}_{i} \leq {n}_{i} \) . It is easy to check that \( K \) is closed and totally bounded and hence compact. Further, \( \mu \left( {X \smallsetminus K}\right) < \epsilon \) .
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Yes
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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 \( \mu \left( {B \smallsetminus K}\right) < \epsilon \) .
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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 \( B \) such that \( \mu \left( {B \smallsetminus K}\right) < \epsilon \) . Since \( K \) is compact with respect to the original topology too, the result follows.
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Yes
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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 \( I,\lambda \left( B\right) = \) \( \mu \left( {{h}^{-1}\left( B\right) }\right) \) .
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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\[ N = \left\{ {y \in I : {F}^{-1}\left( {\{ y\} }\right) }\right. \text{contains more than one point}\} \text{.} \]\n\nSince \( F \) is monotone, \( N \) is countable. If \( N \) is empty, take \( h = F \) . Otherwise, we take an uncountable Borel set \( M \subset I \smallsetminus N \) of Lebesgue measure 0, e.g., \( \mathcal{C} \smallsetminus N \) . So, \( \mu \left( {{F}^{-1}\left( M\right) }\right) = 0 \) . Put \( Q = M \cup N \) and \( P = {F}^{-1}\left( Q\right) \) . Both \( P \) and \( Q \) are uncountable Borel sets with \( \mu \left( P\right) = \lambda \left( Q\right) = 0 \) . Fix a Borel isomorphism \( g : P \rightarrow Q \) . Define\n\n\[ h\left( x\right) = \left\{ \begin{array}{lll} g\left( x\right) & \text{ if } & x \in P, \\ F\left( x\right) & \text{ if } & x \in I \smallsetminus P. \end{array}\right. \]\n\nThe map \( h \) has the desired properties.
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Yes
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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.
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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, \( \mathcal{B} \) is a monotone class. As finite disjoint unions of measurable rectangles form an algebra generating \( \mathcal{A} \otimes {\mathcal{B}}_{Y} \), the result follows from 3.1.14.
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Yes
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Proposition 3.5.1 The collection \( \mathcal{D} \) of all subsets of a topological space \( X \) having the Baire property forms a \( \sigma \) -algebra.
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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\[ \left( {\mathop{\bigcup }\limits_{n}{A}_{n}}\right) \Delta \left( {\mathop{\bigcup }\limits_{n}{U}_{n}}\right) \subseteq \mathop{\bigcup }\limits_{n}\left( {{A}_{n}\Delta {U}_{n}}\right) \]\n\nand the union of a sequence of meager sets is meager, \( \mathop{\bigcup }\limits_{n}{A}_{n} \in \mathcal{D} \) .\n\nClosure under complementation: Let \( A \in \mathcal{D} \) and let \( U \) be an open set such that \( {A\Delta U} \) is meager. We have\n\n\[ \left( {X \smallsetminus A}\right) \Delta \operatorname{int}\left( {X \smallsetminus U}\right) \]\n\n\[ \subseteq \;\left( {\left( {X \smallsetminus A}\right) \Delta \left( {X \smallsetminus U}\right) }\right) \bigcup \left( {\left( {X \smallsetminus U}\right) \smallsetminus \operatorname{Int}\left( {X \smallsetminus U}\right) }\right) .\n\n\[ \text{As}\left( {X \smallsetminus A}\right) \Delta \left( {X \smallsetminus U}\right) = {A\Delta U} \]\n\n\[ \left( {X \smallsetminus A}\right) \Delta \operatorname{int}\left( {X \smallsetminus U}\right) \subseteq \left( {A\Delta U}\right) \bigcup \left( {\left( {X \smallsetminus U}\right) \smallsetminus \operatorname{Int}\left( {X \smallsetminus U}\right) }\right) .\n\nSince for any closed set \( F, F \smallsetminus \operatorname{int}\left( F\right) \) is nowhere dense, \( \left( {X \smallsetminus A}\right) \Delta \operatorname{int}(X \smallsetminus \)\n\n\( U) \) is meager.\n\nThe result follows.\n\nThe \( \sigma \) -algebra \( \mathcal{D} \) defined above is called the Baire \( \sigma \) -algebra of \( X \) .
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Yes
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Corollary 3.5.2 Every Borel subset of a metrizable space has the Baire property.
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Null
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No
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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 \( X \) .
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Null
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No
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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 comeager in \( U \) for each \( n \) .\n\n(ii) Let \( A \subseteq X \) be a nonmeager set with \( {BP} \) . Then \( A \) is comeager in \( {U}_{n} \) for some \( n \) .\n\n(iii) A set \( A \) with \( {BP} \) is comeager if and only if \( A \) is nonmeager in each \( {U}_{n} \) .
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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, \( \mathop{\bigcup }\limits_{n}\left( {U \smallsetminus {A}_{n}}\right) = U \smallsetminus \mathop{\bigcap }\limits_{n}{A}_{n} \) is meager in \( U \) . Thus we have proved (i).\n\nTo prove (ii), take \( A \) with BP. Write \( A = {V\Delta I}, V \) open, \( I \) meager. If \( A \) is nonmeager, \( V \) must be nonempty. Then \( A \) is comeager in every \( {U}_{n} \) contained in \( V \) .\n\nWe now prove (iii). Let \( A \) be comeager. Then trivially \( {U}_{n} \smallsetminus A \) is meager for all \( n \) . As \( {U}_{n} \) is open, it follows that \( {U}_{n} \smallsetminus A \) is meager in \( {U}_{n} \) . Since \( X \) is a Baire space, this implies that \( A \) is nonmeager in \( {U}_{n} \) . Conversely, let \( A \) be not comeager; i.e., \( {A}^{c} \) is not meager. So, there is \( {U}_{n} \) such that \( {A}^{c} \) is comeager in \( {U}_{n} \) ; i.e., \( A \) is meager in \( {U}_{n} \) .
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Yes
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Proposition 3.5.7 A topological group is Baire if and only if it is of second category in itself.
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Proof. The \
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No
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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.
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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, \( f \mid \left( {X \smallsetminus I}\right) \) is continuous.
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Yes
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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.
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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\[ J = \left( {{g}^{-1} \cdot I}\right) \bigcup \mathop{\bigcup }\limits_{k}\left( {{g}_{k}^{-1} \cdot I}\right) . \]\n\nBy 2.4.7, \( J \) is meager. Since \( G \) is completely metrizable, it is of second category in itself by 2.5.6. In particular, \( J \neq G \) . Take any \( h \in G \smallsetminus J \) . Then, \( {g}_{k} \cdot h, g \cdot h \) are all in \( G \smallsetminus I \) . Further, \( {g}_{k} \cdot h \rightarrow g \cdot h \) as \( k \rightarrow \infty \) . Since \( \varphi \mid \left( {G \smallsetminus I}\right) \) is continuous, \( \varphi \left( {{g}_{k} \cdot h}\right) \rightarrow \varphi \left( {g \cdot h}\right) \) ; i.e., \( \varphi \left( {g}_{k}\right) \cdot \varphi \left( h\right) \rightarrow \varphi \left( g\right) \cdot \varphi \left( h\right) \) . Multiplying by \( {\left( \varphi \left( h\right) \right) }^{-1} \) from the right, we have \( \varphi \left( {g}_{k}\right) \rightarrow \varphi \left( g\right) \) .
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Yes
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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 \( {\mathbb{Q}}^{ + } \) is countable, \( \varphi \) is trivially Borel. It is not continuous. To see this, take \( {q}_{n} = 1 - {2}^{-n} \) . Then \( \varphi \left( {q}_{n}\right) = - n \) . As \( {q}_{n} \) converges and \( \varphi \left( {q}_{n}\right) \) does not, \( \varphi \) is not continuous.
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Null
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No
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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 \( {\mathrm{H}}^{-1}\mathrm{H} \) .
