The descriptive set theory of the Lebesgue density theorem. joint with A. Andretta

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1 The descriptive set theory of the Lebesgue density theorem joint with A. Andretta

2 The density function Let (X, d, µ) be a Polish metric space endowed with a Borel probability measure giving positive measures to open sets.

3 The density function Let (X, d, µ) be a Polish metric space endowed with a Borel probability measure giving positive measures to open sets. Given a measurable A X, define the density function: µ(a B ε (x)) D A (x) = lim. ε 0 + µ(b ε (x))

4 The density function Let (X, d, µ) be a Polish metric space endowed with a Borel probability measure giving positive measures to open sets. Given a measurable A X, define the density function: µ(a B ε (x)) D A (x) = lim. ε 0 + µ(b ε (x)) Let also Φ(A) = {x X D A (x) = 1} be the set of points with density 1 in A.

5 If A, B are measure equivalent, in symbols A B, then D A = D B, thus Φ(A) = Φ(B). Thus Φ induces a function Φ : MALG(X ) MALG(X ) on the measure algebra of X.

6 The problem What can be said, in general, about Φ?

7 The problem What can be said, in general, about Φ? Apparently not much. Very simple properties can be established, like: Φ( ) =, Φ(X ) = X A B Φ(A) Φ(B) Φ(A B) = Φ(A) Φ(B) i I Φ(A i) Φ( i I A i) Φ( A) Φ(A) A Φ(A) for open A

8 The problem What can be said, in general, about Φ? Apparently not much. Very simple properties can be established, like: Φ( ) =, Φ(X ) = X A B Φ(A) Φ(B) Φ(A B) = Φ(A) Φ(B) i I Φ(A i) Φ( i I A i) Φ( A) Φ(A) A Φ(A) for open A Thus {A MEAS(X ) A Φ(A)} is a topology (the density topology) extending the topology of X.

9 Lebesgue density theorem There are Polish spaces such that, for all measurable subsets A, A Φ(A).

10 Lebesgue density theorem There are Polish spaces such that, for all measurable subsets A, A Φ(A). This means that map Φ : MALG MALG is the identity

11 Lebesgue density theorem There are Polish spaces such that, for all measurable subsets A, A Φ(A). This means that map Φ : MALG MALG is the identity and Φ : MEAS Bor is a selector for the equivalence relation.

12 Lebesgue density theorem There are Polish spaces such that, for all measurable subsets A, A Φ(A). This means that map Φ : MALG MALG is the identity and Φ : MEAS Bor is a selector for the equivalence relation. Remark. At least in these cases the inclusion Φ( A) Φ(A) cannot be replaced by equality. Otherwise Φ : MEAS Bor would be a homomorphism of Bolean algebras such that A Φ(A) A. The existence of such Borel liftings is independent of ZFC (Shelah).

13 Examples of spaces satisfying LDT R n Polish ultrametric spaces (B. Miller)

14 Examples of spaces satisfying LDT R n Polish ultrametric spaces (B. Miller) Problem. Which Polish spaces satisfy Lebesgue density theorem?

15 Examples of spaces satisfying LDT R n Polish ultrametric spaces (B. Miller) Problem. Which Polish spaces satisfy Lebesgue density theorem? Here I will focus on Cantor space 2 N with the usual distance and Lebesgue (coin-tossing) measure.

16 In 2 N the density function becomes and µ(a N x n ) D A (x) = lim = lim n µ(n x n ) n 2n µ(a N x n ) x Φ(A) ε n m > n 2 m µ(a N x m ) > 1 ε, so that Φ(A) is Π 0 3.

17 In 2 N the density function becomes and µ(a N x n ) D A (x) = lim = lim n µ(n x n ) n 2n µ(a N x n ) x Φ(A) ε n m > n 2 m µ(a N x m ) > 1 ε, so that Φ(A) is Π 0 3. Sets of the form Φ(A) will be called regular. Since Φ 2 = Φ, a set A is regular iff Φ(A) = A.

