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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Weak Mixing and the Chacon Transformation

1 · Prerequisites

2 · Summary

Weak mixing has three equivalent descriptions here: vanishing Cesàro averages of absolute centered correlations, ergodicity of the product system, and absence of nonconstant complex eigenfunctions. The correlation criterion also gives convergence outside a density-zero set, with that exceptional set allowed to depend on the pair of functions. Pairings are linear in the first variable, and probability normalization is essential when constants are separated from centered functions.

The proof develops its Hilbert-space ingredients first. Orthogonal projection yields mean ergodic convergence. Rectangle approximation turns square-integrable kernels into compact operators; an invariant kernel produces a compact intertwiner. A positive compact operator then supplies a finite-dimensional eigenspace, and complex finite-dimensional algebra supplies an eigenvector. Each of these steps is proved before the equivalence theorem.

The Chacon construction uses three equal cuts and one spacer over the middle subcolumn. Explicit compatible partial translations extend on a conull invariant space to an invertible probability-preserving transformation. Tower levels approximate measurable sets. Their common invariant-set density proves ergodicity, while the two return times differing by one force every eigenvalue to equal one. Weak mixing follows from the earlier equivalence. Strong mixing fails on the fixed interval $[0, 2 / 9)$ : along tower heights its self-correlation is at least $2 / 27$ , exceeding the independent value $4 / 81$ by $2 / 81$ .

AC is stated where countable selections or measure-theoretic suppliers require it. The finite tower recurrences themselves require no choice. The Chacon source-verification and independent-proof obligations are recorded as resolved in this batch's coverage: the unavailable Katok–Thouvenot and Creutz backing is waived against the seven complete local constructions, and the separate independent-source requirement is satisfied by the owner's reading of Varju's complete general-eigenfunction proof (evidence records research/phase-2-next-20-chacon-source-alternative-review.json and research/phase-2-next-20-chacon-varju-source-review.json). Ordinary mathematical review of the authored items remains required and is not claimed here.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)Open item page →

Eigenfunction for a probability system

Definition

Let $(X, A, μ, T)$ be a measure-preserving probability system. Work in complex $L^{2} (μ)$ with the pairing $⟨ f, g ⟩ = \int f \overline{g}$ , linear in its first variable, as in The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz. Set $U_{T} f = f \circ T$ on almost-everywhere classes and $H_{0} = {f \in L^{2} (μ) : \int f = 0}$ .

An eigenfunction is a nonzero class $f \in L^{2} (μ)$ for which $U_{T} f = λ f$ for some $λ \in C$ . It is nonconstant when $f \notin C 1$ . Here equality and constancy always mean equality almost everywhere.

We verify the integral interface used here directly. For a nonnegative simple function, augment any finite disjoint display by the measurable complement of its displayed sets, with coefficient $0$ . Given two such augmented displays, their pairwise intersections partition $X$ , and equality of the functions forces the two coefficients to agree on every nonempty intersection. Finite additivity, with $0 \cdot (+ \infty) = 0$ , therefore proves representation independence. On a common augmented refinement the simple integral is monotone and additive; homogeneity is direct when the scalar is $0$ and termwise when it is positive. Now let $0 \leq g_{j} ↑ g$ and put $L = \sup_{j} \int g_{j}$ . For a simple $s \leq g$ and $0 < c < 1$ , the sets $A_{j} = {g_{j} \geq c s}$ increase to $X$ , including on the zero level of $s$ . Since $A \mapsto \int_{A} s$ is the finite sum of the measures of the nonzero level sets, continuity from below gives $c \int s = \lim_{j} c \int_{A_{j}} s \leq L$ . Letting $c ↑ 1$ and then taking the supremum over $s$ proves monotone convergence from the definitions in The integral of a nonnegative simple function and The nonnegative Lebesgue integral.

Applying this monotone convergence result to sums of increasing simple approximants gives nonnegative additivity. Positive/negative and real/imaginary decompositions then give finite complex $L^{1}$ linearity. Applying nonnegative integral invariance to those four parts gives $\int h \circ T = \int h$ for every complex $h \in L^{1} (μ)$ .

For a measure-preserving $T$ , canonical level sets give $\int s \circ T = \int s$ for every nonnegative simple $s$ . Choose $s_{j} ↑ g$ using Every nonnegative measurable function is the increasing limit of simple measurable functions and apply the preceding monotone-convergence argument to $s_{j}$ and $s_{j} \circ T$ ; hence $\int g \circ T = \int g$ for every nonnegative measurable $g$ . Applying this to $g = ∣ f ∣^{2}$ proves directly that the pullback of The Koopman operator is an $L^{2}$ isometry. Thus $∥ f ∥_{2} = ∣ λ ∣ ∥ f ∥_{2}$ forces $∣ λ ∣ = 1$ . For a system invertible modulo null sets in Invertible measure-preserving systems, pullback by the inverse is an inverse isometry, so $U_{T}$ is unitary. The name does not assume invertibility for every probability system.

Cauchy–Schwarz with $1$ , whose norm is one, gives $∣ \int f ∣ \leq ∥ f ∥_{2}$ and integrability of $f$ . Thus $H_{0}$ is a closed linear subspace. The locally proved integral invariance gives $(λ - 1) \int f = 0$ . If $λ \neq 1$ , the eigenfunction already lies in $H_{0}$ . If $λ = 1$ and $f$ is nonconstant, $f - (\int f) 1$ is a nonzero eigenfunction in $H_{0}$ .

In this page's spectral criterion, a completed Lebesgue probability space has the usual interval-and-atoms model modulo null sets, with completed measure; no classification theorem for arbitrary probability spaces is used. The Chacon model below is the completed unit interval. These definitions and the displayed finite calculations make no simultaneous choices of representatives.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)Open item page →

L two operator conventions for weak mixing

Definition

Let $H$ be a closed complex $L^{2}$ subspace with the first-variable-linear pairing of The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz. All operators below map $H$ to $H$ and are complex-linear. An operator $A$ is bounded if $∥ A f ∥ \leq C ∥ f ∥$ for some finite $C \geq 0$ and all $f \in H$ ; its norm is $∥ A ∥ = \sup_{∥ f ∥ \leq 1} ∥ A f ∥$ . It is compact if it is bounded and every bounded sequence $(f_{n})$ has a subsequence $(f_{n_{j}})$ for which $(A f_{n_{j}})$ converges in norm to an element of $H$ .

An adjoint $A^{*}$ is a bounded operator satisfying $⟨ A f, g ⟩ = ⟨ f, A^{*} g ⟩$ for all $f, g \in H$ . Such an operator, if it exists, is unique: subtract two proposed identities and set $f$ equal to the difference of their values at $g$ . Positive definiteness makes that difference zero. Existence is not assumed by this definition.

The operator $A$ is self-adjoint if $⟨ A f, g ⟩ = ⟨ f, A g ⟩$ for all $f, g$ . It is positive if $⟨ A f, f ⟩$ is real and nonnegative for every $f$ . Later positive self-adjoint assertions impose both conditions explicitly.

An isometry preserves the norm; for linear operators it also preserves the pairing. Indeed expansion of $∥ f + g ∥^{2}$ gives $2 Re ⟨ f, g ⟩ = ∥ f + g ∥^{2} - ∥ f ∥^{2} - ∥ g ∥^{2}$ , and expansion of $∥ f + i g ∥^{2}$ gives $2 Im ⟨ f, g ⟩ = ∥ f + i g ∥^{2} - ∥ f ∥^{2} - ∥ g ∥^{2}$ . Applying both identities before and after the isometry proves the assertion. A unitary is a surjective linear isometry.

A linear subspace $E$ is invariant for $U$ if $U (E) \subseteq E$ . Write $f ⊥ E$ when $⟨ f, e ⟩ = 0$ for every $e \in E$ , and $E^{⊥} = {f \in H : f ⊥ E}$ . These conventions allow $H = {0}$ , $E = {0}$ and the zero operator. In the zero space the operator norm is zero because the unit ball is ${0}$ . No infinite selection or assertion of an orthonormal basis enters these definitions.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Closed l two subspaces have orthogonal projections

Statement

Assume AC. Let $H$ be complex $L^{2} (μ)$ or a closed linear subspace of it, and let $M$ be a closed linear subspace of $H$ . There is a unique linear contraction $P_{M} : H \to H$ such that, for every $f \in H$ , $P_{M} f \in M$ and $f - P_{M} f ⊥ M$ . Moreover, $∥ f - P_{M} f ∥ = \inf_{g \in M} ∥ f - g ∥, H = M \oplus M^{⊥} .$ Here the orthogonal complement is taken inside $H$ .

Facts & Assumptions

[F1]

The complex pairing is positive definite and sesquilinear, with norm $∥ f ∥_{2}$ and Cauchy–Schwarz The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F2]

Complex $L^{2}$ is complete under countable choice Complex Lp completeness and almost-everywhere subsequences.

[F3]

AC supplies a choice function on a family of nonempty sets The Axiom of Choice.

[F4]

Orthogonality and contraction use the local operator conventions L two operator conventions for weak mixing.

Proof

Given: AC, $H$ , $M$ and $f$ as in the statement.

1.1

Since $0 \in M$ , $d = \inf_{g \in M} ∥ f - g ∥$ exists in $[0, ∥ f ∥]$ . For every $n \in N$ the set ${g \in M : ∥ f - g ∥^{2} < d^{2} + 1 / (n + 1)}$ is nonempty by the defining property of the infimum. AC selects $g_{n}$ in these sets, including $g_{0}$ . It also supplies the countable choice assumed in complex completeness.