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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 will then follow that \( V \subseteq {H}^{-1}H \) , and the proof will be complete.\n\nLet \( h \in H \) . Note that\n\n\[ \left( {U\bigcap {Uh}}\right) \Delta \left( {H\bigcap {Hh}}\right) \subseteq \left( {U\Delta H}\right) \bigcup \left( {\left( {U\Delta H}\right) h}\right) .\n\]\n\n\( \left( *\right) \)\n\nSo, \( \left( {U\bigcap {Uh}}\right) \Delta \left( {H\bigcap {Hh}}\right) \) is meager. As \( {gV} \subseteq U\bigcap {Uh} \) and \( G \) is Baire, \( U \cap {Uh} \) is nonmeager. Therefore, \( H \cap {Hh} \) is nonmeager by \( \left( \star \right) \) .
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Yes
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Corollary 3.5.13 Every nonmeager Borel subgroup \( H \) of a Polish group \( G \) is clopen.
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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.
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Yes
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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.
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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\[ {W}_{n} = {\pi }_{X}\left( {U\bigcap \left( {X \times {V}_{n}}\right) }\right) \]\n\nit is open. Also, \( {W}_{n} \) is dense. Suppose not. Then \( \left( {X \smallsetminus \operatorname{cl}\left( {W}_{n}\right) }\right) \times {V}_{n} \) is a nonempty open set disjoint from \( U \) . As \( U \) is dense, this is a contradiction.\n\nSince for any \( x \in X \) ,\n\n\( {A}_{x} \) is nowhere dense \( \Leftrightarrow {U}_{x} \) is dense, it follows that\n\n\[ \left\{ {x \in X : {A}_{x}}\right. \text{is nowhere dense}\} = \mathop{\bigcap }\limits_{n}{W}_{n}\text{.} \]\n\nSince \( X \) is a Baire space, the result follows.
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Yes
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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.
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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 \( U \) . Since \( U \) is nonmeager, \( {A}^{*V} \) is nonmeager. In particular, \( {A}^{\Delta X} \) is not meager; i.e.,(ii) is false.
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Yes
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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(iii) \( \left\{ {y \in Y : {A}^{y}}\right. \) is meager (comeager) \( \} \) is comeager.
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Null
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No
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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 \( \\left. U\\right\\} \) are in \( \\mathcal{A} \) .
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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 \( A = B \\times V, B \\in \\mathcal{A} \), and \( V \) open in \( Y \). Then \( {A}^{\\Delta U} \) equals \( B \) if \( U \\cap V \\neq \\varnothing \). Otherwise it is empty. Hence, \( A \\in \\mathcal{B} \). Our proof will be complete if we show that \( \\mathcal{B} \) is closed under countable unions and complementation. For every sequence \( \\left( {A}_{n}\\right) \) of subsets \( X \\times Y \), \[ {\\left( \\mathop{\\bigcup }\\limits_{n}{A}_{n}\\right) }^{\\Delta U} = \\mathop{\\bigcup }\\limits_{n}{A}_{n}^{\\Delta U} \] So, \( \\mathcal{B} \) is closed under countable unions. Let \( A \\in \\mathcal{B} \) and \( U \) open in \( Y \). Let \( x \\in X \). We have \( {\\left( {A}^{c}\\right) }_{x} \) is meager in \( U \\Leftrightarrow {A}_{x} \) is comeager in \( U \) \[ \\Leftrightarrow \\forall {U}_{n} \\subseteq U\\left( {A}_{x}\\right. \\text{is nonmeager in}\\left. {U}_{n}\\right) \\text{.} \] Therefore, \[ {\\left( {A}^{c}\\right) }^{\\Delta U} = {\\left( \\mathop{\\bigcap }\\limits_{{{U}_{n} \\subseteq U}}{A}^{\\Delta {U}_{n}}\\right) }^{c} \] Hence, \( {A}^{c} \\in \\mathcal{B} \). Step 2. Let \( A \\in \\mathcal{A}\\bigotimes {\\mathcal{B}}_{Y} \) and \( U \) be open in \( Y \). Then \[ {A}^{*U} = \\mathop{\\bigcap }\\limits_{{{U}_{n} \\subseteq U}}{A}^{\\Delta {U}_{n}} \] Therefore, \( {A}^{*U} \\in \\mathcal{A} \) by step 1. The remaining part of the result follows easily.
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Yes
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Every \( \sigma \) -finite complete measure space is Marczewski complete.
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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 the general case, write \( A = \bigcup {A}_{n} \) such that \( {\mu }^{ * }\left( {A}_{n}\right) < \infty \) . Since \( \mu \) is \( \sigma \) -finite, this is possible. Take \( \widehat{A} = \bigcup {\widehat{A}}_{n} \) .
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No
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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.
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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 \( W \) . Now note that\n\n\[ \n{A}^{ * } \smallsetminus A \subseteq \left( {{A}^{ * } \smallsetminus W}\right) \bigcup \left( {W \smallsetminus A}\right) \subseteq \left( {\operatorname{cl}\left( W\right) \smallsetminus W}\right) \bigcup \left( {W \smallsetminus A}\right) .\n\]\n\nThis shows that \( {A}^{ * } \smallsetminus A \) is meager. Let \( B \) be any meager \( {F}_{\sigma } \) set containing \( {A}^{ * } \smallsetminus A \) . Take \( \widehat{A} = {A}^{ * } \cup B \) .
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Yes
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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.
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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 \( s \in {\mathbb{N}}^{ < \mathbb{N}} \), let\n\n\[ \n{B}^{s} = \mathop{\bigcup }\limits_{{\{ \alpha : s \prec \alpha \} }}\mathop{\bigcap }\limits_{n}{B}_{\alpha \mid n} \subseteq {B}_{s} \n\]\n\nNote that \( {B}^{e} = B \) and \( {B}^{s} = \mathop{\bigcup }\limits_{n}{B}^{s \uparrow n} \) for all \( s \) . For each \( s \in {\mathbb{N}}^{ < \mathbb{N}} \), choose a minimal \( \mathcal{B} \) -cover \( {\widehat{B}}^{s} \) of \( {B}^{s} \) . Since \( {B}^{s} \subseteq {B}_{s} \), by replacing \( {\widehat{B}}^{s} \) by \( {B}_{s}\bigcap {\widehat{B}}^{s} \) we may assume that \( {\widehat{B}}^{s} \subseteq {B}_{s} \) . Further, by replacing \( {\widehat{B}}^{s} \) by \( \mathop{\bigcap }\limits_{{t \preccurlyeq s}}{\widehat{B}}^{t} \), we can assume that \( \left\{ {{\widehat{B}}^{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) is regular. Let\n\n\[ \n{C}_{s} = {\widehat{B}}^{s} \smallsetminus \mathop{\bigcup }\limits_{n}{\widehat{B}}^{s}{}^{rn}. \n\]\n\nSince \( {B}^{s} = \mathop{\bigcup }\limits_{n}{B}^{{s}^{ \frown }n} \subseteq \mathop{\bigcup }\limits_{n}{\widehat{B}}^{{s}^{ \frown }n} \), every subset of \( {C}_{s} \) is in \( \mathcal{B} \) . Let \( C = \mathop{\bigcup }\limits_{s}{C}_{s} \) .\n\nClaim: \( {\widehat{B}}^{e} \smallsetminus C \subseteq B \) .\n\nAssuming the claim, we complete the proof as follows. Since \( {\widehat{B}}^{e} \smallsetminus B \subseteq C \) and since every subset of \( C \) is in \( \mathcal{B} \), it follows that \( {\widehat{B}}^{e} \smallsetminus B \in \mathcal{B} \) . As \( B = \) \( {\widehat{B}}^{e} \smallsetminus \left( {{\widehat{B}}^{e} \smallsetminus B}\right) \), it belongs to \( \mathcal{B} \) .\n\nProof of the claim. Let \( x \in {\widehat{B}}^{e} \smallsetminus C \) . Since \( x \notin C, x \notin {C}_{e} \) . Since \( x \in {\widehat{B}}^{e} \), there is \( \alpha \left( 0\right) \in \mathbb{N} \) such that \( x \in {\widehat{B}}^{\alpha \left( 0\right) } \) . Suppose \( n > 0 \) and \( \alpha \left( 0\right) ,\alpha \left( 1\right) ,\ldots ,\alpha \left( {n - 1}\right) \) have been defined such that \( x \in {\widehat{B}}^{s} \), where \( s = \) \( \left( {\alpha \left( 0\right) ,\alpha \left( 1\right) ,\ldots ,\alpha \left( {n - 1}\right) }\right) \) . Since \( x \notin {C}_{s} \), there is \( \alpha \left( n\right) \in \mathbb{N} \) such that \( x \in \) \( {\widehat{B}}^{s \land \alpha \left( n\right) } \) . Since \( {\widehat{B}}^{\alpha \mid n} \subset {B}_{\alpha \mid n} \) for all \( n \), we conclude that \( x \in B \) .
