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

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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: DA(x ) = lim

ε→0⁺

µ(A ∩ B_ε(x )) µ(Bε(x )) .

Let also

Φ(A) = {x ∈ X | DA(x ) = 1} be the set of points with density 1 in A.

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The density function

Given a measurable A ⊆ X , define the density function:

DA(x ) = lim

ε→0⁺

µ(A ∩ B_ε(x )) µ(Bε(x )) .

Let also

Φ(A) = {x ∈ X | DA(x ) = 1} be the set of points with density 1 in A.

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The density function

Given a measurable A ⊆ X , define the density function:

DA(x ) = lim

ε→0⁺

µ(A ∩ B_ε(x )) µ(Bε(x )) .

Let also

Φ(A) = {x ∈ X | DA(x ) = 1}

be the set of points with density 1 in A.

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If A, B are measure equivalent, in symbols A ≡ B, then DA = DB, thus Φ(A) = Φ(B).

Thus Φ induces a function Φ : MALG (X ) → MALG (X ) on the measure algebra of X .

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The problem

What can be said, in general, about Φ?

Apparently not much. Very simple properties can be established, like:

I Φ(∅) = ∅, Φ(X ) = X

I A ⊆ B ⇒ Φ(A) ⊆ Φ(B)

I Φ(A ∩ B) = Φ(A) ∩ Φ(B)

I S

i ∈IΦ(A_i) ⊆ Φ(S

i ∈IA_i)

I Φ(¬A) ⊆ ¬Φ(A)

I A ⊆ Φ(A) for open A

Thus {A ∈ MEAS (X ) | A ⊆ Φ(A)} is a topology (the density topology) extending the topology of X .

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The problem

I Φ(∅) = ∅, Φ(X ) = X

I A ⊆ B ⇒ Φ(A) ⊆ Φ(B)

I Φ(A ∩ B) = Φ(A) ∩ Φ(B)

I S

i ∈IA_i)

I Φ(¬A) ⊆ ¬Φ(A)

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The problem

I Φ(∅) = ∅, Φ(X ) = X

I A ⊆ B ⇒ Φ(A) ⊆ Φ(B)

I Φ(A ∩ B) = Φ(A) ∩ Φ(B)

I S

i ∈IA_i)

I Φ(¬A) ⊆ ¬Φ(A)

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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).

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Lebesgue density theorem

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

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Lebesgue density theorem

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

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Lebesgue density theorem

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

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Examples of spaces satisfying LDT

I Rⁿ

I Polish ultrametric spaces (B. Miller)

I Problem. Which Polish spaces satisfy Lebesgue density theorem? Here I will focus on Cantor space 2^Nwith the usual distance and Lebesgue (coin-tossing) measure.

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Examples of spaces satisfying LDT

I Rⁿ

I Problem. Which Polish spaces satisfy Lebesgue density theorem?

Here I will focus on Cantor space 2^Nwith the usual distance and Lebesgue (coin-tossing) measure.

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Examples of spaces satisfying LDT

I Rⁿ

I Problem. Which Polish spaces satisfy Lebesgue density theorem?

Here I will focus on Cantor space 2^Nwith the usual distance and Lebesgue (coin-tossing) measure.

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In 2^Nthe density function becomes DA(x ) = lim

n→∞

µ(A ∩ N_{x |}_n) µ(N_{x |}_n) = lim

n→∞2ⁿµ(A ∩ N_{x |}_n) and

x ∈ Φ(A) ⇔ ∀ε ∃n ∀m > n 2^mµ(A ∩ N_{x |}_m) > 1 − ε, so that Φ(A) is Π⁰₃.

Sets of the form Φ(A) will be called regular. Since Φ²= Φ, a set A is regular iff Φ(A) = A.

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In 2^Nthe density function becomes DA(x ) = lim

n→∞

µ(A ∩ N_{x |}_n) µ(N_{x |}_n) = lim

n→∞2ⁿµ(A ∩ N_{x |}_n) and

x ∈ Φ(A) ⇔ ∀ε ∃n ∀m > n 2^mµ(A ∩ N_{x |}_m) > 1 − ε, so that Φ(A) is Π⁰₃.

Sets of the form Φ(A) will be called regular. Since Φ²= Φ, a set A is regular iff Φ(A) = A.

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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.

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Wadge hierarchy on Cantor space

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,

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Wadge hierarchy on Cantor space

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,

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Wadge hierarchy on Cantor space

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.

