6 Borel sets in the light of analytic sets


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1 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 86 6 Borel sets i the light of aalytic sets 6a Separatio theorem b Borel bijectios c A oborel aalytic set of trees d Borel ijectios Hits to exercises Idex Aalytic sets shed ew light o Borel sets. Stadard Borel spaces are somewhat similar to compact topological spaces. 6a Separatio theorem The first step toward deeper theory of Borel sets. 6a1 Theorem. 1 For every pair of disjoit aalytic 2 subsets A,B of a coutably separated measurable space (X,A) there exists C A such that A C ad B X \C. Rather itriguig: (a) the Borel complexity of C caot be bouded apriori; (b) the give A,B give o clue to ay Borel complexity. How could it be proved? We say that A is separated from B if A C ad B X \ C for some C A. 6a2 Core exercise. If A is separated from B for each = 1,2,... the A 1 A 2... is separated from B. 6a3 Core exercise. IfA m isseparatedfromb forallm,thea 1 A 2... is separated from B 1 B The first separatio theorem for aalytic sets, or the Lusi separatio theorem ; see Srivastava, Sect. 4.4 or Kechris, Sect. 14.B. To some extet, it is cotaied implicitly i the earlier Sousli s proof of Theorem 6a6. 2 Recall 5d9.
2 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 87 6a4 Core exercise. It is sufficiet to prove Theorem 6a1 for X = R, A = B(R). Proof of Theorem 6a1. Accordig to 6a4 we assume that X = R, A = B(R). By 5d10 we take Polish spaces Y,Z ad cotiuous maps f : Y R, g : Z R such that A = f(y), B = g(z). Similarly to the proof of 4c9 we choose a compatible metric o Y ad a coutable base E 2 Y cosistig of bouded sets; ad similarly F 2 Z. Assume the cotrary: f(y) = A is ot separated from g(z) = B. Usig 6a3 we fid U 1 E, V 1 F such that f(u 1 ) is ot separated from g(v 1 ). Usig 6a3 agai we fid U 2 E, V 2 F such that U 2 U 1, diamu 2 0.5diamU 1, V 2 V 1, diamv 2 0.5diamV 1, ad f(u 2 ) is ot separated from g(v 2 ). Ad so o. We get U 1 U 1 U 2 U 2... ad diamu 0; by completeess, U 1 U 2 = {y}for some y Y. Similarly, V 1 V 2 = {z} for some z Z. We ote that f(y) g(z) (sice f(y) A ad g(z) B) ad take ε > 0 such that ( f(y) ε,f(y)+ε ) ( g(z) ε,g(z)+ε ) =. Usig cotiuity we take such that f(u ) ( f(y) ε,f(y)+ε ) ad g(v ) ( g(z) ε,g(z)+ε ) ; the f(u ) is separated from g(v ), a cotradictio. 6a5 Corollary. Let (X, A) be a coutably separated measurable space, ad A X a aalytic set. If X \A is also a aalytic set the A A. Proof. Follows immediately from Theorem 6a1. 6a6 Theorem. (Sousli) Let (X,A) be a stadard Borel space. The followig two coditios o a set A X are equivalet: 1 (a) A A; (b) both A ad X \A are aalytic. Proof. (b)= (a): by 6a5; (a)= (b): by 5d9 ad 2b11(a). 6b Borel bijectios A ivertible homomorphism is a isomorphism, which is trivial. A ivertible Borel map is a Borel isomorphism, which is highly otrivial. 6b1 Core exercise. Let (X,A) be a stadard Borel space, (Y,B) a coutably separated measurable space, ad f : X Y a measurable bijectio. The f is a isomorphism (that is, f 1 is also measurable). 2 1 See also Footote 1 o page A topologicalcouterpart: a cotiuousbijectio from a compact space to a Hausdorff space is a homeomorphism (that is, the iverse map is also cotiuous).