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Yes
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Proposition 3.6.1 (i) For every \( 1 \leq \alpha < {\omega }_{1} \), \[ {\mathbf{\sum }}_{\alpha }^{0},{\mathbf{\Pi }}_{\alpha }^{0} \subseteq {\mathbf{\Delta }}_{\alpha + 1}^{0} \]
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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 \( \alpha \) .
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Yes
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Proposition 3.6.3 Every set of additive class \( \alpha > 2 \) is a countable disjoint union of multiplicative class \( < \alpha \) sets.
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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 \( {B}_{n} = \mathop{\bigcup }\limits_{k}{B}_{k}^{n} \), where the \( {B}_{k}^{n} \) ’s are pairwise disjoint ambiguous class \( < \alpha \) sets. This is possible since \( \alpha > 2 \) . We have\n\n\[ A = {A}_{0}\bigcup \left( {{A}_{1} \cap {B}_{0}}\right) \bigcup \left( {{A}_{2} \cap {B}_{1}}\right) \cup \cdots \]\n\n\[ = \;{A}_{0}\bigcup \mathop{\bigcup }\limits_{{n \geq 1}}\mathop{\bigcup }\limits_{k}\left( {{A}_{n}\bigcap {B}_{k}^{n - 1}}\right) ,\]\n\nand the result follows.
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Yes
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Proposition 3.6.3 Every set of additive class \( \alpha > 2 \) is a countable disjoint union of multiplicative class \( < \alpha \) sets.
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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 \( {B}_{n} = \mathop{\bigcup }\limits_{k}{B}_{k}^{n} \), where the \( {B}_{k}^{n} \) ’s are pairwise disjoint ambiguous class \( < \alpha \) sets. This is possible since \( \alpha > 2 \) . We have\n\n\[ A = {A}_{0}\bigcup \left( {{A}_{1} \cap {B}_{0}}\right) \bigcup \left( {{A}_{2} \cap {B}_{1}}\right) \cup \cdots \]\n\n\[ = \;{A}_{0}\bigcup \mathop{\bigcup }\limits_{{n \geq 1}}\mathop{\bigcup }\limits_{k}\left( {{A}_{n}\bigcap {B}_{k}^{n - 1}}\right) ,\]\n\nand the result follows.
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Yes
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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 \( U \in {\mathbf{\Gamma }}_{\alpha }\left( {{\mathbb{N}}^{\mathbb{N}} \times Y}}\right) \) such that \[ A \in {\mathbf{\Gamma }}_{\alpha }\left( Y\right) \Leftrightarrow \left( {\exists x \in {\mathbb{N}}^{\mathbb{N}}}\right) \left( {A = {U}_{x}}\right) . \] We call such a set \( U \) universal for \( {\mathbf{\Gamma }}_{\alpha }\left( Y\right) \) .
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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 \[ \left( {x, y}\right) \in U \Leftrightarrow y \in \mathop{\bigcup }\limits_{n}{V}_{x\left( n\right) } \] Evidently, \( A \) is open in \( Y \) if and only if \( A = {U}_{x} \) for some \( x \) . It remains to show that \( U \) is open. Let \( \left( {{x}_{0},{y}_{0}}\right) \in U \) . Then there is an \( n \) such that \( {y}_{0} \in {V}_{{x}_{0}\left( n\right) } \) . Then \[ \left( {{x}_{0},{y}_{0}}\right) \in \left\{ {x \in {\mathbb{N}}^{\mathbb{N}} : x\left( n\right) = {x}_{0}\left( n\right) }\right\} \times {V}_{{x}_{0}\left( n\right) } \subseteq U. \] Thus \( U \) is open. Let \( W = {U}^{c} \), where \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times Y \) is universal for open sets. Clearly, \( W \) is universal for closed sets. The result for \( \alpha = 1 \) is proved. Suppose \( \alpha > 1 \) and the result has been proved for all \( \beta < \alpha \) . Case 1: \( \alpha \) is a limit ordinal Fix a sequence of countable ordinals \( \left( {\alpha }_{n}\right) ,1 < {\alpha }_{n} < \alpha \), such that \( \alpha = \sup {\alpha }_{n} \) . Let \( {U}_{n} \) be universal for multiplicative class \( {\alpha }_{n}, n \in \mathbb{N} \) . For \( x \in {\mathbb{N}}^{\mathbb{N}} \) and \( n \in \mathbb{N} \), define \( {x}_{n} \in {\mathbb{N}}^{\mathbb{N}} \) by \[ {x}_{n}\left( m\right) = x\left( {{2}^{n}\left( {{2m} + 1}\right) - 1}\right) . \] \( \left( *\right) \) For each \( n, x \rightarrow {x}_{n} \) is a continuous function. Define \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times Y \) by \[ \left( {x, y}\right) \in U \Leftrightarrow \left( {\exists n}\right) \left( {\left( {{x}_{n}, y}\right) \in {U}_{n}}\right) \] It is routine to check that \( U \) is universal for \( {\mathbf{\sum }}_{\alpha }^{0}\left( Y\right) \) . Case 2: \( \alpha = \beta + 1 \), a successor ordinal Fix a universal \( {\mathbf{\Pi }}_{\beta }^{0} \) set \( P \subseteq {\mathbb{N}}^{\mathbb{N}} \times Y \) . Define \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times Y \) by \[ \left( {x, y}\right) \in U \Leftrightarrow \left( {\exists n}\right) \left( {\left( {{x}_{n}, y}\right) \in P}\right) , \] where \( {x}_{n} \) is as defined in \( \left( \star \right) \) . Clearly, \( U \) is universal for \( {\mathbf{\sum }}_{\alpha }^{0}\left( Y\right) \) . Having defined a universal \( {\mathbf{\sum }}_{\alpha }^{0} \) set \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times Y \), note that \( {U}^{c} \) is universal for \( {\Pi }_{\alpha }^{0}\left( Y\right) \) .
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Yes
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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 \( {\mathbf{\Gamma }}_{\alpha }\left( X\right) \) .
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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 \( V \in {\mathbf{\Gamma }}_{\alpha }\left( {X \times X}\right) \) . The set \( V \) is universal for \( {\mathbf{\Gamma }}_{\alpha }\left( X\right) \) .
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Yes
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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 \( \alpha \) .
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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 \( A \) is not of multiplicative class \( \alpha \) . To the contrary, suppose \( A \) is of multiplicative class \( \alpha \) . Choose \( {x}_{0} \in X \) such that \( {A}^{c} = {U}_{{x}_{0}} \) . Then\n\n\[ {x}_{0} \in {A}^{c} \Leftrightarrow \left( {{x}_{0},{x}_{0}}\right) \in U \Leftrightarrow {x}_{0} \in A. \]\n\nThis is a contradiction.
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Yes
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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 \( \mathbf{\Delta }\left( X\right) \) .