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How it looks

∅ 2^N

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How it looks

∅

∆⁰₁\ {∅, 2^N} 2^N

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How it looks

∅ Σ⁰₁\ Π⁰₁

∆⁰₁\ {∅, 2^N}

2^N Π⁰₁\ Σ⁰₁

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How it looks

∅ Σ⁰₁\ Π⁰₁ • • •

∆⁰₁\ {∅, 2^N} • • · · · •

2^N Π⁰₁\ Σ⁰₁ • • •

· · ·

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How it looks

∅ Σ⁰₁\ Π⁰₁ • • •

∆⁰₁\ {∅, 2^N} • • · · · •

2^N Π⁰₁\ Σ⁰₁ • • •

· · ·

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 ∆²₀is ω₁. The length up to ∆⁰₃is ω₁^ω¹.

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How it looks

∅ Σ⁰₁\ Π⁰₁ • • •

∆⁰₁\ {∅, 2^N} • • · · · •

2^N Π⁰₁\ Σ⁰₁ • • •

· · ·

The length of this hierarchy up to ∆²₀is ω1.

The length up to ∆⁰₃is ω₁^ω¹.

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How it looks

∅ Σ⁰₁\ Π⁰₁ • • •

∆⁰₁\ {∅, 2^N} • • · · · •

2^N Π⁰₁\ Σ⁰₁ • • •

· · ·

The length of this hierarchy up to ∆²₀is ω1. The length up to ∆⁰₃is ω₁^ω¹.

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Question

Which of these Π⁰₃ Wadgre degrees, arranged in ω^ω₁¹+ 1 levels, contain some regular sets?

Remark. If A is clopen, then Φ(A) = A.

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Question

Which of these Π⁰₃ Wadgre degrees, arranged in ω^ω₁¹+ 1 levels, contain some regular sets?

Remark. If A is clopen, then Φ(A) = A.

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Climbing Wadge hierarchy of ∆

⁰₃

sets

In order to prove that in every degree of Wadge rank < ω^ω₁¹ there is a regular set, it would be desirable to have operations on the degrees

◦i(A₀, A₁, . . .) such that:

1. starting with ∆⁰₁sets, they generate all ∆⁰₃ sets 2. Φ ◦_i(A₀, A₁, . . .) = ◦_i(Φ(A₀), Φ(A₁), . . .)

Mimicking some of the Wadge’s constructions on the Baire space, there are candidate operations performing task 1.

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Climbing Wadge hierarchy of ∆

⁰₃

sets

◦i(A₀, A₁, . . .) such that:

1. starting with ∆⁰₁sets, they generate all ∆⁰₃ sets

2. Φ ◦_i(A₀, A₁, . . .) = ◦_i(Φ(A₀), Φ(A₁), . . .)

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Climbing Wadge hierarchy of ∆

⁰₃

sets

◦i(A₀, A₁, . . .) such that:

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Climbing Wadge hierarchy of ∆

⁰₃

sets

◦i(A₀, A₁, . . .) such that:

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Operations generating ∆

⁰₂

(2

^N

)

I A 7→ ¬A.

I (A, B) 7→ A ⊕ B.

If A is non-self dual, then A ⊕ ¬A is a self dual immediate successor of A.

I (An) 7→ (An)^∇, (An) 7→ (An)^◦

If A is self-dual, then ( ~A^∇, ~A^◦) is a pair of non-self dual sets immediate successors of A.

If ∀n An<W An+1, then (A^∇_n, A^◦_n) is the least non-self dual pair immediately above all An.

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Operations generating ∆

⁰₂

(2

^N

)

I A 7→ ¬A.

I (A, B) 7→ A ⊕ B.

I (An) 7→ (An)^∇, (An) 7→ (An)^◦

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Operations generating ∆

⁰₂

(2

^N

)

I A 7→ ¬A.

I (A, B) 7→ A ⊕ B.

I (An) 7→ (An)^∇, (An) 7→ (An)^◦

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The definitions

• A ⊕ B = 0^aA ∪ 1^aB

• A^∇_n =S

n0ⁿ1^aAn

• A^◦_n= A^∇_n ∪ {0^∞}

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The definitions

• A ⊕ B = 0^aA ∪ 1^aB

• A^∇_n =S

n0ⁿ1^aAn

• A^◦_n= A^∇_n ∪ {0^∞}

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Reaching all of ∆

⁰₃

(2

^N

)

I (A, B) 7→ A + B.