3 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 88 6b2 Corollary. A measurable bijectio betwee stadard Borel spaces is a isomorphism. 6b3 Corollary. Let (X,A) be a stadard Borel space ad B A a coutably separated subσalgebra; the B = A. 1 2 Thus, stadard σalgebras are ever comparable. 3 6b4 Core exercise. Let R 1,R 2 be Polish topologies o X. (a) If R 2 is stroger tha R 1 the B(X,R 1 ) = B(X,R 2 ) (that is, the correspodig Borel σalgebras are equal). (b) If R 1 ad R 2 are stroger tha some metrizable (ot ecessarily Polish) topology the B(X,R 1 ) = B(X,R 2 ). A lot of comparable Polish topologies appeared i Sect. 3c. Now we see that the correspodig Borel σalgebras must be equal. Aother example: the strog ad weak topologies o the uit ball of a separable ifiitedimesioal Hilbert space. This is istructive: the structure of a stadard Borel space is cosiderably more robust tha a Polish topology. I particular, we upgrade Theorem 3c12 (as well as 3c15 ad 3d1). 6b5 Theorem. For every Borel subset B of the Cator set X there exists a Polish topology R o X, stroger tha the usual topology o X, such that B is clope i (X,R), ad B(X,R) is the usual B(X). Here is aother useful fact. 6b6 Core exercise. Let (X,A) be a stadard Borel space. The followig two coditios o A 1,A 2, A are equivalet: (a) the sets A 1,A 2,... geerate A; (b) the sets A 1,A 2,... separate poits. The graph of a map f : X Y is a subset {(x,f(x)) : x X} of X Y. Is measurability of the graph equivalet to measurability of f? 6b7 Propositio. Let(X, A),(Y, B) be measurable spaces, (Y, B) coutably separated, ad f : X Y measurable; the the graph of f is measurable. 1 A topological couterpart: if a Hausdorff topology is weaker tha a compact topology the these two topologies are equal. 2 See also the footote to 5d7. 3 Similarly to compact Hausdorff topologies.
4 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 89 6b8 Core exercise. It is sufficiet to prove Prop. 6b7 for Y = R, B = B(R). Proof of Prop. 6b7. Accordig to 6b8 we assume that Y = R, B = B(R). The map X R (x,y) f(x) y R is measurable (sice the map R R (z,y) z y R is). Thus, {(x,y) : f(x) y = 0} is measurable. Here is aother proof, ot usig 6b8. Proof of Prop. 6b7 (agai). WetakeB 1,B 2, B thatseparatepoitsad ote that y = f(x) (x,y) ( (f ) ( 1 (B ) B (X\f 1 (B )) (Y \B ) )) sice y = f(x) if ad oly if ( y B f(x) B ). 6b9 Extra exercise. If a measurable space(y, B) is ot coutably separated thethereexist ameasurable space(x,a)adameasurable mapf : X Y whose graph is ot measurable. 6b10 Propositio. Let (X, A),(Y, B) be stadard Borel spaces ad f : X Y a fuctio. If the graph of f is measurable the f is measurable. Proof. The graph G X Y is itself a stadard Borel space by 2b11. The projectio g : G X, g(x,y) = x, is a measurable bijectio. By 6b2, g is a isomorphism.thus, f 1 (B) = g ( G (X B) ) A for B B. Here is a stroger result. 6b11 Propositio. Let (X, A),(Y, B) be coutably separated measurable spaces ad f : X Y a fuctio. If the graph of f is aalytic 1 the f is measurable. Proof. Deote the graph by G. Let B B, the G (X B) is aalytic (thik, why), thereforeitsprojectiof 1 (B)isaalytic. Similarly, f 1 (Y\B) is aalytic. We ote that f 1 (Y \ B) = X \ f 1 (B), apply 6a5 ad get f 1 (B) A. 6b12 Extra exercise. Give a example of a omeasurable fuctio with measurable graph, betwee coutably separated measurable spaces. 1 As defied by 5d9, takig ito accout that X Y is coutably separated by 1d24.