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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, \( A \in \mathbf{\Delta } \) . As \( \mathbf{\Delta } \) is closed under complementation, \( {A}^{c} \in \mathbf{\Delta } \) . Let \( {A}^{c} = {U}_{{x}_{0}} \) for some \( {x}_{0} \in X \) . Then\n\n\[ {x}_{0} \in {A}^{c} \Leftrightarrow \left( {{x}_{0},{x}_{0}}\right) \in U \Leftrightarrow {x}_{0} \in A. \]\n\nThis is a contradiction.
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Yes
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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 pairwise disjoint sets of additive class \( \alpha \), and\n\n(b) \( \mathop{\bigcup }\limits_{n}{A}_{n} = \mathop{\bigcup }\limits_{n}{B}_{n} \) .\n\n(See Figure 3.1.) Consequently the \( {B}_{n} \) ’s are of ambiguous class \( \alpha \) if \( \mathop{\bigcup }\limits_{n}{A}_{n} \) is so.\n\nThe result is also true for \( \alpha = 1 \) if \( X \) is zero-dimensional and second countable.
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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 \( \left\{ {{C}_{nm} : n, m \in \mathbb{N}}\right\} \) in a single sequence, say \( \left( {D}_{i}\right) \) . Let\n\n\[ \n{E}_{i} = {D}_{i} \smallsetminus \mathop{\bigcup }\limits_{{j < i}}{D}_{j} \n\]\n\nTake\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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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 \( {B}_{n} \supseteq {A}_{n} \) with \( \bigcap {B}_{n} = \varnothing \) .
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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 \( {B}_{n} = {C}_{n}^{c} \) .
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Yes
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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 \( U \subseteq {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \) a universal \( {\mathbf{\sum }}_{\gamma }^{0} \) set. Define\n\n\[ \n{U}_{i} = \left\{ {\left( {\alpha ,\beta }\right) : \left( {{\alpha }_{i},\beta }\right) \in U}\right\} ,\;i = 0\text{ or }1.\n\]\n\nIt easy quite easy to check that \( {U}_{0},{U}_{1} \) are additive class \( \gamma \) sets such that for every pair \( \left( {{A}_{0},{A}_{1}}\right) \) of additive class \( \gamma \) sets, there exists an \( \alpha \in {\mathbb{N}}^{\mathbb{N}} \) such that \( {\left( {U}_{i}\right) }_{\alpha } = {A}_{i}, i = 0 \) or 1 . Such a pair of sets \( {U}_{0},{U}_{1} \) will be called a universal pair for additive class \( \gamma \) .
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Null
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No
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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 \( A = \lim {A}_{n} \) .
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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 \( {D}_{n + 1} \subseteq {D}_{n} \) . By 3.6.11, there is a set \( {A}_{n} \) of ambiguous class \( < \alpha \) such that\n\n\[ {C}_{n} \subseteq {A}_{n} \subseteq {D}_{n} \]\n\nThen \( A = \lim {A}_{n} \) as we now show. Let \( x \in \lim \sup {A}_{n} \) . Thus, \( x \in {A}_{n} \) for infinitely many \( n \) . Then \( x \in {D}_{n} \) for infinitely many \( n \) and hence for all \( n \) . Therefore,\n\n\[ \lim \sup {A}_{n} \subseteq A\text{.} \]\n\n(1)\n\nNow let \( x \in A \) . Then \( x \in {C}_{n} \) for all but finitely many \( n \) . Since \( {C}_{n} \subseteq {A}_{n} \) for all \( n \), \n\n\[ A \subseteq \liminf {A}_{n}\text{.} \]\n\n(2)\n\nThe result follows from (1) and (2).
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Yes
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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 \( {\mathbf{\sum }}_{\alpha }^{0}\left( X\right) \) with \( \lim \sup {B}_{n} = \varnothing \) .
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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, \( \lim \sup {A}_{n} = \varnothing \) . Suppose there exist \( {B}_{n} \supseteq {A}_{n} \) in \( {\mathbf{\sum }}_{\alpha }^{0}\left( X\right) \) with \( \lim \sup {B}_{n} = \) \( \varnothing \) . We shall get a contradiction.\n\nBy 3.6.11, there is a set \( {C}_{n} \in {\mathbf{\Delta }}_{\alpha }^{0}\left( X\right) \) such that \( {A}_{n} \subseteq {C}_{n} \subseteq {B}_{n} \) . Note that \( \lim \sup {C}_{n} = \varnothing \) . As the sets \( {A}_{n} \) are in \( {\mathbf{\Delta }}_{\alpha + 1}^{0}\left( X\right) \), by 3.6.13 there are sets \( {A}_{n}^{k} \in {\mathbf{\Delta }}_{\alpha }^{0}\left( X\right) \) such that \( {A}_{n} = \mathop{\lim }\limits_{k}{A}_{n}^{k} \) . Now define\n\n\[ \n{D}_{k} = \left( {{A}_{1}^{k}\bigcap {C}_{1}}\right) \bigcup \left( {{A}_{2}^{k}\bigcap {C}_{2}}\right) \bigcup \cdots \bigcup \left( {{A}_{k}^{k}\bigcap {C}_{k}}\right) .\n\]\n\nThen \( {D}_{k} \in {\mathbf{\Delta }}_{\alpha }^{0}\left( X\right) \) . It is now fairly easy to check that \( \lim \sup {D}_{k} \subseteq A \subseteq \) \( \liminf {D}_{k} \), so \( A = \lim {D}_{k} \) . This implies that \( A \in {\mathbf{\Delta }}_{\alpha + 1}^{0}\left( X\right) \), and we have arrived at a contradiction.
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Yes
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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 that \( {f}_{n} \rightarrow f \) pointwise.
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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 all \( x \)
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No
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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.
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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 \( \epsilon \) -net in \( Y \) . Set\n\n\[ \n{A}_{i} = {f}^{-1}\left( {B\left( {{y}_{i},\epsilon }\right) }\right) \n\]\n\nThe sets \( {A}_{1},{A}_{2},\ldots ,{A}_{n} \) are of additive class \( \alpha \) with union \( X \) . By 3.6.10, there are pairwise disjoint ambiguous class \( \alpha \) sets \( {B}_{1},{B}_{2},\ldots ,{B}_{n} \) such that\n\n\[ \n{B}_{1} \subseteq {A}_{1},{B}_{2} \subseteq {A}_{2},\ldots ,{B}_{n} \subseteq {A}_{n} \n\]\n\nand\n\n\[ \n\bigcup {B}_{i} = \bigcup {A}_{i} = X \n\]\n\nDefine \( g : X \rightarrow Y \) by\n\n\[ \ng\left( x\right) = {y}_{i}\text{ if }x \in {B}_{i} \n\]\n\nThen \( d\left( {f\left( x\right), g\left( x\right) }\right) < \epsilon \) for all \( x \) .
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Yes
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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 \( E \) .
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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 \( < \alpha \) such that \( {A}_{i} = \) \( \mathop{\lim }\limits_{m}{A}_{im} \) . Fix \( m \) . Let\n\n\[ \n{B}_{1}^{m} = {A}_{1m},{B}_{2}^{m} = {A}_{2m} \smallsetminus {A}_{1m},\ldots ,{B}_{n}^{m} = {A}_{nm} \smallsetminus \mathop{\bigcup }\limits_{{j < n}}{A}_{jm} \n\]\n\nand\n\n\[ \n{B}_{n + 1}^{m} = X \smallsetminus \mathop{\bigcup }\limits_{{j \leq n}}{A}_{jm} \n\]\n\nEvidently, the sets \( {B}_{1}^{m},{B}_{2}^{m},\ldots ,{B}_{n + 1}^{m} \) are pairwise disjoint and of ambiguous class less than \( \alpha \) with union \( X \) . So there is a function \( {f}_{m} : X \rightarrow Y \) of class \( < \alpha \) satisfying\n\n\[ \n{f}_{m}\left( x\right) = {y}_{i},\text{ if }x \in {B}_{i}^{m},\;1 \leq i \leq n. \n\]\n\nWe claim that \( {f}_{m}\left( {x}_{0}\right) \rightarrow f\left( {x}_{0}\right) \) for all \( {x}_{0} \in X \) . Assume that \( {x}_{0} \in {A}_{i} \) . So, \( f\left( {x}_{0}\right) = {y}_{i} \) . Since \( {x}_{0} \notin \lim \mathop{\sup }\limits_{m}{A}_{jm} \) for all \( j \neq i \), there is an integer \( M \) such that \( {x}_{0} \notin {A}_{jm} \) for \( m > M \) and \( j \neq i \) . Since \( {x}_{0} \in \lim \mathop{\inf }\limits_{m}{A}_{im} \), we can further assume that \( {x}_{0} \in {A}_{im} \) for all \( m > M \) . Thus, \( {f}_{m}\left( {x}_{0}\right) = {y}_{i} \) for all \( m > M \) . Hence, \( {f}_{m} \rightarrow f \) pointwise.