18 Wadge hierarchy on Cantor space For A, B 2 N, set A W B there is a continuous f : 2 N 2 N such that A = f 1 (B).

19 Wadge hierarchy on Cantor space For A, B 2 N, set A W B there is a continuous f : 2 N 2 N such that A = f 1 (B). A W B A W B W A.

20 Wadge hierarchy on Cantor space For A, B 2 N, set A W B there is a continuous f : 2 N 2 N such that A = f 1 (B). A W B A W B W A. A (equivalently, its Wadge degree [A] W ) is self dual iff A W A.

21 Wadge hierarchy on Cantor space For A, B 2 N, set A W B there is a continuous f : 2 N 2 N such that A = f 1 (B). A W B A W B W A. A (equivalently, its Wadge degree [A] W ) is self dual iff A W A. (Wadge; Martin) For all Borel A, B, A W B B W A; moreover, W is well founded on Borel sets.

22 How it looks 2 N

23 How it looks 2 N 0 1 \ {, 2 N }

24 How it looks Σ 0 1 \ Π \ {, 2 N } 2 N Π 0 1 \ Σ 0 1

25 How it looks Σ 0 1 \ Π \ {, 2 N } 2 N Π 0 1 \ Σ 0 1

26 How it looks Σ 0 1 \ Π \ {, 2 N } 2 N Π 0 1 \ Σ 0 1 Self dual degrees and non-self dual pairs of degrees alternate. At all limit levels there is a non-self dual pair. Each degree is assigned a rank, according to its position in the hierarchy (starting from 1).

27 How it looks Σ 0 1 \ Π \ {, 2 N } 2 N Π 0 1 \ Σ 0 1 Self dual degrees and non-self dual pairs of degrees alternate. At all limit levels there is a non-self dual pair. Each degree is assigned a rank, according to its position in the hierarchy (starting from 1). The length of this hierarchy up to 2 0 is ω 1.

28 How it looks Σ 0 1 \ Π \ {, 2 N } 2 N Π 0 1 \ Σ 0 1 Self dual degrees and non-self dual pairs of degrees alternate. At all limit levels there is a non-self dual pair. Each degree is assigned a rank, according to its position in the hierarchy (starting from 1). The length of this hierarchy up to 2 0 is ω 1. The length up to 0 3 is ω ω1 1.

29 Question Which of these Π 0 3 Wadgre degrees, arranged in ω ω levels, contain some regular sets?

30 Question Which of these Π 0 3 Wadgre degrees, arranged in ω ω levels, contain some regular sets? Remark. If A is clopen, then Φ(A) = A.

31 Climbing Wadge hierarchy of 0 3 sets In order to prove that in every degree of Wadge rank < ω ω1 1 there is a regular set, it would be desirable to have operations on the degrees i (A 0, A 1,...) such that:

32 Climbing Wadge hierarchy of 0 3 sets In order to prove that in every degree of Wadge rank < ω ω1 1 there is a regular set, it would be desirable to have operations on the degrees i (A 0, A 1,...) such that: 1. starting with 0 1 sets, they generate all 0 3 sets

33 Climbing Wadge hierarchy of 0 3 sets In order to prove that in every degree of Wadge rank < ω ω1 1 there is a regular set, it would be desirable to have operations on the degrees i (A 0, A 1,...) such that: 1. starting with 0 1 sets, they generate all 0 3 sets 2. Φ i (A 0, A 1,...) = i (Φ(A 0 ), Φ(A 1 ),...)

34 Climbing Wadge hierarchy of 0 3 sets In order to prove that in every degree of Wadge rank < ω ω1 1 there is a regular set, it would be desirable to have operations on the degrees i (A 0, A 1,...) such that: 1. starting with 0 1 sets, they generate all 0 3 sets 2. Φ i (A 0, A 1,...) = i (Φ(A 0 ), Φ(A 1 ),...) Mimicking some of the Wadge s constructions on the Baire space, there are candidate operations performing task 1.