5 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 90 6c A oborel aalytic set of trees A example, at last... We adapt the otio of a tree to our eeds as follows. 6c1 Defiitio. (a)atree cosists ofaatmostcoutable set T of odes, a ode 0 T called the root, ad a biary relatio o T such that for every s T there exists oe ad oly oe fiite sequece (s 0,...,s ) T T 2 T 3... such that 0 T = s 0 s 1 s 1 s = s. (b) A ifiite brach of a tree T is a ifiite sequece (s 0,s 1,...) T such that 0 T = s 0 s 1...; the set [T] of all ifiite braches is called the body of T. (c) A tree T is prued if every ode belogs to some (at least oe) ifiite brach. (Or equivaletly, s T t T s t.) We edow the body [T] with a metrizable topology, compatible with the metric ρ ( (s ),(t ) ) = 2 if{:s t }. The metric is separable ad complete (thik, why); thus, [T] is Polish. 6c2 Example. The full biary tree {0,1} < = =0,1,2,... {0,1} : Its body is homeomorphic to the Cator set {0,1}. 6c3 Example. The full ifiitely splittig tree: {1,2,...} <. Its body is homeomorphic to {1,2,...}, as well as to [0,1]\Q (the space of irratioal umbers), sice these two spaces are homeomorphic: {1,2,...} (k 1,k 2,...) 1 1 k 1 + k Let T be a tree ad T 1 T a oempty subset such that s T t T 1 (s t = s T 1 ). The T 1 is itself a tree, a subtreee of T. Clearly, [T 1 ] [T] is a closed subset. 6c4 Defiitio. (a) A regular scheme o a set X is a family (A s ) s T of subsets of X idexed by a tree T, satisfyig A s A t wheever s t. (b) A regular scheme (A s ) s T o a metric space X, idexed by a prued tree T, has vaishig diameter if diam(a s ) 0 (as ) for every (s ) [T].
6 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 91 6c5 Example. Dyadic itervals [ k, k+1 ] [0,1], aturally idexed by the 2 2 full biary tree, are a vaishig diameter scheme. For every (s ) [T] we have A s {x} for some x [0,1], which gives a cotiuous map from the Cator set oto [0,1]. Note that the map is ot oetooe. Let (A s ) s T be a regular scheme o X, ad x X. The set T x = {s T : A s x}, if ot empty, is a subtree of T. The followig two coditios o the scheme are equivalet (thik, why): (6c6a) (6c6b) T x is a prued tree for every x X; A 0T = X, ad A s = A t for all s T. t:s t Let X be a complete metric space, ad (F s ) s T a vaishig diameter scheme of closed sets o X. The for every (s ) [T] we have F s {x} for some x X. We defie the associated map f : [T] X by F s {f ( (s ) ) }. This map is cotiuous (thik, why), adf 1 ({x}) = [T x ] for all x X (look agai at 6c5); here, if x / F 0T the T x =, ad we put [ ] =. If the scheme satisfies (6c6) the f([t]) = X (sice [T x ] for all x). 6c7 Core exercise. O every compact metric space there exists a vaishig diameter scheme of closed sets, satisfyig (6c6), idexed by a fiitely splittig tree (that is, the set {t : s t} is fiite for every s). It follows easily that every compact metrizable space is a cotiuous image of the Cator set. 6c8 Core exercise. O every complete separable metric space there exists a vaishig diameter scheme of closed sets, satisfyig (6c6). It follows easily that every Polish space is a cotiuous image of the space of irratioal umbers. Ad therefore, every aalytic set (i a Polish space) is also a cotiuous image of the space of irratioal umbers!
7 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 92 A aalytic set A i a Polish space Y is the image of some Polish space X uder some cotiuous map ϕ : X Y; A = ϕ(x) Y. We choose complete metrics o X ad Y. Accordig to 6c8 we take o X a vaishig diameter scheme of closed sets (F s ) s T satisfyig (6c6). 6c9 Core exercise. The family ( ϕ(f s ) ) s T is a vaishig diameter scheme of closed sets o Y. (Here ϕ(f s ) is the closure of the image.) We cosider the associated maps f : [T] X ad g : [T] Y. Clearly, ϕ f = g, f([t]) = X, ad therefore g([t]) = A. We coclude. 6c10 Propositio. For every aalytic set A i a Polish space X there exists a vaishig diameter scheme (F s ) s T of closed sets o X whose associated map f satisfies f([t]) = A. Ad further... 6c11 Propositio. A subset A of a Polish space X is aalytic if ad oly if A = (s ) [T] for some regular scheme (F s ) s T of closed sets F s X idexed by a prued tree T. 1 Proof. Oly if : follows from 6c10. If : A is the projectio of the Borel set of pairs ( (s ),x ) [T] X satisfyig x F s for all. Wereturto6c10. Therelatiof([T]) = A,icombiatiowithf 1 ({x}) = [T x ], gives A = {x : [T x ] }, that is, F s A = {x : T x IF(T)} where IF(T) is the set of all subtrees of T that have (at least oe) ifiite brach. (Such trees are called illfouded.) Thus, every aalytic set A X is the iverse image of IF(T) uder the map x T x for some regular scheme of closed sets. The set Tr(T) of all subtrees of T (plus the empty set) is a closed subset of the space 2 T homeomorphic to the Cator set (uless T is fiite). Thus IF(T) Tr(T) is a subset of a compact metrizable space. 1 I other words: a set is aalytic if ad oly if it ca be obtaied from closed sets by the socalled Sousli operatio; see Srivastava, Sect or Kechris, Sect. 25.C.