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Yes
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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 \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) whose range is \( A \) .\n\n(iv) There is a closed subset \( C \) of \( X \times {\mathbb{N}}^{\mathbb{N}} \) whose projection is \( A \) .\n\n(v) For every uncountable Polish space \( Y \) there is a \( {G}_{\delta } \) set \( B \) in \( X \times Y \) whose projection is \( A \) .
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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 \( {\mathbb{N}}^{\mathbb{N}} \) onto \( B \) . Take \( f = {\pi }_{X} \circ g \) . Since the range of \( f \) is \( A \) ,(ii) implies (iii).\n\nSince the graph of a continuous map \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) is a closed subset of \( {\mathbb{N}}^{\mathbb{N}} \times X \) with projection \( A \) ,(iii) implies (iv).\n\nBy 2.6.5, every uncountable Polish space \( Y \) contains a homeomorph of \( {\mathbb{N}}^{\mathbb{N}} \), which is necessarily a \( {G}_{\delta } \) set in \( Y \) . Therefore,(iv) implies (v).\n\n(i) trivially follows from (v).
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Yes
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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.
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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\[ C = \{ \left( {z, x}\right) \in Z \times X : \left( {f\left( z\right), x}\right) \in B\} . \]\n\nThe set \( C \) is Borel, and \( {\pi }_{X}\left( C\right) = {f}^{-1}\left( A\right) \) . So \( {f}^{-1}\left( A\right) \) is analytic.\n\nClosure under countable unions and countable intersections: Let \( {A}_{0},{A}_{1},{A}_{2},\ldots \) be analytic subsets of \( X \) . By 4.1.1, there are Borel subsets \( {B}_{0},{B}_{1},{B}_{2},\ldots \) of \( X \times {\mathbb{N}}^{\mathbb{N}} \) whose projections are \( {A}_{0},{A}_{1},{A}_{2},\ldots \) respectively. Take\n\n\[ C = \left\{ {\left( {x,\alpha }\right) \in X \times {\mathbb{N}}^{\mathbb{N}} : \left( {x,{\alpha }^{ * }}\right) \in {B}_{\alpha \left( 0\right) }}\right\} \]\n\nand\n\n\[ D = \left\{ {\left( {x,\alpha }\right) \in X \times {\mathbb{N}}^{\mathbb{N}} : \left( {x,{f}_{i}\left( \alpha \right) }\right) \in {B}_{i}\text{ for every }i}\right\} , \]\n\nwhere \( {\alpha }^{ * }\left( i\right) = \alpha \left( {i + 1}\right) \) and \( \left( {{f}_{0},{f}_{1},{f}_{2},\ldots }\right) : {\mathbb{N}}^{\mathbb{N}} \rightarrow {\left( {\mathbb{N}}^{\mathbb{N}}\right) }^{\mathbb{N}} \) is a continuous surjection. Note that the map \( \alpha \rightarrow {\alpha }^{ * } \) is also continuous. Hence, the sets \( C \) and \( D \) are Borel with projections \( \mathop{\bigcup }\limits_{i}{A}_{i} \) and \( \mathop{\bigcap }\limits_{i}{A}_{i} \) respectively. We have shown that \( {\mathbf{\sum }}_{1}^{1} \) is closed under countable unions and countable intersections. The closure properties of \( {\mathbf{\Pi }}_{1}^{1} \) follow.
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Yes
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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 \( {\mathbf{\sum }}_{1}^{1}\left( X\right) \) .
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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\[ U = \left\{ {\left( {\alpha, x}\right) \in {\mathbb{N}}^{\mathbb{N}} \times X : \left( {\alpha, x,\beta }\right) \in C\text{ for some }\beta }\right\} .\n\]\n\nAs \( U = {\exists }^{{\mathbb{N}}^{\mathbb{N}}}C \), it follows that \( U \in {\mathbf{\sum }}_{1}^{1} \) . Let \( A \subseteq X \) be \( {\mathbf{\sum }}_{1}^{1} \) . Choose a closed set \( F \subseteq X \times {\mathbb{N}}^{\mathbb{N}} \) whose projection is \( A \) (4.1.1). Let \( \alpha \in {\mathbb{N}}^{\mathbb{N}} \) be such that \( F = {C}_{\alpha } \) . Then \( A = {U}_{\alpha } \) .
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Yes
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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.
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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 \( {\mathbf{\sum }}_{1}^{1} \) is closed under continuous preimages, \( A \in {\mathbf{\sum }}_{1}^{1} \) . We claim that \( A \) is not coanalytic and hence not Borel. Suppose not. Then \( {A}^{c} \) analytic. Take an \( {x}_{0} \in X \) such that \( {A}^{c} = {U}_{{x}_{0}} \) . Then\n\n\[ {x}_{0} \in A \Leftrightarrow \left( {{x}_{0},{x}_{0}}\right) \in U \Leftrightarrow {x}_{0} \in {A}^{c}.\n\]\n\nWe have arrived at a contradiction.
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Yes
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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 \( {\mathbf{\sum }}_{n}^{1} \) is closed under projections \( {\exists }^{Y} \), and \( {\mathbf{\Pi }}_{n}^{1} \) is closed under coprojections \( {\forall }^{Y}, Y \) Polish.
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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 \( {\mathbf{\sum }}_{n}^{1} \) also has the stated closure properties. Since \( {\mathbf{\Pi }}_{n}^{1} = \neg {\mathbf{\sum }}_{n}^{1} \), the remaining part of the result follows.
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No
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Proposition 4.1.9 For every \( n \geq 1 \) ,\n\n\[ \n{\mathbf{\sum }}_{n}^{1}\bigcup {\mathbf{\Pi }}_{n}^{1} \subseteq {\mathbf{\Delta }}_{n + 1}^{1} \n\]
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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 under continuous preimages, the set \( C = A \times X \) is analytic. Since\n\n\[ \nA = {\forall }^{X}C, \n\]\n\n\( A \) is in \( {\mathbf{\Pi }}_{2}^{1} \) . Hence \( A \in {\mathbf{\Delta }}_{2}^{1} \) . The rest of the result now follows fairly easily by induction.
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Yes
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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 \( A = {U}_{\alpha } \) for some \( \alpha \) ; i.e., \( U \) is universal for \( \mathbf{\Gamma }\left( X\right) \) .
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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 \( C \subseteq {\mathbb{N}}^{\mathbb{N}} \times (X \times \) \( {\mathbb{N}}^{\mathbb{N}} \) ) be universal for \( {\mathbf{\Pi }}_{n}^{1}\left( {X \times {\mathbb{N}}^{\mathbb{N}}}\right) \) . As in 4.1.4, we see that \( {\exists }^{{\mathbb{N}}^{\mathbb{N}}}C \) is universal for \( {\mathbf{\sum }}_{n + 1}^{1}\left( X\right) \), and its complement is universal for \( {\mathbf{\Pi }}_{n + 1}^{1}\left( X\right) \) .
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No
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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 \( X \) that is in \( {\mathbf{\sum }}_{n}^{1}\left( X\right) \) but not in \( {\mathbf{\Pi }}_{n}^{1}\left( X\right) \) .