35 Operations generating 0 2(2 N ) A A.

36 Operations generating 0 2(2 N ) A A. (A, B) A B. If A is non-self dual, then A A is a self dual immediate successor of A.

37 Operations generating 0 2(2 N ) A A. (A, B) A B. If A is non-self dual, then A A is a self dual immediate successor of A. (A n ) (A n ), (A n ) (A n ) If A is self-dual, then ( A, A ) is a pair of non-self dual sets immediate successors of A. If n A n < W A n+1, then (A n, A n) is the least non-self dual pair immediately above all A n.

38 The definitions A B = 0 A 1 B

39 The definitions A B = 0 A 1 B A n = n 0n 1 A n A n = A n {0 }

40 Reaching all of 0 3(2 N ) (A, B) A + B. If A is self-dual, then A + B W = A W + B W.

41 Reaching all of 0 3(2 N ) (A, B) A + B. If A is self-dual, then A + B W = A W + B W. (A n ) (A n ), (A n ) (A n ). If A is self-dual, then A, A are non-self dual, A W A and their Wadge rank is A W ω 1.

42 The definition For s = (s 0, s 1,...) 2 ω, let s = (s 0, s 0, s 1, s 1,...). Also, Ā = {ā} a A. A + B = { sta s 2 <ω, t {01, 10}, a A} B.

43 The definition For s = (s 0, s 1,...) 2 ω, let s = (s 0, s 0, s 1, s 1,...). Also, Ā = {ā} a A. A + B = { sta s 2 <ω, t {01, 10}, a A} B. A = { s 1 t 1... s n t n ā s i 2 <ω, t i {01, 10}, a A} A = A {x 2 N n x(2n) x(2n + 1)}

44 The definition For s = (s 0, s 1,...) 2 ω, let s = (s 0, s 0, s 1, s 1,...). Also, Ā = {ā} a A. A + B = { sta s 2 <ω, t {01, 10}, a A} B. A = { s 1 t 1... s n t n ā s i 2 <ω, t i {01, 10}, a A} A = A {x 2 N n x(2n) x(2n + 1)} Warning: Something suspicious here: sets of the form Ā have measure 0.

45 A serious obstacle Except for, the above operations do not commute with Φ and they do not take regular sets to regular sets. To fix this, they need to be replaced by more suitable ones.

46 A serious obstacle Except for, the above operations do not commute with Φ and they do not take regular sets to regular sets. To fix this, they need to be replaced by more suitable ones. The case of and. Let f : N N \ {0} be such that lim n f (n) = +.

47 A serious obstacle Except for, the above operations do not commute with Φ and they do not take regular sets to regular sets. To fix this, they need to be replaced by more suitable ones. The case of and. Let f : N N \ {0} be such that lim n f (n) = +. Definition. Rake(f, A n ) = n 0n 1 f (n) A n Rake + (f, A n ) = Rake(f, A n ) {0 } n,t {N 0 n t lh(t) = f (n), t 0 f (n), 1 f (n) }

48 Theorem for Rake A n W Rake(f, A n ). Φ(Rake(f, A n )) = Rake(f, Φ(A n )). In particular, Rake(f, A n ) is regular if all A n are. In fact: If n A n ran(φ Π 0 1 ) then Rake(f, A n ) ran(φ Π 0 1 ). If n A n ran(φ Σ 0 1 ) then Rake(f, A n ) ran(φ Σ 0 1 ).

49 Theorem for Rake + Similarly, A n W Rake + (f, A n ) Φ(Rake + (f, A n )) = Rake + (f, Φ(A n )). In particular, Rake + (f, A n ) is regular if all A n are. In fact: If n A n ran(φ Π 0 1 ) then Rake + (f, A n ) ran(φ Π 0 1 ). If n A n ran(φ Σ 0 1 ) then Rake + (f, A n ) ran(φ Σ 0 1 ).