F3given

2.1

Expanding the pairing gives $∥ u + v ∥^{2} + ∥ u - v ∥^{2} = 2 ∥ u ∥^{2} + 2 ∥ v ∥^{2}$ : the two cross terms cancel. Apply this to $u = f - g_{n}$ , $v = f - g_{m}$ . Since $(g_{n} + g_{m}) / 2 \in M$ , it follows that $∥ g_{n} - g_{m} ∥^{2} \leq 2 ∥ f - g_{n} ∥^{2} + 2 ∥ f - g_{m} ∥^{2} - 4 d^{2} < 2 / (n + 1) + 2 / (m + 1)$ . Thus $(g_{n})_{n \in N}$ is Cauchy.

F1step 1.1

3.1

Completeness gives a norm limit in $L^{2}$ . Closedness of $H$ , then of $M$ in $H$ , puts this limit $g$ in $M$ . The triangle inequality implies $∣ ∥ f - g_{n} ∥ - ∥ f - g ∥ ∣ \leq ∥ g_{n} - g ∥$ , so $∥ f - g ∥ = d$ . This reasoning applies equally when $H$ is the full space.

F2step 2.1given

4.1

Set $z = f - g$ . For $h \in M$ and real $t$ , minimality gives $0 \leq ∥ z - t h ∥^{2} - ∥ z ∥^{2} = - 2 t Re ⟨ z, h ⟩ + t^{2} ∥ h ∥^{2}$ . Dividing separately for positive and negative $t$ and letting $t \to 0$ makes the real part zero. Replacing $h$ by $i h \in M$ makes the imaginary part zero because $⟨ z, i h ⟩ = - i ⟨ z, h ⟩$ . Hence $z ⊥ M$ .

F1F4step 3.1

5.1

If $g^{'}$ also lies in $M$ with $f - g^{'} ⊥ M$ , then $g - g^{'} \in M \cap M^{⊥}$ ; its squared norm is zero, so $g = g^{'}$ . Define $P_{M} f = g$ . Conversely any such orthogonal decomposition minimizes distance: for $h \in M$ , expansion yields $∥ f - h ∥^{2} = ∥ f - g ∥^{2} + ∥ g - h ∥^{2} \geq ∥ f - g ∥^{2}$ . Thus it has exactly the required distance property.

F1step 4.1

6.1

For $a, b \in C$ and $f_{1}, f_{2} \in H$ , the vector $a P_{M} f_{1} + b P_{M} f_{2}$ lies in $M$ , and its difference from $a f_{1} + b f_{2}$ is orthogonal to $M$ by sesquilinearity. Uniqueness gives linearity. Orthogonal expansion gives $∥ f ∥^{2} = ∥ P_{M} f ∥^{2} + ∥ f - P_{M} f ∥^{2}$ , proving contraction. Pairing with each fixed $h \in M$ is continuous by Cauchy–Schwarz, so $M^{⊥}$ is closed; it is a subspace by linearity. Each $f$ has the displayed decomposition, and its uniqueness follows from $M \cap M^{⊥} = {0}$ . When $M = {0}$ this gives $P_{M} = 0$ , and when $M = H$ it gives $P_{M} = I$ , including the zero-space case.

F1F4step 5.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Hilbert cesaro averages converge to the fixed subspace

Statement

Assume AC. Let $U$ be a linear isometry on a closed complex $L^{2}$ subspace $H$ , and put $F = \ker (I - U)$ . For each $f \in H$ , $A_{N} f = \frac{1}{N} \sum_{n = 0}^{N - 1} U^{n} f ⟶ P_{F} f in norm as N \to \infty .$ Also $R = U (H)$ is closed, and $V = U^{- 1} P_{R}$ is a linear contraction satisfying $⟨ U f, g ⟩ = ⟨ f, V g ⟩$ and $V U = I$ . Here $U^{- 1}$ means the inverse from $R$ to $H$ , not a surjectivity assumption on $U$ .

Facts & Assumptions

[F1]

Closed subspaces have unique orthogonal projections and orthogonal decompositions under AC Closed l two subspaces have orthogonal projections.

[F2]

Isometries preserve the pairing and the norm; adjoint and invariant-subspace conventions are fixed locally L two operator conventions for weak mixing.

[F3]

The complex pairing is sesquilinear and satisfies Cauchy–Schwarz The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F4]

Assume AC The Axiom of Choice, as required for the projections and completeness used in their proof.

[F5]

Under countable choice, complex $L^{2}$ is complete Complex Lp completeness and almost-everywhere subsequences.

Proof

Given: $H$ , $U$ and AC as stated; $N$ is a positive integer.

1.1

F5 and AC make the ambient complex $L^{2}$ complete. Hence its closed subspace $H$ is complete: an $H$ -valued Cauchy sequence converges in the ambient space by F5, and closedness puts its limit in $H$ . If $U f_{n}$ converges in $H$ , then $∥ f_{n} - f_{m} ∥ = ∥ U f_{n} - U f_{m} ∥$ makes $(f_{n})$ Cauchy. Its limit $f \in H$ satisfies $U f_{n} \to U f$ by isometry, so $R$ is closed. Isometry makes $U$ injective; its inverse on $R$ is linear and isometric. The projection $P_{R}$ therefore defines the linear contraction $V = U^{- 1} P_{R}$ .

F1F2F4F5

1.2

Let $M = \overline{(I - U) H}$ . This is a closed subspace: sums and scalar multiples of limits remain limits by the norm inequalities. If $h ⊥ M$ , then $⟨ h, h - U h ⟩ = 0$ , whence $⟨ h, U h ⟩ = ∥ h ∥^{2}$ . Expansion and isometry give $∥ h - U h ∥^{2} = 2 ∥ h ∥^{2} - 2 Re ⟨ h, U h ⟩ = 0$ , so $U h = h$ . Conversely, if $U h = h$ , then for each $g \in H$ , $⟨ h, U g ⟩ = ⟨ U h, U g ⟩ = ⟨ h, g ⟩$ . Thus $h$ is orthogonal to $(I - U) H$ , and Cauchy–Schwarz extends orthogonality to its closure. Consequently $M^{⊥} = F$ .

F2F3

2.1

Write $P_{R} g = U h$ . Orthogonality gives $⟨ U f, g ⟩ = ⟨ U f, U h ⟩ = ⟨ f, h ⟩ = ⟨ f, V g ⟩$ . Since $P_{R} U f = U f$ , we have $V U f = f$ . This proves the adjoint identity without a representation theorem or an inverse of $U$ on all of $H$ .

F1F2step 1.1

2.2

By orthogonal decomposition, $H = M \oplus F$ . The subspace $F$ is closed, either as $M^{⊥}$ or directly by continuity of $I - U$ . In the decomposition $f = m + h$ , $m \in M$ , $h \in F$ , the vector $m$ is orthogonal to $F$ , so uniqueness of projection gives $h = P_{F} f$ .

F1step 1.2

2.3

Isometry and the triangle inequality give $∥ A_{N} ∥ \leq 1$ . For $g \in H$ , cancellation of the finite sum gives $A_{N} (I - U) g = (g - U^{N} g) / N$ , of norm at most $2 ∥ g ∥ / N$ . For $m \in M$ and $ε > 0$ , choose one $g$ with $∥ m - (I - U) g ∥ < ε$ . Hence ${lim sup}_{N} ∥ A_{N} m ∥ \leq ε$ . As $ε$ is arbitrary, $A_{N} m \to 0$ . No sequence of such approximants is needed.

F2step 1.2

3.1

For $h \in F$ , every $U^{n} h = h$ , so $A_{N} h = h$ . Applying this and the previous limit to $f = m + h$ gives $A_{N} f \to h = P_{F} f$ . For $N = 1$ the average is the identity; zero vectors and $H = {0}$ obey every formula without division by a vector norm. AC is inherited from the projection/completeness argument in step 1.1 and the projections in step 2.2.

F4step 1.1step 2.2step 2.3∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Product rectangle kernels are dense in complex l two

Statement

Assume AC. On the completed product of two finite measure spaces, finite complex linear combinations of rectangle indicators $1_{E} (x) 1_{F} (y)$ are dense in complex $L^{2}$ . Consequently a kernel pairing to zero against every rectangle indicator is the zero $L^{2}$ class.

Facts & Assumptions

[F1]

Finite disjoint rectangle unions form an algebra generating the product sigma-algebra Finite disjoint unions of measurable rectangles form an algebra generating the product sigma-algebra.

[F2]

In a finite measure space, every measurable set is approximable in symmetric difference by a generating algebra Approximation in symmetric difference by a generating algebra.

[F3]

Complex finite simple functions are dense in finite-exponent $L^{p}$ Complex finite-simple and smooth compact-support density for finite p.

[F4]

Under countable choice every completion-measurable real function has a base-measurable a.e. representative A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra.

[F5]

The complex pairing satisfies Cauchy–Schwarz and induces the $L^{2}$ norm The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F6]

AC supplies countable choice The Axiom of Choice.

Proof

Given: Finite measure spaces $(X, A, μ)$ and $(Y, B, ν)$ , a complex kernel $k$ in their completed product $L^{2}$ , and $ε > 0$ .

1.1

Apply the representative theorem separately to the real and imaginary parts of $k$ . The resulting base-measurable functions equal those components a.e.; their sets of infinite values are base-measurable and null. Replacing their infinite values by zero and combining them gives a finite complex product-measurable representative $k_{0}$ of $k$ . Its norm is unchanged. AC supplies the countable choice required in this step.

F4F6

2.1

The uncompleted product has finite mass $μ (X) ν (Y)$ . On it choose a finite simple function $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ with $∥ k_{0} - s ∥_{2} < ε / 2$ . Terms with zero coefficient may be removed. If no terms remain, the zero rectangle combination already approximates $k$ within $ε$ .