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Proof. The result is proved in exactly the same way as 4.1.5.
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No
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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 \( X \) such that for every \( \alpha \in {\mathbb{N}}^{\mathbb{N}} \) diameter \( \left( {F}_{\alpha \mid n}\right) \rightarrow 0 \) and \( A = \mathcal{A}\left( \left\{ {F}_{s}\right\} \right) \).\n\n(iii) There is a system \( \left\{ {{F}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) of closed subsets of \( X \) such that \( A = \mathcal{A}\left( \left\{ {F}_{s}\right\} \right) \)
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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) .\n\nLet\n\n\[ C = \left\{ {\left( {x,\alpha }\right) \in X \times {\mathbb{N}}^{\mathbb{N}} : \forall n\left( {x \in {F}_{\alpha \mid n}}\right) }\right\} .\n\nAs\n\n\[ C = \mathop{\bigcap }\limits_{n}\mathop{\bigcup }\limits_{{\{ s : \left| s\right| = n\} }}\left( {{F}_{s} \times \sum \left( s\right) }\right) \]\n\n\( C \) is closed. Since \( A \) is the projection of \( C \), it is analytic.\n\n(i) \( \Rightarrow \) (ii): Let \( A \subseteq X \) be analytic. By 4.1.1, there is a continuous map \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \) whose range is \( A \). Take\n\n\[ {F}_{s} = \operatorname{cl}\left( {f\left( {\sum \left( s\right) }\right) }\right) .\n\nClearly, the system of closed sets \( \left\{ {{F}_{s} : s \in {\mathbb{N}}^{ < \mathbb{N}}}\right\} \) is regular. Since \( f \) is continuous, diameter \( \left( {F}_{\alpha \mid n}\right) \) converges to 0 as \( n \rightarrow \infty \).\n\nLet \( x = f\left( \alpha \right) \in A \). Then for all \( n, x \in {F}_{\alpha \mid n} \). Thus \( A \subseteq \mathcal{A}\left( \left\{ {F}_{s}\right\} \right) \).\n\nTo show the reverse inclusion, take any \( x \in \mathcal{A}\left( \left\{ {F}_{s}\right\} \right) \). Let\n\n\[ x \in {F}_{\alpha \mid n} = \operatorname{cl}\left( {f\left( {\sum \left( {\alpha \mid n}\right) }\right) }\right) \]\n\nfor all \( n \). Choose \( {\alpha }_{n} \in \sum \left( {\alpha \mid n}\right) \) such that \( d\left( {x, f\left( {\alpha }_{n}\right) }\right) < {2}^{-n} \). So, \( f\left( {\alpha }_{n}\right) \rightarrow x \). Since \( {\alpha }_{n} \rightarrow \alpha \) and \( f \) is continuous, \( f\left( {\alpha }_{n}\right) \rightarrow f\left( \alpha \right) \). Hence, \( x \in A \), and the result follows.
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Yes
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Theorem 4.1.14 The pointclass \( {\mathbf{\sum }}_{1}^{1} \) is closed under the Souslin operation.
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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 the family of closed sets, the result follows.
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Yes
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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\n\[ \Leftrightarrow \;T\left\lbrack \alpha \right\rbrack \text{ is well-ordered with respect to }{ \leq }_{KB}\text{. } \]
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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 \( {\pi }_{1} : {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \rightarrow {\mathbb{N}}^{\mathbb{N}} \) is the projection onto the first coordinate space. The existence of such a set follows from 4.1.1. By 2.2.13, there is a tree \( T \) on \( \mathbb{N} \times \mathbb{N} \) such that \( \left\lbrack T\right\rbrack = C \) . Now note that\n\n\[ \alpha \in {A}^{c} \Leftrightarrow \exists \beta \left( {\left( {\alpha ,\beta }\right) \in \left\lbrack T\right\rbrack }\right) \]\n\n\[ \Leftrightarrow \;\exists \beta \left( {\beta \in \left\lbrack {T\left( \alpha \right) }\right\rbrack }\right) \]\n\n\[ \Leftrightarrow T\left\lbrack \alpha \right\rbrack \text{is not well-founded} \]\n\nThus (ii) follows from (i).\n\n(ii) \( \Rightarrow \) (i): Let \( A \subseteq {\mathbb{N}}^{\mathbb{N}} \) satisfy (ii). Then \( {A}^{c} \) is the projection of \( \left\lbrack T\right\rbrack \), and so \( A \) is coanalytic.
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Yes
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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 analytic set \( A \subseteq \mathbb{R} \) that is not Borel. Suppose \( B \subseteq \mathbb{R} \times \mathbb{R} \) is a Borel set whose projection is \( A \) . Take \( g = {\chi }_{B} \).
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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 \( \left\{ {{r}_{m} : m \in \mathbb{N}}\right\} \) of the set of all rational numbers. Let \[ A = \{ \left( {x, y}\right) : f\left( x\right) > y\} . \] Since \[ A = \mathop{\bigcup }\limits_{m}\left\{ {\left( {x, y}\right) \in \mathbb{R} \times \mathbb{R} : f\left( x\right) > {r}_{m} > y}\right\} \] and \( f \) is an A-function, \( A \) is analytic. By 4.1.1, there is a Borel set \( B \subseteq \) \( \left( {\mathbb{R} \times \mathbb{R}}\right) \times \mathbb{R} \) whose projection is \( A \) . Define \( h : {\mathbb{R}}^{3} \rightarrow \mathbb{R} \) by \[ h\left( {x, y, z}\right) = \left\{ \begin{array}{ll} y & \text{ if }\left( {x, y, z}\right) \in B, \\ n & \text{ if }x \in {B}_{n}\& \left( {x, y, z}\right) \in {\mathbb{R}}^{3} \smallsetminus B. \end{array}\right. \] The function \( h \) is Borel, and \[ f\left( x\right) = \mathop{\sup }\limits_{\left( y, z\right) }h\left( {x, y, z}\right) . \] Let \( u : \mathbb{R} \rightarrow {\mathbb{R}}^{2} \) be a Borel isomorphism. Such a map exists by the Borel isomorphism theorem. Define \( g \) by \[ g\left( {x, y}\right) = h\left( {x, u\left( y\right) }\right) . \]
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Yes
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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 \( f\left( x\right) = \mathop{\sup }\limits_{y}g\left( {x, y}\right) \) .
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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 \( \left\{ {{r}_{m} : m \in \mathbb{N}}\right\} \) of the set of all rational numbers. Let\n\n\[ \nA = \{ \left( {x, y}\right) : f\left( x\right) > y\} .\n\] \n\nSince\n\n\[ \nA = \mathop{\bigcup }\limits_{m}\left\{ {\left( {x, y}\right) \in \mathbb{R} \times \mathbb{R} : f\left( x\right) > {r}_{m} > y}\right\} \n\] \n\nand \( f \) is an A-function, \( A \) is analytic. By 4.1.1, there is a Borel set \( B \subseteq \) \( \left( {\mathbb{R} \times \mathbb{R}}\right) \times \mathbb{R} \) whose projection is \( A \) . Define \( h : {\mathbb{R}}^{3} \rightarrow \mathbb{R} \) by\n\n\[ \nh\left( {x, y, z}\right) = \left\{ \begin{array}{ll} y & \text{ if }\left( {x, y, z}\right) \in B, \\ n & \text{ if }x \in {B}_{n}\& \left( {x, y, z}\right) \in {\mathbb{R}}^{3} \smallsetminus B. \end{array}\right. \n\] \n\nThe function \( h \) is Borel, and\n\n\[ \nf\left( x\right) = \mathop{\sup }\limits_{\left( y, z\right) }h\left( {x, y, z}\right) .\n\] \n\nLet \( u : \mathbb{R} \rightarrow {\mathbb{R}}^{2} \) be a Borel isomorphism. Such a map exists by the Borel isomorphism theorem. Define \( g \) by\n\n\[ \ng\left( {x, y}\right) = h\left( {x, u\left( y\right) }\right) .\n\]
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Yes
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We show that \( {WF} \) is \( {\mathbf{\Pi }}_{1}^{1} \) -complete.
