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 o-borel 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 sub-σ-algebra; the B = A. 1 2 Thus, stadard σ-algebras 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 σ-algebras 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 σ-algebras 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 o-borel 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 oe-to-oe. 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 ill-fouded.) 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 so-called 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 o-borel subset of Tr(T). Thus, Tr(T) cotais a o-borel 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 o-borel 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 oe-to-oe 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 Lusi-Sousli; 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 oe-to-oe 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 oe-tooe. 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 ill-fouded, 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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