F3step 1.1

3.1

Otherwise put $B = \sum_{j = 1}^{m} ∣ c_{j} ∣ > 0$ and $δ = (ε / (2 B))^{2}$ . By the generating-algebra approximation choose, for each of these finitely many $j$ , a finite disjoint rectangle union $R_{j}$ with $(μ \otimes ν) (E_{j} △ R_{j}) < δ$ . Since $∥ 1_{E_{j}} - 1_{R_{j}} ∥_{2}$ is the square root of that measure, the triangle inequality gives $∥ s - \sum_{j} c_{j} 1_{R_{j}} ∥_{2} < B \sqrt{δ} = ε / 2$ . Each $1_{R_{j}}$ is a finite sum of disjoint rectangle indicators. Combined with step 2.1 this proves density, on the completion as well because the norms of base-measurable functions agree with their completed norms.

F1F2F5step 2.1

4.1

If $⟨ k, 1_{E \times F} ⟩ = 0$ for every rectangle, conjugate-linearity gives $⟨ k, R ⟩ = 0$ for every finite complex rectangle combination $R$ . For each $ε > 0$ choose such $R$ with $∥ k - R ∥_{2} < ε$ by step 3.1. Then $∥ k ∥_{2}^{2} = ∣ ⟨ k, k - R ⟩ ∣ \leq ∥ k ∥_{2} ε$ . If $∥ k ∥_{2} > 0$ , division and arbitrarily small $ε$ give a contradiction; thus $k = 0$ . When either factor has zero total measure, every class is zero and all assertions hold with $R = 0$ . No exchange of a universal test-function quantifier with an a.e. section quantifier is used.

F5step 3.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Square integrable kernels define bounded compact integral operators

Statement

Assume AC. Let $(X, A, μ)$ be a probability space and $k \in L^{2}$ of its completed square. The formula $K f (x) = \int k (x, y) f (y) d μ (y)$ , interpreted a.e., defines a representative-independent bounded compact linear operator on complex $L^{2} (μ)$ , with $∥ K ∥ \leq ∥ k ∥_{2}$ . It is an operator-norm limit of finite-rank rectangle-kernel operators. If $K = 0$ , then $k = 0$ as an $L^{2}$ class. For incomplete factors the integrals can first be computed with product-measurable representatives and then interpreted as classes on the original factors.

Facts & Assumptions

[F1]

Rectangle combinations are dense in the completed product $L^{2}$ , and zero pairing against every rectangle forces the zero class Product rectangle kernels are dense in complex l two.

[F2]

Completed-product Tonelli and Fubini apply to sigma-finite factors, with a.e. section assertions Tonelli and Fubini for the completed product, with only almost-everywhere section measurability.

[F3]

Cauchy–Schwarz holds for the complex $L^{2}$ pairing The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F4]

Complex $L^{2}$ is complete under countable choice Complex Lp completeness and almost-everywhere subsequences.

[F5]

Every bounded real sequence has a convergent subsequence Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence.

[F6]

Compactness means subsequential norm convergence on bounded sequences L two operator conventions for weak mixing.

[F7]

Assume AC The Axiom of Choice.

[F8]

Under countable choice, a real function measurable for a completed measure has a base-measurable almost-everywhere representative A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra.

[F9]

On an uncompleted sigma-finite product, Fubini gives integrable section-integral functions on the original factors after zero extension on exceptional parameter sets Fubini's theorem for L^1 functions on a sigma-finite product.

Proof

Given: The probability space, kernel and AC in the statement.

1.1

Apply F8 separately to the real and imaginary parts of k on the completed product. Replacing their infinite values on the resulting measurable null sets by zero and recombining gives a finite product-measurable representative of k. Tonelli on $∣ k ∣^{2}$ gives square-integrable sections for almost every x. For such x, Cauchy–Schwarz gives $∣ K f (x) ∣^{2} \leq (\int ∣ k (x, y) ∣^{2} d μ (y)) ∥ f ∥_{2}^{2}$ . Also the product-square function $f (y)$ has norm $∥ f ∥_{2}$ , since $μ (X) = 1$ ; hence $\iint ∣ k (x, y) f (y) ∣ \leq ∥ k ∥_{2} ∥ f ∥_{2} < \infty$ . Apply F9 to the product-measurable integrable function $k (x, y) f (y)$ , using an original-factor measurable representative of $f$ . It gives an original-factor measurable section integral after assigning zero on its measurable null exceptional parameter set. Integrating the squared inequality proves $∥ K f ∥_{2} \leq ∥ k ∥_{2} ∥ f ∥_{2}$ . Thus the output is a class on the original factor even when that factor is incomplete.

F2F3F7F8F9

2.1

Changing $k$ on a product-null set changes its sections only on factor-null sets for a.e. $x$ , by Tonelli on a measurable null cover. Changing $f$ on a factor-null set likewise leaves the integrals unchanged for a.e. $x$ . Thus $K$ is well-defined on classes; integral linearity gives complex linearity. Applied to $k - l$ , step 1.1 gives $∥ K_{k} - K_{l} ∥ \leq ∥ k - l ∥_{2}$ .

F2step 1.1

3.1

A rectangle kernel $\sum_{j = 1}^{m} a_{j} 1_{E_{j}} (x) 1_{F_{j}} (y)$ maps $f$ to $\sum_{j} a_{j} 1_{E_{j}} \int_{F_{j}} f$ , so its range lies in the span of finitely many indicators. Delete dependent vectors from this finite list. Successively subtract from each remaining vector its components along previous normalized vectors, then normalize the nonzero residual. Pairing expansion gives a finite orthonormal basis of that span. For a bounded sequence of images each basis coefficient is bounded by Cauchy–Schwarz. Apply real Bolzano–Weierstrass successively to their finitely many real and imaginary coordinates. The resulting common subsequence has all coordinates convergent, hence its finite basis sum converges in norm. In dimension zero every image is zero. Thus every rectangle-kernel operator is compact.

F3F5F6step 2.1

4.1

By F1 and AC choose rectangle kernels $k_{m}$ with $∥ k - k_{m} ∥_{2} < 2^{- m}$ . For a sequence $∥ f_{n} ∥ \leq B$ , successively extract nested subsequences whose $K_{k_{m}}$ images converge, using step 3.1 and AC. The diagonal subsequence, with strictly increasing original indices, is eventually a subsequence of every chosen one. For two sufficiently late diagonal terms $u, v$ , step 2.1 gives $∥ K u - K v ∥ \leq 2 B 2^{- m} + ∥ K_{k_{m}} u - K_{k_{m}} v ∥$ . First fix $m$ to make the first term small, then choose the two indices large to make the second small. The images are Cauchy and converge in $L^{2}$ by completeness. This proves compactness and the claimed finite-rank norm approximation; if $B = 0$ , all terms were zero already.

F1F4F7step 2.1step 3.1

5.1

Finally, if $K = 0$ , then for each measurable $E, F$ , Fubini gives $0 = ⟨ K 1_{F}, 1_{E} ⟩ = \int_{E \times F} k = ⟨ k, 1_{E \times F} ⟩$ . The zero-pairing conclusion of rectangle density gives $k = 0$ . Each test is a separate equality of integrals; no common exceptional set for all tests is required.

F1F2step 4.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Conjugate transpose kernels give adjoints

Statement

Assume AC. For a square-integrable kernel $k$ on a completed probability square, the adjoint of its operator $K$ is the kernel operator of $k^{*} (x, y) = \overline{k (y, x)}$ . Both are compact, $(K^{*})^{*} = K$ , and $∥ K^{*} ∥ = ∥ K ∥$ .

Facts & Assumptions

[F1]

Square-integrable kernels define bounded compact operators with operator norm at most kernel norm Square integrable kernels define bounded compact integral operators.

[F2]

Adjoint means the first-variable-linear pairing identity and is unique if it exists L two operator conventions for weak mixing.

[F3]

Completed-product Tonelli and Fubini give both iterated integrals Tonelli and Fubini for the completed product, with only almost-everywhere section measurability.

[F4]

The complex pairing is sesquilinear and satisfies Cauchy–Schwarz The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F5]

Assume AC The Axiom of Choice.

Proof

Given: $k$ , its operator $K$ , and AC as in the statement.

1.1

Factor swap is measurable on the product sigma-algebra because the inverse image of $E \times F$ is $F \times E$ . For a nonnegative product-measurable $q$ , Tonelli applied in both orders gives $\iint q (y, x) d μ (y) d μ (x) = \iint q (x, y) d μ (y) d μ (x)$ . In particular it preserves null sets. Thus it is measurable and measure-preserving on the completion as well: a completed measurable set differs from a product-measurable set by a subset of a product-null set, whose swapped set is still null. Consequently $k^{*}$ is a well-defined completed $L^{2}$ class with $∥ k^{*} ∥_{2} = ∥ k ∥_{2}$ . F1 gives its compact bounded operator $L$ . AC supplies the countable-choice hypotheses here and in F1.

F1F3F5

2.1

For $f, g \in L^{2}$ , Tonelli gives $∥ f (y) \overline{g (x)} ∥_{L^{2} (X^{2})} = ∥ f ∥_{2} ∥ g ∥_{2}$ . Product Cauchy–Schwarz bounds the absolute integral of $k (x, y) f (y) \overline{g (x)}$ by $∥ k ∥_{2} ∥ f ∥_{2} ∥ g ∥_{2}$ . Hence Fubini applies, and conjugating the inner integral gives $⟨ K f, g ⟩ = \int f (y) \overline{\int \overline{k (x, y)} g (x) d μ (x)} d μ (y) = ⟨ f, L g ⟩$ . Therefore $L = K^{*}$ by adjoint uniqueness. The a.e. section conventions are those of F1.

F2F3F4step 1.1

3.1

Applying the formula twice gives $(k^{*})^{*} = k$ and hence $(K^{*})^{*} = K$ . For every $z \in L^{2}$ , Cauchy–Schwarz gives $\sup_{∥ g ∥ \leq 1} ∣ ⟨ z, g ⟩ ∣ \leq ∥ z ∥$ ; equality follows by $g = z / ∥ z ∥$ if $z \neq 0$ , and both sides are zero if $z = 0$ . Thus $∥ K f ∥ = \sup_{∥ g ∥ \leq 1} ∣ ⟨ f, K^{*} g ⟩ ∣ \leq ∥ f ∥ ∥ K^{*} ∥$ . Taking the supremum over the unit ball gives $∥ K ∥ \leq ∥ K^{*} ∥$ . Apply this inequality to $K^{*}$ and use its double adjoint to obtain the reverse inequality. Compactness of both operators was supplied by F1 and step 1.1.