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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\[ E = \left\{ {\left( {T,\beta }\right) \in {2}^{{\mathbb{N}}^{ < \mathbb{N}}} \times {\mathbb{N}}^{\mathbb{N}} : T \in \operatorname{Tr}\& \exists n\left( {T\left( {\beta \mid n}\right) = 0}\right) }\right\} .\n\]\n\nIt is quite easy to see that the set \( E \) is Borel. Hence, \( {WF} \) is coanalytic.\n\nNow take any coanalytic set \( C \) in \( {\mathbb{N}}^{\mathbb{N}} \) . By 4.1.20, there is a tree \( T \) on \( \mathbb{N} \times \mathbb{N} \) such that\n\n\[ \alpha \in C \Leftrightarrow T\left\lbrack \alpha \right\rbrack \text{is well-founded. }\n\]\n\nDefine \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow \operatorname{Tr} \) by\n\n\[ f\left( \alpha \right) = T\left\lbrack \alpha \right\rbrack \n\]\n\nthe section of \( T \) at \( \alpha \) . The map \( f \) is continuous: Take any \( s \in {\mathbb{N}}^{ < \mathbb{N}} \) and note that\n\n\[ f\left( \alpha \right) \left( s\right) = 1 \Leftrightarrow T\left( {\alpha \left| \right| s \mid, s}\right) = 1.\n\]\n\nThus \( {\pi }_{s} \circ f \) is continuous for all \( s \), and so \( f \) is continuous.\n\nAs \( C = {f}^{-1}\left( {WF}\right) \), by the Borel isomorphism theorem it follows that \( {WF} \) is \( {\mathbf{\Pi }}_{1}^{1} \) -complete.
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Yes
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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 \( {WF} = {R}^{-1}\left( {WO}\right) \) .
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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\[ \vee \left( {u\left( k\right) \in T\& u\left( l\right) \notin T}\right) \]\n\n\[ \vee \left( {u\left( k\right), u\left( l\right) \in T\& \;u\left( k\right) { \leq }_{KB}u\left( l\right) }\right) \]\n\nIt is easy to check that \( T \rightarrow R\left( T\right) \) is a continuous map from \( {Tr} \) to \( {2}^{\mathbb{N} \times \mathbb{N}} \) . Since a tree \( T \) on \( \mathbb{N} \) is well-founded if and only if \( { \leq }_{KB} \) is a well-order on \( T \) (1.10.10.), \( {WF} = {R}^{-1}\left( {WO}\right) \) .
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Yes
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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.
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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\( K \) is perfect \( \Leftrightarrow \forall n\left( {K\bigcap {V}_{n} \neq \varnothing }\right.\)\n\n\[ \Rightarrow \exists k\exists l\left( {{V}_{k},{V}_{l} \subseteq {V}_{n}}\right. \]\n\n\[ \text{&}{V}_{k} \cap {V}_{l} = \varnothing \text{&}K \cap {V}_{k}, K \cap {V}_{l} \neq \varnothing )\text{).} \]\n\nSo,\n\n\[ P\left( X\right) = \mathop{\bigcap }\limits_{n}\left\lbrack {{A}_{n}^{c}\bigcup \mathop{\bigcup }\limits_{{\left( {k, l}\right) \in {S}_{n}}}\left( {{A}_{k}\bigcap {A}_{l}}\right) }\right\rbrack \]\n\nwhere\n\n\[ {A}_{n} = \left\{ {K \in K\left( X\right) : K\bigcap {V}_{n} \neq \varnothing }\right\} \]\n\nand\n\n\[ {S}_{n} = \left\{ {\left( {k, l}\right) : {V}_{k} \subseteq {V}_{n}\& {V}_{l} \subseteq {V}_{n}\& {V}_{k}\bigcap {V}_{l} = \varnothing }\right\} . \]\n\nHence, \( P\left( X\right) \) is Borel. Let \( K \in K\left( X\right) \) . By 2.6.3,\n\n\( K \) is uncountable \( \Leftrightarrow \left( {\exists P \in K\left( X\right) }\right) \left( {P \in P\left( X\right) \& P \subseteq K}\right) \) .\n\nBy 2.4.11, the set\n\n\[ \{ \left( {K, L}\right) \in K\left( X\right) \times K\left( X\right) : K \subseteq L\} \]\n\nis closed. Hence, \( U\left( X\right) \in {\mathbf{\sum }}_{1}^{1} \) .\n\nIt remains to show that \( U\left( X\right) \) is \( {\mathbf{\sum }}_{1}^{1} \) -complete. Since every uncountable Polish space contains a \( {G}_{\delta } \) set homeomorphic to \( {\mathbb{N}}^{\mathbb{N}} \), it is sufficient to prove the result for \( X = {\mathbb{N}}^{\mathbb{N}} \) . Let \( N \) be as in 4.2.3. Define \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow K\left( {\mathbb{N}}^{\mathbb{N}}\right) \) by\n\n\[ f\left( \alpha \right) = \left\{ {\beta \in {\mathbb{N}}^{\mathbb{N}} : \beta \leq \alpha \text{ pointwise }}\right\} . \]\n\nThen \( f \) is continuous. Further,\n\n\[ \alpha \in N \Leftrightarrow f\left( \alpha \right) \text{is uncountable.} \]\n\nNow consider the map \( g : K\left( {K\left( {\mathbb{N}}^{\mathbb{N}}\right) }\right) \rightarrow K\left( {\mathbb{N}}^{\mathbb{N}}\right) \) defined by\n\n\[ g\left( \mathcal{K}\right) = \bigcup \mathcal{K},\;\mathcal{K} \in K\left( {K\left( {\mathbb{N}}^{\mathbb{N}}\right) }\right) . \]\n\nThe map \( g \) is continuous (2.4.11). Define\n\n\[ h\left( K\right) = g\left( {f\left( K\right) }\right) ,\;K \in K\left( {\mathbb{N}}^{\mathbb{N}}\right) . \]\n\nThe map \( h \) is continuous, and\n\n\[ I{F}^{ * } = {h}^{-1}\left( \left\{ {K \in K\left( {\mathbb{N}}^{\mathbb{N}}\right) : K\text{ is uncountable }}\right\} \right) . \]\n\nThe result follows from 4.2.3.
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Yes
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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.
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Null
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No
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Theorem 4.3.2 Every analytic subset of a Polish space has the Baire property.
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Null
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No
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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.
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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\( U \in \sigma \left( \left\{ {{C}_{i} \times {D}_{i} : i \in \mathbb{N}}\right\} \right) \) . Let \( Y \) be an uncountable Borel subset of \( X \) such that each \( {D}_{i}\bigcap Y \) is Borel. In particular, every section \( {\left( U\bigcap \left( X \times Y\right) \right) }_{x} \) , \( x \in X \), is Borel. Let \( E \) be an analytic non-Borel set contained in \( Y \) . Since \( U \) is universal,\n\n\[ E = {U}_{{x}_{0}} = {\left( U\bigcap \left( X \times Y\right) \right) }_{{x}_{0}} \]\n\nfor some \( {x}_{0} \in X \) . We have arrived at a contradiction.
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Yes
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Theorem 4.3.5 Every uncountable analytic set contains a homeomorph of the Cantor set and hence is of cardinality \( \mathfrak{c} \) .