8 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 93 6c12 Core exercise. IF(T) is a aalytic subset of Tr(T). We retur to a regular scheme of closed sets ad the correspodig map X x T x Tr(T). 6c13 Core exercise. Let B 2 T ad A = {x : T x B} X. (a) If B is clope the A belogs to Π 2 Σ 2 (that is, both G δ ad F σ ). (b) If B Π the A Π +2. If B Σ the A Σ +2. (Here = 1,2,...) 6c14 Propositio. Let T = {1,2,...} < bethefull ifiitely splittig tree. The the subset IF(T) of Tr(T) does ot belog to the algebra Σ. Proof. By the hierarchy theorem (see Sect. 1c), there exists a Borel subset A of the Cator set such that A / Σ. By Prop. 3e2, A is aalytic. By Prop. 6c10, A = {x : T x IF(T 1 )} for some tree T 1 ad some scheme. Applyig 6c13 to B = IF(T 1 ) we get IF(T 1 ) / Σ i 2 T, therefore itr(t). It remais to embed T 1 ito T. A similar argumet applied to the trasfiite Borel hierarchy shows that IF(T) is a oborel subset of Tr(T). Thus, Tr(T) cotais a oborel aalytic set. The same holds for the Cator set (sice Tr(T) embeds ito 2 T ) ad for [0,1] (sice the Cator set embeds ito [0,1]). 1 6c15 Extra exercise. Takig for grated that IF(T) is ot a Borel set (for T = {1,2,...} < ), prove that the real umbers of the form 1 1 k 1 + k such that some ifiite subsequece (k i1,k i2,...) of the sequece (k 1,k 2,...) satisfies the coditio: each elemet is a divisor of the ext elemet, are a oborel aalytic subset of R. 2 1 I fact, the same holds for all ucoutable Polish spaces, as well as all ucoutable stadard Borel spaces (these are mutually isomorphic). 2 Lusi 1927.
9 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 94 6d Borel ijectios The secod step toward deeper theory of Borel sets. 6d1 Theorem. 1 Let X,Y be Polish spaces ad f : X Y a cotiuous map. If f is oetooe the f(x) is Borel measurable. If a tree has a ifiite brach the, of course, this tree is ifiite ad moreover, of ifiite height (that is, for every there exists a elemet brach). The coverse does ot hold i geeral (thik, why), but holds for fiitely splittig trees ( Köig s lemma ). I geeral the coditio T x IF(T) (that is, [T x ] ) caot be rewritte i the form T x R (for some R T), sice i this case the set {x : T x R } = {x : s R s T x } = s R F s must be a F σδ set (give a regular scheme of closed sets), while theset {x : T x IF(T)} = A, beig just aalytic, eed ot be F σδ. But if each T x is fiitely splittig the Köig s lemma applies ad so, A is Borel measurable (give a regular scheme of closed sets). I particular, this is the case if each T x does ot split at all, that is, is a brach! (Thus, we eed oly the trivial case of Köig s lemma.) I terms of the scheme (B s ) s T it meas that (6d2) B t1 B t2 = wheever s t 1, s t 2, t 1 t 2. We coclude. 6d3 Lemma. Let (B s ) s T be a regular scheme of Borel sets satisfyig (6d2). The the followig set is Borel measurable: B = {x : T x IF(T)} = B s. Ideed, B = s R B s, (s ) [T] where R is the th level of T (that is, s i 0 T = s 0 s 1 s rus over R ). 6d4 Core exercise. Foreveryregularscheme(A s ) s T ofborelsetssatisfyig (6c6) there exists a regular scheme (B s ) s T of Borel sets B s A s, satisfyig (6c6) ad (6d2). 1 LusiSousli; see Srivastava, Th or Kechris, Th. (15.1).