F1F4step 2.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Invariant square integrable kernel produces a compact intertwiner

Statement

Assume AC. Suppose $T$ preserves a completed Lebesgue probability space and $k \in L^{2} (X^{2})$ satisfies $k (T x, T y) = k (x, y)$ a.e. Its compact kernel operator satisfies $K U = U K$ and $K^{*} U = U K^{*}$ , where $U = U_{T}$ , even if $U$ is not surjective. If both marginal integrals of $k$ vanish a.e., then $K 1 = K^{*} 1 = 0$ , both operators preserve $H_{0}$ , and $k \neq 0$ implies $K ∣_{H_{0}} \neq 0$ .

Facts & Assumptions

[F1]

Kernel operators are compact and the kernel-to-operator map is bounded and injective Square integrable kernels define bounded compact integral operators.

[F2]

Conjugate-transpose kernels give adjoints Conjugate transpose kernels give adjoints.

[F3]

A linear isometry has the explicit adjoint $V$ with $V U = I$ Hilbert cesaro averages converge to the fixed subspace.

[F4]

Finite rectangle combinations are dense in product $L^{2}$ Product rectangle kernels are dense in complex l two.

[F5]

The local canonical-simple and monotone-convergence argument proves that Koopman pullback is an isometry on complex $L^{2}$ Eigenfunction for a probability system.

[F6]

Preservation on generating rectangles implies preservation on the product sigma-algebra Measure preservation can be checked on a generating pi-system.

[F7]

Completed-product Fubini applies with a.e. sections Tonelli and Fubini for the completed product, with only almost-everywhere section measurability.

[F8]

The centered space is $H_{0} = {f : \int f = 0}$ Eigenfunction for a probability system.

[F9]

Assume AC The Axiom of Choice.

Proof

Given: $T, k$ and AC as in the statement.

1.1

The preimage under $T \times T$ of $E \times F$ is $T^{- 1} E \times T^{- 1} F$ , of the same measure. These rectangles generate the product sigma-algebra and include the whole probability square, so F6 applies. Preimages of completed null subsets lie in the preimages of product-measurable null covers, hence are measurable and null in the completion. Thus $T \times T$ preserves the completed product. Its Koopman operator $W$ and the factor operator $U$ are isometries. Obtain $V = U^{*}$ and $V U = I$ from F3, under AC.

F3F5F6F9

2.1

For a rectangle tensor $l (x, y) = a (x) b (y)$ , the adjoint identity and its conjugate give $\int b (T y) f (y) d μ (y) = ⟨ f, U \overline{b} ⟩ = ⟨ V f, \overline{b} ⟩ = \int b (y) V f (y) d μ (y)$ . Multiplying by $a (T x)$ yields $K_{W l} = U K_{l} V$ . Finite linear combinations obey the same identity. For any $l \in L^{2}$ , approximate by such combinations in kernel norm; isometry of $W$ and F1 make both sides converge in operator norm. Hence the identity holds for every kernel.

F1F4step 1.1

3.1

Since $W k = k$ , step 2.1 gives $K = U K V$ , and multiplying on the right by $U$ gives $K U = U K$ . The swapped conjugate kernel obeys $W k^{*} = k^{*}$ too: F2 makes conjugate transpose a well-defined operation on completed $L^{2}$ kernel classes, algebraically $W (k^{*}) = (W k)^{*}$ , and $W k = k$ . Applying the same identity to $k^{*}$ yields $K^{*} U = U K^{*}$ . Compactness of both operators comes from F1–F2.

F1F2step 2.1

4.1

Vanishing $y$ -marginal gives $K 1 = 0$ . Vanishing $x$ -marginal gives $K^{*} 1 = 0$ after conjugation. For $f \in H_{0}$ , $⟨ K f, 1 ⟩ = ⟨ f, K^{*} 1 ⟩ = 0$ , and similarly for $K^{*}$ using its double adjoint. Thus both preserve $H_{0}$ . If $k \neq 0$ , F1 gives $K \neq 0$ . Every $f$ splits as $(\int f) 1 + f_{0}$ with $f_{0} \in H_{0}$ ; since $K 1 = 0$ , a vector with $K f \neq 0$ supplies $K f_{0} \neq 0$ . Therefore the restriction is nonzero. All marginal equalities are a.e. equalities of integrable sections, justified by F7.

F1F2F7F8step 3.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Nonzero compact kernel operators yield nonzero positive compact k star k

Statement

Assume AC. For a nonzero compact kernel operator $K$ with $K 1 = K^{*} 1 = 0$ , $S = K^{*} K$ on $H_{0}$ is bounded, compact, self-adjoint, positive and nonzero. If $K$ and $K^{*}$ commute with a Koopman isometry $U$ , then $S U = U S$ on $H_{0}$ .

Facts & Assumptions

[F1]

Kernel adjoints are bounded, satisfy the pairing identity, and have double adjoint $K$ Conjugate transpose kernels give adjoints.

[F2]

Positivity, self-adjointness and compactness use the local operator conventions L two operator conventions for weak mixing.

[F3]

The complex pairing is positive definite The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F4]

Assume AC The Axiom of Choice.

Proof

Given: $K$ and its two zero images of $1$ , under AC.

1.1

If $h \in H_{0}$ , the adjoint identities and the two zero images give $⟨ K h, 1 ⟩ = ⟨ h, K^{*} 1 ⟩ = 0$ and $⟨ K^{*} h, 1 ⟩ = ⟨ h, K 1 ⟩ = 0$ . Thus both operators preserve $H_{0}$ . Their restrictions include a nonzero K: choose f with $K f \neq 0$ and write $f = (\int f) 1 + f_{0}$ ; then $f_{0} \in H_{0}$ and $K f_{0} = K f \neq 0$ . Thus $S$ is a well-defined bounded endomorphism of $H_{0}$ , with $∥ S ∥ \leq ∥ K^{*} ∥ ∥ K ∥$ . AC supplies the inherited kernel results.

F1F3F4

2.1

For $f, g \in H_{0}$ , the two adjoint identities give $⟨ S f, g ⟩ = ⟨ K f, K g ⟩ = ⟨ f, S g ⟩$ . Also $⟨ S f, f ⟩ = ∥ K f ∥^{2} \geq 0$ . Therefore $S$ is self-adjoint and positive. For the $f_{0}$ in step 1.1, this quantity is strictly positive; hence $S f_{0} \neq 0$ and $S \neq 0$ .

F1F2F3step 1.1

3.1

For any bounded sequence in $H_{0}$ , compactness of $K$ gives a subsequence of its images convergent in $L^{2}$ . The limit lies in $H_{0}$ because it is closed. Applying the bounded, hence continuous, operator $K^{*}$ shows the corresponding $S$ images converge in $H_{0}$ . This is compactness of $S$ . A Koopman operator fixes $1$ ; since the isometry U preserves the pairing, $⟨ U h, 1 ⟩ = ⟨ U h, U 1 ⟩ = ⟨ h, 1 ⟩$ , so U preserves $H_{0}$ . If both factors commute with U, then $S U = K^{*} K U = K^{*} U K = U K^{*} K = U S$ on $H_{0}$ . No eigenvalue of K itself has been asserted.

F1F2step 2.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The positive norm eigenvalue of a nonzero positive compact self-adjoint operator has a nonzero finite-dimensional eigenspace

Statement

Assume AC. Let $H$ be a nonzero closed complex $L^{2}$ subspace and $S : H \to H$ a nonzero bounded positive self-adjoint compact operator. Then $α = ∥ S ∥ > 0$ is an eigenvalue, and $\ker (S - α I)$ is nonzero, closed and finite-dimensional.

Facts & Assumptions

[F1]

Operator norms, positivity, self-adjointness and sequential compactness have the local conventions L two operator conventions for weak mixing.

[F3]

The complex pairing is positive definite and satisfies Cauchy–Schwarz The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F4]

Assume AC The Axiom of Choice.

Proof

Given: $H, S$ as stated and AC.

1.1

Write $B (x, y) = ⟨ S x, y ⟩$ and $q (x) = B (x, x)$ . Self-adjointness makes $B$ Hermitian and positivity gives $q \geq 0$ . For $q (y) > 0$ , expand $q (x - t y)$ with $t = B (x, y) / q (y)$ to obtain $0 \leq q (x) - ∣ B (x, y) ∣^{2} / q (y)$ . If $q (y) = 0$ and $B (x, y) \neq 0$ , taking $t = R B (x, y)$ with arbitrarily large positive real $R$ makes $q (x - t y) = q (x) - 2 R ∣ B (x, y) ∣^{2} < 0$ , a contradiction. Thus in all cases $∣ B (x, y) ∣^{2} \leq q (x) q (y)$ .

F1F3

2.1

Put $a = \sup_{∥ x ∥ = 1} q (x)$ . This is finite and nonnegative, since $q (x) \leq ∥ S ∥$ on the nonempty unit sphere. For any $z$ , the norm formula $∥ z ∥ = \sup_{∥ y ∥ = 1} ∣ ⟨ z, y ⟩ ∣$ follows from Cauchy–Schwarz and testing $y = z / ∥ z ∥$ when $z \neq 0$ ; when $z = 0$ both sides vanish. Step 1.1 consequently gives $∥ S x ∥^{2} \leq a q (x)$ for all $x$ . On unit vectors this is at most $a^{2}$ , hence $∥ S ∥ \leq a$ . The reverse inequality follows from the definition of $a$ , so $a = ∥ S ∥ = α$ . If $a = 0$ , the displayed bound would force $S = 0$ , contrary to the hypothesis; thus $α > 0$ .