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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 \( \left| s\right| = \left| t\right| \) and \( s \neq t, f\left( {F}_{s}\right) \cap f\left( {F}_{t}\right) = \varnothing \) . Since the range of \( f \) is uncountable, we get an uncountable \( Z \subseteq {\mathbb{N}}^{\mathbb{N}} \) such that \( f \mid Z \) is one-to-one. By the Cantor - Bendixson theorem (2.6.2), we can further assume that \( Z \) is dense-in-itself. Take a compatible complete metric \( d < 1 \) on \( {\mathbb{N}}^{\mathbb{N}} \) . We define a system \( \left\{ {{U}_{s} : s \in {2}^{ < \mathbb{N}}}\right\} \) of nonempty open subsets of \( {\mathbb{N}}^{\mathbb{N}} \) satisfying the following conditions: (i) \( \operatorname{diameter}\left( {U}_{s}\right) < {2}^{-\left| s\right| } \) ; (ii) \( {U}_{s} \cap Z \neq \varnothing \) ; (iii) \( \operatorname{cl}\left( {U}_{s \cap \epsilon }\right) \subseteq {U}_{s},\epsilon = 0,1 \) ; and (iv) whenever \( \left| s\right| = \left| t\right| \) and \( s \neq t, f\left( {\operatorname{cl}\left( {U}_{s}\right) }\right) \cap f\left( {\operatorname{cl}\left( {U}_{t}\right) }\right) = \varnothing \) . In particular, \( \operatorname{cl}\left( {U}_{s}\right) \cap \operatorname{cl}\left( {U}_{t}\right) = \varnothing . \) We define such a system by induction on \( \left| s\right| \) . Take \( {U}_{e} = X \) . Suppose \( {U}_{s} \) has been defined for some \( s \) . Since \( Z \) is dense-in-itself and \( {U}_{s} \) open, \( {U}_{s} \cap Z \) has at least two distinct points, say \( {x}_{0},{x}_{1} \) . Then \( f\left( {x}_{0}\right) \neq f\left( {x}_{1}\right) \) . Let \( {W}_{0} \) and \( {W}_{1} \) be disjoint open sets containing \( f\left( {x}_{0}\right) \) and \( f\left( {x}_{1}\right) \) respectively. Since \( f \) is continuous, there are open sets \( {U}_{s}{ \hat{} }_{0} \) and \( {U}_{s}{ \hat{} }_{1} \) satisfying the following conditions: (b) \( \operatorname{diameter}\left( {U}_{s \uparrow \epsilon }\right) < \frac{1}{{2}^{\left| s\right| + 1}} \) ; and (c) \( f\left( {\operatorname{cl}\left( {U}_{s \cap \epsilon }\right) }\right) \subseteq {W}_{\epsilon },\epsilon = 0 \) or 1 . (d) In particular, \( f\left( {\operatorname{cl}\left( {U}_{{s}^{ \frown }0}\right) }\right) \cap f\left( {\operatorname{cl}\left( {U}_{{s}^{ \frown }1}\right) }\right) = \varnothing \) . Put \( {F}_{s} = \operatorname{cl}\left( {U}_{s}\right) \) . Let \( C = \mathcal{A}\left( \left\{ {F}_{s}\right\} \right) \) . Then \( C \) is homeomorphic to the Cantor set, and \( f \mid C \), being one-to-one and continuous, is an embedding.
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Yes
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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(iii) There is a Polish space \( Y \) and an analytic set \( B \subseteq X \times Y \) such that\n\n\[ A = \left\{ {x \in X : {B}_{x}}\right. \text{is uncountable}\} \text{.} \]
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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 \( {\pi }_{1}^{-1}\left( \alpha \right) \) uncountable for all \( \alpha \) . Since \( {\mathbb{N}}^{\mathbb{N}} \times {\mathbb{N}}^{\mathbb{N}} \) is homeomorphic to \( {\mathbb{N}}^{\mathbb{N}} \) , this shows that there is a continuous map \( h : {\mathbb{N}}^{\mathbb{N}} \rightarrow {\mathbb{N}}^{\mathbb{N}} \) such that \( {h}^{-1}\left( \alpha \right) \) is uncountable for all \( \alpha \) . Take \( C = \operatorname{graph}\left( {f \circ h}\right) \) .\n\n(iii) is a special case of (ii).\n\n(iii) \( \Rightarrow \) (i): By (4.3.6), we have the following: Let \( P, Q \) be Polish spaces and \( f : P \rightarrow Q \) a continuous map. The range of \( f \) is uncountable if and only if there is a countable dense-in-itself subset \( Z \) of \( P \) such that \( f \mid Z \) is one-to-one.\n\nNote also that the set\n\n\[ D = \left\{ {\left( {x}_{n}\right) \in {\left( {\mathbb{N}}^{\mathbb{N}}\right) }^{\mathbb{N}} : \left\{ {{x}_{n} : n \in \mathbb{N}}\right\} \text{ is dense-in-itself }}\right\} \]\n\nis a \( {G}_{\delta } \) set in \( {\left( {\mathbb{N}}^{\mathbb{N}}\right) }^{\mathbb{N}} \) .\n\nNow let \( X, Y \) and \( B \) be as in (iii). Let \( f : {\mathbb{N}}^{\mathbb{N}} \rightarrow X \times Y \) be a continuous map with range \( B \) . By (a),\n\n\[ {B}_{x}\text{ is uncountable } \Leftrightarrow \left( {\exists \left( {z}_{n}\right) \in D}\right) \left( {\forall i\forall j\left( {i \neq j \Rightarrow f\left( {z}_{i}\right) \neq f\left( {z}_{j}\right) }\right. }\right) ,\n\n\left. {\& \forall k\left( {{\pi }_{X}\left( {f\left( {z}_{k}\right) }\right) = x}\right) }\right) ,\n\nwhere \( {\pi }_{X} : X \times Y \rightarrow X \) is the projection map. The result follows from (b).
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Yes
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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.
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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 \( \left| Z\right| \leq \mathfrak{c} \) . Let \( {X}^{\prime } = {f}^{-1}\left( Z\right) \) . Note that \( {X}^{\prime } \) is Borel. Now take any \( A \subseteq \mathbb{R} \) of the same cardinality as \( Z \) that does not contain any uncountable closed set. We have proved the existence of such a set in 3.2.8. Let \( g \) be any one-to-one map from \( Z \) onto \( A \) . Since \( Z \) is discrete, \( g \) is continuous. Clearly, \( g \circ f \) is Borel. As \( A = g\left( {f\left( {X}^{\prime }\right) }\right), A \) is an uncountable analytic set not containing a perfect set. This contradicts 4.3.5.
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Yes
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Corollary 4.3.9 Every Borel homomorphism \( \varphi : G \rightarrow H \) from a completely metrizable group \( G \) to a metrizable group \( H \) is continuous.
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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.
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No
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Proposition 4.3.10 (i) Every countable set of reals has strong measure zero.
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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 \( \mathbb{N} \) whose union is \( \mathbb{N} \) . For each \( n \) choose open intervals \( {I}_{m}^{n}, m \in {I}_{n} \), such that \( \left| {I}_{m}^{n}\right| \leq {a}_{m} \) and \( {A}_{n} \subseteq \mathop{\bigcup }\limits_{{m \in {I}_{n}}}{I}_{m}^{n} \) . Note that\n\n\[ \mathop{\bigcup }\limits_{n}{A}_{n} \subseteq \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}\mathop{\bigcup }\limits_{{m \in {I}_{n}}}{I}_{m}^{n} \]\n\nThe proof of (iii) is clearly seen now.
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No
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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.
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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 continuous, for each \( n \) there is a positive real number \( {b}_{n} \) such that whenever \( X \subseteq \left\lbrack {0,1}\right\rbrack \) is of diameter at most \( {b}_{n} \) , the diameter of \( f\left( X\right) \) is at most \( {a}_{n} \) . Since \( A \) has strong measure zero, there are open intervals \( {I}_{n}, n \in \mathbb{N} \), such that \( \left| {I}_{n}\right| \leq {b}_{n} \) and \( A \subseteq \mathop{\bigcup }\limits_{n}{I}_{n} \) . Take \( {J}_{n} = f\left( {I}_{n}\right) \) .
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Yes
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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 zero set. It follows that not all measure zero sets have strong measure zero.
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Null
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No
|
Proposition 4.3.13 No set of reals containing a perfect set has strong measure zero.
|
Null
|
No
|
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