10 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 95 We combie it with 6c8. 6d5 Lemma. O every complete separable metric space there exists a vaishig diameter scheme of Borel sets, satisfyig (6c6) ad (6d2). Give X,Y,f asi6d1, we take (B s ) s T ox accordig to 6d5, itroduce A s = f(b s ) Y ad get A s = ( ) f(b s ) = f B s = f(x), (s ) [T] (s ) [T] (s ) [T] (A s ) s T beig a vaishig diameter scheme o Y satisfyig (6d2) (thik, why). However, are A s Borel sets? For ow we oly kow that they are aalytic. I spite of the vaishig diameter, it may happe that A s A s (sice A s may be empty); evertheless, (6d6) A s = A s = f(x), (s ) [T] (s ) [T] sice (for some x X) A s = f(b s ) f(b s ) = f ( B s ) = f({x}) = {f(x)} f(x). (The ecessarily A s = A s for aother brach (s ) [T].) However, (A s ) s T eed ot satisfy (6d2). By (6d6) ad 6d3, Theorem 6d1 is reduced to the followig. 6d7 Lemma. For every regular scheme (A s ) s T of aalytic sets, satisfyig (6d2), there exists a regular scheme (B s ) s T of Borel sets, satisfyig (6d2) ad such that A s B s A s for all s T. 6d8 Core exercise. Let A 1,A 2,... be disjoit aalytic sets. The there exist disjoit Borel sets B 1,B 2,... such that A B for all. We ca get more: A B A for all (just by replacig B with B A ). Proof of Lemma 6d7. First, we use 6d8 for costructig B s for s R 1 (the first level of T), that is, 0 T s. The, for every s 1 R 1, we do the same for s such that s 1 s (stayig withi B s1 ); thus we get B s for s R 2. Ad so o. Theorem 6d1 is thus proved.
11 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 96 6d9 Core exercise. If (X,A) is a stadard Borel space, (Y,B) a coutably separated measurable space, ad f : X Y a measurable oetooe map the f(x) B. 1 6d10 Corollary. If a subset of a coutably separated measurable space is itself a stadard Borel space the it is a measurable subset d11 Corollary. A subset of a stadard Borel space is itself a stadard Borel space if ad oly if it is Borel measurable. 1 The topological couterpart is ot quite similar: a cotiuous image of a compact topological space i a Hausdorff topological space is closed, eve if the map is ot oetooe. 2 A topological couterpart: if a subset of a Hausdorff topological space is itself a compact topological space the it is a closed subset. 3 See also the foototes to 4c12, 4d10 ad 5d7.
12 Tel Aviv Uiversity, 2012 Measurability ad cotiuity 97 Hits to exercises 6a4: recall the proof of 5d11. 6b1: use 6a5 ad 6a6. 6b4: use 6b3, 4d7 ad 3c6. 6b6: use 6b1 ad 1d32. 6b8: similar to 6a4. 6c9: be careful: ϕ eed ot be uiformly cotiuous. 6c12: recall the proof of 6c11. 6c13: recall 1b, 1c. 6d4: A 1 A 2 = A 1 (A 2 \A 1 )... 6d8: first, apply Theorem 6a1 to A 1 ad A 2 A d9: similar to 6a4. Idex associated map, 91 body, 90 brach, 90 fiitely splittig, 91 full biary tree, 90 full ifiitely splittig tree, 90 graph, 88 illfouded, 92 prued, 90 regular scheme, 90 root, 90 separated, 86 subtreee, 90 tree, 90 vaishig diameter, 90 0 T, 90 [T], 90, 90 IF(T), 92 Tr(T), 92
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