If A is self-dual, then ||A + B||W = ||A||W + ||B||W.

I (An) 7→ (An)^\, (An) 7→ (An)^[.

If A is self-dual, then A^\, A^[are non-self dual, A^\≡W ¬A^[and their Wadge rank is ||A||Wω1.

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Reaching all of ∆

⁰₃

(2

^N

)

I (A, B) 7→ A + B.

If A is self-dual, then ||A + B||W = ||A||W + ||B||W.

I (An) 7→ (An)^\, (An) 7→ (An)^[.

If A is self-dual, then A^\, A^[are non-self dual, A^\≡W ¬A^[and their Wadge rank is ||A||Wω1.

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The definition

For s = (s0, s1, . . .) ∈ 2^≤ω, let ¯s = (s0, s0, s1, s1, . . .). Also, ¯A = {¯a}a∈A.

• A + B = {¯sta | s ∈ 2^<ω, t ∈ {01, 10}, a ∈ A} ∪ ¯B.

• A^\= { ¯s1t1. . . ¯sntna | s¯ i ∈ 2^<ω, ti ∈ {01, 10}, a ∈ A}

• A^[= A^\∪ {x ∈ 2^N| ∃^∞n x (2n) 6= x (2n + 1)} Warning:

Something suspicious here: sets of the form ¯A have measure 0.

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The definition

• A + B = {¯sta | s ∈ 2^<ω, t ∈ {01, 10}, a ∈ A} ∪ ¯B.

• A^[= A^\∪ {x ∈ 2^N| ∃^∞n x (2n) 6= x (2n + 1)}

Warning:

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The definition

• A + B = {¯sta | s ∈ 2^<ω, t ∈ {01, 10}, a ∈ A} ∪ ¯B.

• A^[= A^\∪ {x ∈ 2^N| ∃^∞n x (2n) 6= x (2n + 1)}

Warning:

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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.

I Rake(f , An) =S

n0ⁿ1^{f (n)a}An

I Rake⁺(f , A_n) = Rake(f , A_n) ∪ {0^∞} ∪S

n,t{N₀nt | lh(t) = f (n), t 6= 0^{f (n)}, 1^{f (n)}}

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A serious obstacle

Let f : N → N \ {0} be such that lim^n→∞f (n) = +∞.

Definition.

I Rake(f , An) =S

n0ⁿ1^{f (n)a}An

n,t{N₀nt | lh(t) = f (n), t 6= 0^{f (n)}, 1^{f (n)}}

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A serious obstacle

Let f : N → N \ {0} be such that lim^n→∞f (n) = +∞.

Definition.

I Rake(f , An) =S

n0ⁿ1^{f (n)a}An

n,t{N0ⁿt | lh(t) = f (n), t 6=

0^{f (n)}, 1^{f (n)}}

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Theorem for Rake

I A^∇_n ≡W Rake(f , An).

I 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:

I If ∀n An∈ ran(Φ|_Π0

1) then Rake(f , An) ∈ ran(Φ|_Π0 1).

I If ∀n A_n∈ ran(Φ|_Σ0

1) then Rake(f , A_n) ∈ ran(Φ|_Σ0 1).

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Theorem for Rake

⁺

Similarly,

I A^◦_n≡_W Rake⁺(f , A_n)

I Φ(Rake⁺(f , An)) = Rake⁺(f , Φ(An)). In particular, Rake⁺(f , An) is regular if all An are. In fact:

1) then Rake⁺(f , An) ∈ ran(Φ|_Π0 1).

I If ∀n A_n∈ ran(Φ|_Σ⁰

1) then Rake⁺(f , A_n) ∈ ran(Φ|_Σ0 1).

Corollary. Given any Wadge degree d ⊆ ∆⁰₂(2^N) there is a regular set A ∈ d such that A = Φ(C ) = Φ(U) for some C closed and U open.

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Theorem for Rake

⁺

Similarly,

I A^◦_n≡_W Rake⁺(f , A_n)

I Φ(Rake⁺(f , An)) = Rake⁺(f , Φ(An)). In particular, Rake⁺(f , An) is regular if all An are. In fact:

1) then Rake⁺(f , An) ∈ ran(Φ|_Π0 1).

I If ∀n A_n∈ ran(Φ|_Σ⁰

1) then Rake⁺(f , A_n) ∈ ran(Φ|_Σ0 1).