50 Theorem for Rake + Similarly, A n W Rake + (f, A n ) Φ(Rake + (f, A n )) = Rake + (f, Φ(A n )). In particular, Rake + (f, A n ) is regular if all A n are. In fact: If n A n ran(φ Π 0 1 ) then Rake + (f, A n ) ran(φ Π 0 1 ). If n A n ran(φ Σ 0 1 ) then Rake + (f, A n ) ran(φ Σ 0 1 ). Corollary. Given any Wadge degree d 0 2 (2N ) there is a regular set A d such that A = Φ(C) = Φ(U) for some C closed and U open.

51 The other operations It is possible to define operations Sum(A, B), Nat(A), Flat(A) such that: Sum(A, B) W A + B, Nat(A) W A, Flat(A) W A. Sum(A, B), Nat(A), Flat(A) are regular when A, B are. If A, B are image under Φ of a closed set, the same holds for Sum(A, B), Nat(A), Flat(A). If A, B are image under Φ of an open set, the same holds for Sum(A, B), Nat(A), Flat(A).

52 Did you ask for a definition? Here you go for Sum(A, B):

53 Did you ask for a definition? Here you go for Sum(A, B): Fix an increasing sequence r n of positive real numbers converging to 1. Sum(A, B) = B { st O(2 length(s) r length(s) µ(a N s )) s 2 <ω, t {01, 10}} e E(B) e B,

54 Did you ask for a definition? Here you go for Sum(A, B): Fix an increasing sequence r n of positive real numbers converging to 1. Sum(A, B) = B { st O(2 length(s) r length(s) µ(a N s )) s 2 <ω, t {01, 10}} e E(B) e B, where O(r) = 2 N \ N 0 min{h>0 r 1 2 h} 1 E(B) = { st0 min{h>0 2length(s) r length(s) µ(b N s ) 1 2 h} 1 s 2 <ω, t {01, 10}}

55 The result for 0 3 degrees Every 0 3 Wedge degree contains a regular set. In fact, every 0 3 Wadge degree contains a set A such that A = Φ(C) = Φ(U) for some closed C and open U.

56 Regular Π 0 3-complete sets Where everything began: There are a closed set C and an open set U such that Φ(C) = Φ(U) is Π 0 3-complete.

57 Regular Π 0 3-complete sets Where everything began: There are a closed set C and an open set U such that Φ(C) = Φ(U) is Π 0 3-complete. In fact there are many regular Π 0 3-complete sets: Theorem. If a regular non-empty set has empty interior, then it is Π 0 3-complete.

58 Regular Π 0 3-complete sets Where everything began: There are a closed set C and an open set U such that Φ(C) = Φ(U) is Π 0 3-complete. In fact there are many regular Π 0 3-complete sets: Theorem. If a regular non-empty set has empty interior, then it is Π 0 3-complete. Corollary. If C is a closed set of positive measure with empty interior, then Φ(C) is Π 0 3-complete.

59 Most regular sets are Π 0 3-complete Each element of the measure algebra MALG can be assigned as a colour the Wedge degree of the unique regular set belonging to it: let W d = {[A] MALG Φ(A) d}.

60 Most regular sets are Π 0 3-complete Each element of the measure algebra MALG can be assigned as a colour the Wedge degree of the unique regular set belonging to it: let W d = {[A] MALG Φ(A) d}. Theorem. Except for d = { }, {2 N }, all W d are topologically dense in MALG. W Π 0 3 \ 0 3 W Π 0 3 \ 0 3 is comeagre in MALG. the unique W d that is dense in the sense of the forcing (MALG, ).

61 A corollary of the construction Fact. Every measurable set can be approximated by an F σ set from the inside and a G δ set from the outside. Hence each element of the measure algebra MALG contains both an F σ and a G δ set.

62 A corollary of the construction Fact. Every measurable set can be approximated by an F σ set from the inside and a G δ set from the outside. Hence each element of the measure algebra MALG contains both an F σ and a G δ set. This cannot be improved: Theorem. The elements of MALG that contain a 0 2 member form a meagre subset of MALG.

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