F1F3step 1.1

3.1

By AC choose unit vectors $x_{n}$ for every $n \in N$ with $q (x_{n}) > α - 1 / (n + 1)$ . Expansion, step 2.1, and self-adjointness yield $∥ S x_{n} - α x_{n} ∥^{2} = ∥ S x_{n} ∥^{2} - 2 α q (x_{n}) + α^{2} \leq α (α - q (x_{n})) < α / (n + 1)$ . Compactness gives a subsequence $S x_{n_{j}} \to y \in H$ . Thus $x_{n_{j}} \to x = y / α$ , and $∥ x ∥ = 1$ . Boundedness gives $S x_{n_{j}} \to S x$ , while $S x_{n_{j}} - α x_{n_{j}} \to 0$ ; uniqueness of limits gives $S x = α x$ .

F1F3F4step 2.1

4.1

Let $E = \ker (S - α I)$ . It is a linear subspace and is closed by continuity of $S - α I$ . It contains the unit vector from step 3.1. Suppose it has no finite spanning set. A choice function on the nonempty subsets of $E$ permits the following recursive selection: choose its value on the complement of the span of the finitely many previously obtained orthonormal vectors, subtract its projections onto them and normalize. The residual is nonzero because the chosen vector is not in that span. Pairing expansion shows that the resulting sequence $(e_{n})$ is orthonormal and lies in $E$ . Then for $n \neq m$ , $∥ S e_{n} - S e_{m} ∥ = α ∥ e_{n} - e_{m} ∥ = α \sqrt{2}$ . No subsequence of these images is Cauchy, contradicting compactness. Therefore $E$ has a finite spanning set and is finite-dimensional. The only infinite selections were the maximizing sequence and this hypothetical orthonormal recursion, both under AC.

F1F3F4step 3.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Compact intertwiners produce finite dimensional invariant subspaces

Statement

Assume AC. A nonzero invariant square-integrable kernel with zero marginals for a measure-preserving transformation of a completed Lebesgue probability space produces a nonzero finite-dimensional $U_{T}$ -invariant subspace $E$ of $H_{0}$ . The restriction $U_{T} ∣_{E}$ is unitary, even when $T$ is not invertible.

Facts & Assumptions

[F1]

The associated $S = K^{*} K$ is nonzero, positive, self-adjoint and compact on $H_{0}$ , and commutes with $U_{T}$ Nonzero compact kernel operators yield nonzero positive compact k star k.

[F2]

Such an $S$ has a nonzero finite-dimensional positive eigenspace The positive norm eigenvalue of a nonzero positive compact self-adjoint operator has a nonzero finite-dimensional eigenspace.

[F3]

Invariant zero-marginal kernels give commuting operators preserving $H_{0}$ Invariant square integrable kernel produces a compact intertwiner.

[F4]

Koopman is an isometry and $H_{0}$ is closed Eigenfunction for a probability system.

[F5]

A finite-dimensional linear map satisfies rank-nullity Rank-nullity: $\dim_{F} V = nullity T + rank T$ .

[F6]

Assume AC The Axiom of Choice.

Proof

Given: The kernel in the statement and AC.

1.1

F3 supplies the nonzero restriction of $K$ to the closed space $H_{0}$ , so this space is nonzero. Apply F1 there, then F2, to obtain $α > 0$ and the nonzero finite-dimensional subspace $E = \ker (S - α I) \subseteq H_{0}$ . AC supplies all inherited projection and compactness selections.

F1F2F3F4F6

2.1

For $f \in E$ , commutation gives $S (U_{T} f) = U_{T} (S f) = α U_{T} f$ , so $U_{T} f \in E$ . The restriction is injective because $U_{T}$ preserves norm. Rank-nullity gives image dimension equal to $\dim E$ , and a subspace of a finite-dimensional space with full dimension equals that space: a basis of a proper subspace could be enlarged by a vector outside it, contradicting the dimension. Thus the restriction is surjective. Being a surjective isometry, it is unitary.

F1F4F5step 1.1∎

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Nonzero finite dimensional complex invariant subspaces have unitary eigenvectors

Statement

Every nonzero finite-dimensional complex subspace $E$ invariant under a Koopman isometry $U$ contains a nonzero vector $f$ with $U f = λ f$ and $∣ λ ∣ = 1$ . If $E \subseteq H_{0}$ , this eigenfunction is nonconstant. The assertion for a supplied finite-dimensional $E$ does not require AC; a preceding construction of $E$ may carry that assumption.

Facts & Assumptions

[F1]

Eigenfunctions are nonzero $L^{2}$ classes, and $H_{0}$ is the zero-mean subspace of a probability space Eigenfunction for a probability system.

[F2]

An endomorphism of a nonzero finite-dimensional space over an algebraically closed field has an eigenvalue Every endomorphism of a nonzero finite-dimensional vector space over an algebraically closed field has an eigenvalue.

[F3]

Every nonconstant complex polynomial has a root Fundamental theorem of algebra: every nonconstant complex polynomial has a complex root.

[F4]

The $L^{2}$ norm is positive definite The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

Proof

Given: A nonzero finite-dimensional invariant complex subspace $E$ as stated.

1.1

Invariance makes $U ∣_{E} : E \to E$ a complex-linear endomorphism. The field $C$ satisfies the algebraic-closedness hypothesis of F2 by F3. Since $\dim E \geq 1$ , F2 yields $λ \in C$ and a nonzero $f \in E$ with $U f = λ f$ . This uses a single finite-dimensional eigenvalue assertion; it selects no infinite family of eigenvectors.

F2F3given

2.1

Isometry gives $∥ f ∥ = ∥ U f ∥ = ∣ λ ∣ ∥ f ∥$ . Since $f \neq 0$ , positivity permits division by $∥ f ∥$ , giving $∣ λ ∣ = 1$ . If also $E \subseteq H_{0}$ and $f = c 1$ , then $0 = \int f = c$ because the total measure is one. This would give $f = 0$ , impossible. Thus in that case $f$ is nonconstant. The case $\dim E = 1$ is included; $E = {0}$ is excluded before F2 is applied.

F1F4step 1.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Weak mixing is equivalent to absence of nonconstant eigenfunctions

Statement

Assume AC. For a measure-preserving transformation $T$ on a completed Lebesgue probability space the following are equivalent: (i) weak mixing in the absolute-Cesaro sense; (ii) no nonconstant complex $L^{2}$ eigenfunction; (iii) ergodicity of $T \times T$ on the completed product. Equivalently, for each pair $f, g \in L^{2}$ its centered correlations tend to zero outside a set of integers of natural density zero; the exceptional set may depend on the pair. Invertibility is not required.

Facts & Assumptions

[F1]

Eigenfunctions and the closed zero-mean space $H_{0}$ have the stated conventions Eigenfunction for a probability system.

[F2]

Cesaro averages of an isometry converge to the orthogonal projection onto its fixed space Hilbert cesaro averages converge to the fixed subspace.

[F4]

An invariant kernel gives compact intertwiners commuting with $U$ , and zero marginals give the stated nonzero restriction to $H_{0}$ Invariant square integrable kernel produces a compact intertwiner.

[F5]

A nonzero invariant kernel with zero marginals gives a nonzero finite-dimensional invariant subspace of $H_{0}$ Compact intertwiners produce finite dimensional invariant subspaces.

[F6]

Such a subspace contains a nonconstant eigenfunction Nonzero finite dimensional complex invariant subspaces have unitary eigenvectors.

[F7]

Ergodicity is equivalent to constancy a.e. of invariant measurable complex functions Equivalent invariant-set and invariant-function criteria for ergodicity.

[F8]

Weak mixing is equivalent to absolute-Cesaro convergence of centered complex $L^{2}$ correlations Mixing correlations extend to L2 functions.

[F9]

Weak mixing keeps the absolute value inside the average Strong and weak mixing on a probability space.

[F10]

Completed-product Fubini applies after checking integrability Tonelli and Fubini for the completed product, with only almost-everywhere section measurability.

[F11]

Assume AC The Axiom of Choice.

[F12]

On a finite measure space, preservation on a generating pi-system implies preservation on the generated sigma-algebra Measure preservation can be checked on a generating pi-system.

[F13]

The local canonical-simple and monotone-convergence argument proves that pullback by a measure-preserving map is a linear isometry on complex $L^{2}$ Eigenfunction for a probability system.

[F14]

Every set in a completion has the form $A \cup N$ , where $A$ is base measurable and $N$ is contained in a base null set The completion domain and proposed completed set function of a measure space.

[F15]

A square-integrable kernel defines a compact operator, and the zero operator has only the zero kernel Square integrable kernels define bounded compact integral operators.

Proof

Given: The completed Lebesgue probability system and AC.

1.1

If $f$ is a nonconstant eigenfunction, center it when its eigenvalue is one; when the eigenvalue differs from one its mean already vanishes. This gives a nonzero $f_{0} \in H_{0}$ with $U^{n} f_{0} = λ^{n} f_{0}$ and $∣ λ ∣ = 1$ . Thus $∣ ⟨ U^{n} f_{0}, f_{0} ⟩ ∣ = ∥ f_{0} ∥^{2} > 0$ for every $n$ . Its absolute-Cesaro average cannot tend to zero. F8–F9 therefore prove (i) implies (ii).

F1F8F9

1.2

On every measurable rectangle $E \times F$ , the product map $S = T \times T$ satisfies

(μ \times μ) (S^{- 1} (E \times F)) = μ (T^{- 1} E) μ (T^{- 1} F) = μ (E) μ (F) .