Corollary. Given any Wadge degree d ⊆ ∆⁰₂(2^N) there is a regular set A ∈ d such that A = Φ(C ) = Φ(U) for some C closed and U open.

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The other operations

It is possible to define operations Sum(A, B), Nat(A), Flat(A) such that:

I Sum(A, B) ≡_W A + B, Nat(A) ≡_W A^\, Flat(A) ≡_W A^[.

I Sum(A, B), Nat(A), Flat(A) are regular when A, B are.

I If A, B are image under Φ of a closed set, the same holds for Sum(A, B), Nat(A), Flat(A).

I If A, B are image under Φ of an open set, the same holds for Sum(A, B), Nat(A), Flat(A).

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Did you ask for a definition?

Here you go for Sum(A, B):

Fix an increasing sequence rn of positive real numbers converging to 1. Sum(A, B) = ¯B ∪S{¯st^aO(2^length(s)rlength(s)µ(A ∩ Ns)) | s ∈ 2^<ω, t ∈ {01, 10}} ∪S

e∈E(B)e^aB, where

O(r ) = 2^N\ N₀min{h>0|r ≤1−2−h }1

E(B) = {¯st0min{h>0|2^length(s)r_length(s)µ(B∩Ns)≤1−2^−h}1 | s ∈ 2^<ω, t ∈ {01, 10}}

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Did you ask for a definition?

Fix an increasing sequence rn of positive real numbers converging to 1.

Sum(A, B) = ¯B ∪S{¯st^aO(2^length(s)r_length(s)µ(A ∩ Ns)) | s ∈ 2^<ω, t ∈ {01, 10}} ∪S

e∈E(B)e^aB,

where

O(r ) = 2^N\ N₀min{h>0|r ≤1−2−h }1

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Did you ask for a definition?

Fix an increasing sequence rn of positive real numbers converging to 1.

Sum(A, B) = ¯B ∪S{¯st^aO(2^length(s)r_length(s)µ(A ∩ Ns)) | s ∈ 2^<ω, t ∈ {01, 10}} ∪S

e∈E(B)e^aB, where

O(r ) = 2^N\ N₀min{h>0|r ≤1−2−h }1

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The result for ∆

⁰₃

degrees

Every ∆⁰₃ Wedge degree contains a regular set. In fact, every ∆⁰₃Wadge degree contains a set A such that A = Φ(C ) = Φ(U) for some closed C and open U.

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Regular Π

⁰₃

-complete sets

Where everything began: There are a closed set C and an open set U such that Φ(C ) = Φ(U) is Π⁰₃-complete.

In fact there are many regular Π⁰₃-complete sets:

Theorem. If a regular non-empty set has empty interior, then it is Π⁰₃-complete.

Corollary. If C is a closed set of positive measure with empty interior, then Φ(C ) is Π⁰₃-complete.

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Regular Π

⁰₃

-complete sets

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Regular Π

⁰₃

-complete sets

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Most regular sets are Π

⁰₃

-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

Wd = {[A] ∈ MALG | Φ(A) ∈ d }.

Theorem.

I Except for d = {∅}, {2^N}, all Wd are topologically dense in MALG .

I W_Π0

3\∆⁰₃ is comeagre in MALG .

I W_Π0

3\∆⁰₃ the unique W_d that is dense in the sense of the forcing (MALG , ≤).

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Most regular sets are Π

⁰₃

-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

Wd = {[A] ∈ MALG | Φ(A) ∈ d }.

Theorem.

I Except for d = {∅}, {2^N}, all Wd are topologically dense in MALG .

I W_Π0

3\∆⁰₃ is comeagre in MALG .

I W_Π0

3\∆⁰₃ the unique W_d that is dense in the sense of the forcing (MALG , ≤).

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A corollary of the construction

Fact. Every measurable set can be approximated by an Fσ set from the inside and a Gδ set from the outside. Hence each element of the measure algebra MALG contains both an Fσ and a Gδ set.

This cannot be improved:

Theorem. The elements of MALG that contain a ∆⁰₂ member form a meagre subset of MALG .

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A corollary of the construction

Fact. Every measurable set can be approximated by an Fσ set from the inside and a Gδ set from the outside. Hence each element of the measure algebra MALG contains both an Fσ and a Gδ set.

This cannot be improved:

Theorem. The elements of MALG that contain a ∆⁰₂ member form a meagre subset of MALG .