The inverse images of rectangles are measurable, so the sets whose inverse images are product measurable form a sigma-algebra; hence $S$ is measurable because rectangles generate the product sigma-algebra. Rectangles together with $X^{2}$ form a generating pi-system, and F12 makes $S$ measure preserving on the uncompleted product. If $A \cup N$ is completed measurable as in F14, with $N \subseteq Z$ and $(μ \times μ) (Z) = 0$ , then $S^{- 1} N \subseteq S^{- 1} Z$ and the latter is a base null set. Thus $S^{- 1} (A \cup N)$ is completed measurable and has the same completed measure as $A \cup N$ . Therefore $S$ preserves the completed product, and F13 makes its Koopman operator $W$ a complex $L^{2}$ isometry.

F12F13F14

2.1

Assume (ii). Every invariant indicator is an eigenfunction of eigenvalue one unless zero, so it is constant a.e.; its set is null or conull. Thus $T$ is ergodic. If $W k = k$ and $k$ is nonconstant, subtract its total mean to make $\iint k = 0$ , still with $k \neq 0$ . F15 gives a nonzero compact kernel operator $K$ , while F4 gives $K U = U K$ and $K^{*} U = U K^{*}$ . Since $U 1 = 1$ , the images $K 1$ and $K^{*} 1$ are invariant $L^{2}$ functions. Ergodicity makes both constant. Their means are $\iint k = 0$ and its conjugate, respectively, by Fubini; hence both are zero. These are exactly the two zero-marginal conditions. F5 and F6 would then produce a nonconstant eigenfunction, contradicting (ii). Every $W$ -fixed $L^{2}$ class is consequently constant. In particular each product-invariant indicator is constant, so (iii) follows. The kernel is integrable because its $L^{2}$ norm is finite and the product mass is one.

F1F4F5F6F7F10F11F15step 1.2

2.2

Assume (iii). If a nonconstant eigenfunction existed, center it as in step 1.1 to obtain $f_{0} \neq 0$ of mean zero. Put $F (x, y) = f_{0} (x) \overline{f_{0} (y)}$ . Tonelli gives $∥ F ∥_{2}^{2} = ∥ f_{0} ∥_{2}^{4} > 0$ , and Fubini gives $\iint F = ∣ \int f_{0} ∣^{2} = 0$ . Its pullback is $W F = ∣ λ ∣^{2} F = F$ ; factor-null exceptional sets pull back to null subsets of the square by the completed-product preservation in step 1.2. Product ergodicity and F7 force $F$ to be constant, and its zero mean forces that constant to be zero, a contradiction. Thus (iii) implies (ii).

F1F7F10step 1.1step 1.2

3.1

Under (iii), F7 says the fixed space of $W$ consists exactly of constants. For arbitrary $f, g \in L^{2}$ , let $f_{0} = f - (\int f) 1$ , $F = f_{0} \otimes \overline{f_{0}}$ and $G = g \otimes \overline{g}$ . As in step 2.2, these are in product $L^{2}$ and $F$ has integral zero. Its projection onto constants is therefore zero. Since step 1.2 proves that $W$ is an isometry, F2 yields $N^{- 1} \sum_{n < N} W^{n} F \to 0$ in norm, so its pairing with $G$ tends to zero by Cauchy–Schwarz. Fubini computes $⟨ W^{n} F, G ⟩ = ∣ ⟨ U^{n} f_{0}, g ⟩ ∣^{2}$ : each factor $f_{0} \circ T^{n} \overline{g}$ is integrable by Cauchy–Schwarz, and the absolute double integral is the square of its finite $L^{1}$ norm.

F1F2F7F10F11step 1.2step 2.2

4.1

Finite Cauchy–Schwarz on the $N$ real nonnegative numbers $b_{n} = ∣ ⟨ U^{n} f_{0}, g ⟩ ∣$ gives $N^{- 1} \sum_{n < N} b_{n} \leq (N^{- 1} \sum_{n < N} b_{n}^{2})^{1 / 2} \to 0$ . The correlation here equals $C_{n} (f, g)$ of F8 because $U 1 = 1$ . Thus F8 proves (i). Together with steps 1.1, 2.1 and 2.2 this closes all three implications. If $f_{0} = 0$ or $g = 0$ , both sides of the estimate are zero. The assumed AC supplies every cited projection, compactness and completion result that requires it; no spectral-measure construction is used.

F8F9step 1.1step 1.2step 2.1step 2.2step 3.1

5.1

For completeness let $a_{n} = ∣ C_{n} (f, g) ∣$ , bounded by $M = ∥ f - (\int f) 1 ∥_{2} ∥ g ∥_{2}$ . If its Cesaro mean tends to zero, each $E_{j} = {n : a_{n} > 1 / j}$ has density zero since $∣ E_{j} \cap [0, N) ∣ / N \leq j N^{- 1} \sum_{n < N} a_{n}$ . Choose $N_{j}$ recursively as the least integer larger than $N_{j - 1}$ such that this density is below $1 / j$ for every $N \geq N_{j}$ ; the convergence just proved ensures existence. Put $D = ⋃_{j \geq 1} (E_{j} \cap [N_{j}, N_{j + 1}))$ . For $N_{j} \leq N < N_{j + 1}$ , the increasing property $E_{i} \subseteq E_{j}$ for $i \leq j$ implies $D \cap [0, N) \subseteq E_{j} \cap [0, N)$ , so $D$ has density zero. Off $D$ on its $j$ th interval, $a_{n} \leq 1 / j$ , proving the claimed convergence. Conversely, if $D$ has density zero and $a_{n} \to 0$ off $D$ , choose $N_{0}$ so that $a_{n} < ε$ off $D$ for $n \geq N_{0}$ . Then $N^{- 1} \sum_{n < N} a_{n} \leq M N_{0} / N + M ∣ D \cap [0, N) ∣ / N + ε$ . Its limsup is at most every positive $ε$ , hence zero. This proves the pairwise density-zero equivalent criterion without asserting one exceptional set for all uncountably many pairs. Least integer cutoffs add no choice use.

F8step 4.1∎

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)Open item page →

Chacon three cut one spacer towers

Definition

Work on $[0, 1)$ with the Lebesgue measurable sets and restricted set function introduced in Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ . Under the countable-choice consequence of The Axiom of Choice, Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume makes this a complete measure space. The same assumption supplies the countable choice in A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included used for the actual interval measures below. The finite interval recursion itself is choice-free.

The stage-zero tower is the ordered list consisting of $L_{0, 0} = [0, 2 / 3)$ ; its reservoir is $R_{0} = [2 / 3, 1)$ . Suppose the stage- $r$ list is $(L_{r, 0}, \dots, L_{r, h_{r} - 1})$ , with equal width $w_{r}$ . Cut each physical half-open interval $L_{r, j}$ into its left, middle and right thirds $L_{r, j}^{(0)}, L_{r, j}^{(1)}, L_{r, j}^{(2)}$ . The next ordered list is $(L_{r, 0}^{(0)}, \dots, L_{r, h_{r} - 1}^{(0)}, L_{r, 0}^{(1)}, \dots, L_{r, h_{r} - 1}^{(1)}, J_{r}, L_{r, 0}^{(2)}, \dots, L_{r, h_{r} - 1}^{(2)}),$ where $J_{r} = [1 - 3^{- (r + 1)}, 1 - 3^{- (r + 2)})$ is taken from the left end of $R_{r}$ and retained as a new level. Put $R_{r + 1} = [1 - 3^{- (r + 2)}, 1)$ .

The height and width obey $h_{0} = 1$ , $h_{r + 1} = 3 h_{r} + 1$ and $w_{r} = 2 / 3^{r + 1}$ . Indeed the initial width is $2 / 3$ ; taking thirds divides it by three, and $∣ J_{r} ∣ = 3^{- (r + 1)} - 3^{- (r + 2)} = 2 / 3^{r + 2} = w_{r + 1}$ . The spacer and the remaining reservoir partition the previous reservoir; all old levels partition into their thirds. Thus by finite induction the new intervals are pairwise disjoint and, together with $R_{r + 1}$ , partition $[0, 1)$ .

Induction also gives $h_{r} = (3^{r + 1} - 1) / 2$ : the initial value is one, and $3 (3^{r + 1} - 1) / 2 + 1 = (3^{r + 2} - 1) / 2$ . Consequently the tower union $C_{r} = ⋃_{j < h_{r}} L_{r, j}$ has measure $h_{r} w_{r} = 1 - 3^{- (r + 1)}$ , and the reservoir has measure $3^{- (r + 1)}$ . Every interval is left-closed and right-open; no endpoint belongs to two levels.

Define $T_{r}$ on $C_{r} ∖ L_{r, h_{r} - 1}$ by the unique translation taking $L_{r, j}$ to $L_{r, j + 1}$ . If their left endpoints are $a_{j}, a_{j + 1}$ , that formula is $T_{r} (x) = x + a_{j + 1} - a_{j}$ on $L_{r, j}$ . At stage zero this is the empty partial map. Its range is $C_{r} ∖ L_{r, 0}$ . The existence of an invertible limiting probability transformation is a separate lemma, not part of the finite definition.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Chacon partial maps extend to an invertible map mod null sets

Statement

Assume AC. The partial translations of the normalized Chacon towers determine an invertible Lebesgue-probability-preserving transformation modulo null sets. There is a measurable conull $X_{0} \subseteq [0, 1)$ on which both directions are everywhere defined and measurable and $T (X_{0}) = X_{0}$ . Extending by the identity off $X_{0}$ gives an ambient measure-preserving map.

Facts & Assumptions

[F1]

The towers and consecutive-level translations are defined with height $h_{r}$ and width $w_{r} = 2 / 3^{r + 1}$ Chacon three cut one spacer towers.

[F2]

Translation preserves Lebesgue measurability and measure Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation.

[F3]

Countable unions of measurable null sets are null Finite and countable subadditivity of measures.

[F4]

Increasing measurable unions have measure equal to the supremum Continuity from below for measures.

[F5]

Invertibility modulo null sets means an actual measurable invariant conull restriction with measurable inverse Invertible measure-preserving systems.

[F6]

Assume AC The Axiom of Choice.

Proof

Given: The finite Chacon towers under AC.

1.1

Put $D_{r} = C_{r} ∖ L_{r, h_{r} - 1}$ and $E_{r} = C_{r} ∖ L_{r, 0}$ . On each of its finitely many levels $T_{r}$ is a translation to the next level of the same width. Thus it is a measurable measure-preserving bijection $D_{r} \to E_{r}$ with measurable inverse. At stage zero both sets are empty. At the next stage each old non-top arrow restricts to the three corresponding third-to-third arrows; the remaining new arrows connect column tops to the next bases or spacer. Thus $D_{r} \subseteq D_{r + 1}$ , $E_{r} \subseteq E_{r + 1}$ and $T_{r + 1}$ extends $T_{r}$ , as do their inverses.

F1F2F6

2.1

The complement of either $D_{r}$ or $E_{r}$ has measure $3^{- (r + 1)} + w_{r} = 3^{- r}$ . Hence $D = ⋃_{r} D_{r}$ and $E = ⋃_{r} E_{r}$ are conull by continuity from below. Compatible unions give a bijection $T : D \to E$ . Partition $D$ into the measurable pieces $D_{r} ∖ D_{r - 1}$ (with $D_{- 1} = \emptyset$ ), subdivided by the finitely many level pieces of $T_{r}$ . On each it is a translation; the images are disjoint because the union map is injective. Countable additivity and F2 therefore prove that images and preimages of measurable sets are measurable and have the same measure in the two domains. This also proves measurability of both directions.

F2F4step 1.1

3.1

Define $B_{0} = [0, 1) ∖ (D \cap E)$ and recursively $B_{n + 1} = B_{n} \cup T (B_{n} \cap D) \cup T^{- 1} (B_{n} \cap E)$ . Each $B_{n}$ is measurable and null by step 2.1 and induction. Thus $B = ⋃_{n \geq 0} B_{n}$ is measurable and null. Put $X_{0} = [0, 1) ∖ B \subseteq D \cap E$ . If $x \in X_{0}$ and $T x \in B_{n}$ , then $x \in T^{- 1} (B_{n} \cap E) \subseteq B_{n + 1}$ , impossible. The analogous implication using $T (B_{n} \cap D)$ shows $T^{- 1} x \in X_{0}$ . Hence both directions preserve $X_{0}$ and restrict to measurable inverse bijections there.

F3step 2.1

4.1

On $X_{0}$ measure preservation is inherited from step 2.1. Define the ambient map to be the identity on its measurable null complement. This map and its inverse are measurable by the two-piece definition; preimages differ from their $X_{0}$ preimages only by null subsets of that complement, so it preserves Lebesgue probability. It satisfies exactly F5's conull restriction convention. AC is used through the finite-tower measure assertions and hence the Lebesgue measure properties, with no selection of arbitrary pointwise inverses.

F5F6step 3.1∎

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Chacon levels approximate measurable sets

Statement

Assume AC. Every Lebesgue-measurable $E \subseteq [0, 1)$ can be approximated in symmetric-difference measure by unions of levels of the stage- $r$ Chacon tower, with error tending to zero as $r \to \infty$ . Functions constant on those levels and zero off their tower are dense in complex $L^{2}$ . If $μ (E) > 0$ and $δ > 0$ , every sufficiently late stage has a level $J$ with $μ (E \cap J) > (1 - δ) μ (J)$ .

Facts & Assumptions

[F1]

The physical tower levels and reservoir partition $[0, 1)$ , refining at each stage, with widths and reservoir mass tending to zero Chacon three cut one spacer towers.

[F2]

Finite-measure sets admit symmetric-difference approximation by a generating algebra Approximation in symmetric difference by a generating algebra.

[F3]

Half-open intervals generate the Borel sigma-algebra The sigma-algebra generated by the half-open boxes of $R^{n}$ is the Borel sigma-algebra.

[F4]

Lebesgue measurable sets have Borel representatives modulo null sets under countable choice $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets.

[F5]

Finite complex simple functions are dense in $L^{2}$ Complex finite-simple and smooth compact-support density for finite p.

[F6]

The complex pairing supplies the norm inequalities The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz.

[F7]

Assume AC The Axiom of Choice.

Proof

Given: $E \subseteq [0, 1)$ measurable and AC.

1.1

The algebra of finite unions of intervals in $[0, 1)$ with any endpoint conventions generates the trace Borel sigma-algebra by F3. Under AC, F4 replaces $E$ by a Borel set modulo a null set. Given $ε > 0$ , F2 supplies a finite interval union $V$ with $μ (E △ V) < ε / 2$ . Let $q$ count its finitely many endpoints. The physical partition at stage $r$ has maximum atom length $b_{r} = \max (w_{r}, 3^{- (r + 1)}) \to 0$ . Outside the at most $2 q$ atoms incident to endpoints, every atom is wholly inside or outside $V$ . Taking all atoms wholly inside $V$ therefore gives symmetric-difference error at most $2 q b_{r}$ (endpoint singletons are null). Removing the reservoir adds at most $3^{- (r + 1)}$ . Hence a level union $Q_{r}$ satisfies $μ (E △ Q_{r}) \leq ε / 2 + 2 q b_{r} + 3^{- (r + 1)} < ε$ for all sufficiently large $r$ .

F1F2F3F4F7

2.1

Given $f \in L^{2}$ and $ε > 0$ , choose $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ with $∥ f - s ∥_{2} < ε / 2$ . If $B = \sum ∣ c_{j} ∣ = 0$ , use zero. Otherwise approximate each $E_{j}$ by a level union at one common sufficiently late stage so that every error measure is below $(ε / (2 B))^{2}$ , using step 1.1. Then $s_{r} = \sum_{j} c_{j} 1_{Q_{r, j}}$ is constant on each level and zero off the tower, and $∥ s - s_{r} ∥_{2} \leq \sum_{j} ∣ c_{j} ∣ \sqrt{μ (E_{j} △ Q_{r, j})} < ε / 2$ . Thus $∥ f - s_{r} ∥_{2} < ε$ .

F5F6step 1.1

3.1

For $μ (E) > 0$ take $0 < η < δ μ (E) / (1 + δ)$ and choose, at any sufficiently late stage, a level union $Q$ with $μ (E △ Q) < η$ . Then $μ (Q) > μ (E) - η > 0$ . If every level of $Q$ had $E$ -proportion at most $1 - δ$ , summing over its disjoint levels would give $μ (Q ∖ E) \geq δ μ (Q) > δ (μ (E) - η) > η$ , contradicting the error bound. At least one level has the required strict density. The case $δ \geq 1$ is also covered by this contradiction, and only finitely many levels are compared.

F1step 1.1∎

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Chacon transformation is ergodic

Statement

Assume AC. The normalized Chacon probability transformation is ergodic.

Facts & Assumptions

[F1]

Chacon is an invertible probability transformation agreeing on an invariant conull set with all finite partial translations Chacon partial maps extend to an invertible map mod null sets.

[F2]

Any positive-measure set has arbitrarily late levels of proportion exceeding $1 - δ$ for each $δ > 0$ ; the towers exhaust measure one Chacon levels approximate measurable sets.

[F3]

Ergodicity tests strictly invariant measurable sets Ergodicity relative to an invariant measure.

[F4]

Assume AC The Axiom of Choice.

[F5]

At each stage r, the tower is an ordered list of $h_{r}$ equal-width levels and its partial map translates each level to the next Chacon three cut one spacer towers.

Proof

Given: A strictly invariant measurable set $E$ for Chacon with $μ (E) > 0$ .

1.1

Fix $0 < δ < 1$ . At every sufficiently late stage r, F2 supplies a level J with $μ (E \cap J) > (1 - δ) w_{r}$ . Repeatedly composing F5's consecutive partial translations shows that the finite partial map sends level j to level k after $k - j$ iterates whenever $j \leq k < h_{r}$ . F1 makes the limiting T agree with these arrows on its invariant conull set. Strict invariance implies equality of the measures of E in these levels, since the iterates preserve measure and membership in E. Removing the fixed null complement does not affect these equalities. Hence every level of this tower has E-measure greater than $(1 - δ) w_{r}$ .

F1F2F4F5

2.1

Summing over the disjoint levels gives $μ (E) \geq μ (E \cap C_{r}) > (1 - δ) μ (C_{r})$ . Letting $r \to \infty$ gives $μ (E) \geq 1 - δ$ because $μ (C_{r}) \to 1$ . Since every $0 < δ < 1$ is allowed and $μ (E) \leq 1$ , $μ (E) = 1$ . Sets of zero measure already satisfy the alternative. This is ergodicity by F3. AC is inherited from the measure construction and generating-level approximation, with only one finite-stage level needed at a time.

F3F4step 1.1∎

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Chacon eigenfunctions are constant

Statement

Assume AC. Every complex $L^{2}$ eigenfunction of the Chacon transformation is constant almost everywhere; its eigenvalue is one.

Facts & Assumptions

[F1]

An eigenfunction is a nonzero class and its eigenvalue has modulus one Eigenfunction for a probability system.

[F2]

Chacon is ergodic Chacon transformation is ergodic.

[F3]

Positive-measure sets have levels of arbitrarily high relative density at late stages Chacon levels approximate measurable sets.

[F4]

On an invariant conull set the limiting map agrees with all finite tower arrows Chacon partial maps extend to an invertible map mod null sets.

[F5]

Finite-valued measurable invariant real or complex functions on an ergodic probability system are constant a.e. Equivalent invariant-set and invariant-function criteria for ergodicity.

[F6]

Assume AC The Axiom of Choice.

[F7]

At stage r, the levels have common width $w_{r}$ and height $h_{r}$ ; stage r+1 lists all left thirds, then all middle thirds, then the spacer, then all right thirds, and its partial map translates each listed level to its successor Chacon three cut one spacer towers.

Proof

Given: $f \neq 0$ with $f \circ T = λ f$ a.e.

1.1

By F1, $∣ λ ∣ = 1$ , so $∣ f ∣ \circ T = ∣ f ∣$ a.e. Choose a finite-valued measurable representative of the $L^{2}$ class by setting it to zero on its null exceptional set. F2–F5 make $∣ f ∣$ a constant $c$ a.e. Nonzeroness forces $c > 0$ . Divide by $c$ , so henceforth $∣ f ∣ = 1$ a.e. For each positive integer $k$ , iteration gives $f (T^{k} x) = λ^{k} f (x)$ outside the finite union of preimages of the original exceptional null set. F4's measure preservation makes that union null. These relations may therefore be used for either finite return time below.

F1F2F4F5F6

2.1

Fix $ε > 0$ and $0 < δ < 1 / 6$ . Cover the unit circle by finitely many open disks of radius $ε$ with centers on the circle: equally spaced arguments with spacing less than $ε$ suffice, using $∣ e^{i t} - e^{i s} ∣ \leq ∣ t - s ∣$ . Since $∣ f ∣ = 1$ a.e., at least one disk centered at $a$ , $∣ a ∣ = 1$ , has positive-measure inverse image $E = {x : ∣ f (x) - a ∣ < ε}$ . By F3 choose a level $J = L_{r, j}$ with $μ (J ∖ E) < δ w_{r}$ . F7's next-stage ordering places $J^{(1)}$ exactly $h_{r}$ levels after $J^{(0)}$ and $J^{(2)}$ exactly $h_{r} + 1$ levels after $J^{(1)}$ , because the latter route crosses the one spacer. Together with F4 this gives $T^{h_{r}} : J^{(0)} \to J^{(1)}$ and $T^{h_{r} + 1} : J^{(1)} \to J^{(2)}$ as measure-preserving translations on the invariant conull set.

F3F4F7step 1.1

3.1

For the first route, the set of points $x \in J^{(0)}$ for which either $x \notin E$ or $T^{h_{r}} x \notin E$ has measure at most $2 δ w_{r}$ . Thus a set of measure at least $(1 / 3 - 2 δ) w_{r} > 0$ satisfies both memberships and the eigenfunction iterate relation. At one such point, $∣ λ^{h_{r}} a - a ∣ \leq ∣ λ^{h_{r}} (a - f (x)) ∣ + ∣ f (T^{h_{r}} x) - a ∣ < 2 ε$ . The same argument on $J^{(1)}$ with return time $h_{r} + 1$ gives $∣ λ^{h_{r} + 1} - 1 ∣ < 2 ε$ . Null exceptions from step 1.1 and the conull tower convention do not change positive measure.

F1F4step 1.1step 2.1

4.1

Since $∣ λ ∣ = 1$ , $∣ λ - 1 ∣ = ∣ λ^{h_{r} + 1} - λ^{h_{r}} ∣ \leq ∣ λ^{h_{r} + 1} - 1 ∣ + ∣ λ^{h_{r}} - 1 ∣ < 4 ε$ . Every positive $ε$ is allowed, so $λ = 1$ . Now F5 makes $f$ constant a.e., and undoing the normalization preserves constancy. AC is inherited from the tower and ergodicity inputs; the disk and positive-measure witnesses require only finite choices for each fixed epsilon.

F5F6step 3.1∎

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Chacon tower height correlations obstruct mixing

Statement

Assume AC. For the fixed set $A = L_{1, 0} = [0, 2 / 9)$ and every $r \geq 1$ , $μ (A \cap T^{- h_{r}} A) \geq 2 / 27 > 4 / 81 = μ (A)^{2}$ . Consequently Chacon is not strongly mixing. More generally, every measurable $E \subseteq [0, 1)$ satisfies ${lim inf}_{r \to \infty} μ (E \cap T^{- h_{r}} E) \geq μ (E) / 3$ .

Facts & Assumptions

[F1]

The next Chacon tower stacks left thirds, middle thirds, one spacer and right thirds in that order, with $h_{r} = (3^{r + 1} - 1) / 2$ Chacon three cut one spacer towers.

[F2]

The limiting probability transformation agrees with finite tower arrows off a fixed null set Chacon partial maps extend to an invertible map mod null sets.

[F3]

Every measurable set has tower-level-union approximants with symmetric-difference error tending to zero Chacon levels approximate measurable sets.

[F4]

Strong mixing requires convergence of every fixed set-pair correlation to the product of the measures Strong and weak mixing on a probability space.

[F5]

Assume AC The Axiom of Choice.

Proof

Given: The normalized Chacon towers and their transformation under AC.

1.1

If $Q$ is any union of levels at stage $r$ , let $Q^{(0)}$ be the union of their left thirds. These disjoint thirds have total measure $μ (Q) / 3$ . F1's ordering and F2 show $T^{h_{r}} Q^{(0)}$ is the union of the corresponding middle thirds, modulo the fixed null set. Both unions lie in $Q$ , so $Q^{(0)} \subseteq Q \cap T^{- h_{r}} Q$ modulo null sets, and $μ (Q \cap T^{- h_{r}} Q) \geq μ (Q) / 3$ .

F1F2F5

2.1

The stage-one base is the left third of $[0, 2 / 3)$ , hence $A = [0, 2 / 9)$ with measure $2 / 9$ . At every later stage it is exactly the union of all its descendant levels, since each old level partitions into three retained thirds. Step 1.1 gives the bound $μ (A) / 3 = 2 / 27$ for every $r \geq 1$ . But $μ (A)^{2} = 4 / 81$ and $2 / 27 - 4 / 81 = 2 / 81 > 0$ . The heights $h_{r} \to \infty$ by F1. Thus this one fixed pair $(A, A)$ fails F4's limit, proving failure of strong mixing.

F1F4step 1.1

3.1

For measurable $E$ , F3 gives stage- $r$ level unions $Q_{r}$ with $η_{r} = μ (E △ Q_{r}) \to 0$ . The symmetric difference of $E \cap T^{- h_{r}} E$ and $Q_{r} \cap T^{- h_{r}} Q_{r}$ lies in $(E △ Q_{r}) \cup T^{- h_{r}} (E △ Q_{r})$ . Measure preservation bounds its measure by $2 η_{r}$ , while $μ (Q_{r}) \geq μ (E) - η_{r}$ . Step 1.1 therefore yields $μ (E \cap T^{- h_{r}} E) \geq μ (E) / 3 - (7 / 3) η_{r}$ . Taking the liminf proves the general assertion. AC is inherited from F1–F3 and permits the countable choice of approximants; the estimate remains valid for null or conull $E$ without division.

F2F3F5step 1.1∎

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-5.6-terra)Open item page →

Chacon transformation is weakly mixing but not mixing

Statement

Assume AC. There exists an invertible completed Lebesgue probability system, the normalized three-cut one-spacer Chacon transformation, that is weakly mixing in the absolute-Cesaro sense but is not strongly mixing.

Facts & Assumptions

[F1]

The Chacon partial maps extend to an invertible completed Lebesgue probability system Chacon partial maps extend to an invertible map mod null sets.

[F2]

Its complex $L^{2}$ eigenfunctions are constant Chacon eigenfunctions are constant.

[F3]

On such a probability space absence of nonconstant complex $L^{2}$ eigenfunctions is equivalent to absolute-Cesaro weak mixing Weak mixing is equivalent to absence of nonconstant eigenfunctions.

[F4]

The fixed set $A = [0, 2 / 9)$ has correlations along $h_{r}$ at least $2 / 27$ , exceeding $μ (A)^{2} = 4 / 81$ Chacon tower height correlations obstruct mixing.

[F5]

Assume AC The Axiom of Choice.

Proof

Given: AC and the normalized Chacon construction.

1.1

Take the ambient transformation furnished by F1, or its invariant conull restriction. It preserves completed Lebesgue probability and has a measurable inverse modulo null sets, meeting all F3 hypotheses. F2 excludes every nonconstant complex $L^{2}$ eigenfunction, so F3 gives weak mixing with the absolute value inside the Cesaro average.

F1F2F3F5

2.1

In the same system, F4 gives a fixed measurable $A$ of measure $2 / 9$ and unbounded heights $h_{r}$ with $μ (A \cap T^{- h_{r}} A) - μ (A)^{2} \geq 2 / 81 > 0$ . These correlations cannot tend to zero, so this system is not strongly mixing. The null-set modification of F1 leaves these measures unchanged. Together with step 1.1 this supplies the claimed witness, with AC inherited from every local construction and spectral input.

F4F5step 1.1∎

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Weak mixing implies strong mixing

Statement

False claim: Every weakly mixing probability-preserving transformation is strongly mixing.

Facts & Assumptions

[F1]

Under AC, normalized Chacon is an invertible completed Lebesgue probability system that is weakly mixing but not strongly mixing Chacon transformation is weakly mixing but not mixing.

[F2]

Weak mixing uses absolute-Cesaro correlations, whereas strong mixing requires pointwise convergence in time for each fixed measurable pair Strong and weak mixing on a probability space.

[F3]

Assume AC The Axiom of Choice.

[F4]

The fixed Chacon set $A = [0, 2 / 9)$ satisfies $d_{h_{r}} (A, A) \geq 2 / 81$ for every $r \geq 1$ Chacon tower height correlations obstruct mixing, and $h_{r} = (3^{r + 1} - 1) / 2 \to \infty$ Chacon three cut one spacer towers.

Refutation

Given: AC.

1.1

Use the single probability system of F1. Its weak-mixing conclusion is precisely the premise in F2.

F1F2F3

2.1

In the system of step 1.1 the fixed pair $A = B = [0, 2 / 9)$ satisfies $d_{h_{r}} (A, A) \geq 2 / 81 > 0$ for all $r \geq 1$ by F4. Since $h_{r} \to \infty$ , these correlations do not converge to zero. Thus the conclusion of the false claim fails while its hypothesis holds. The example assumes AC exactly as F1 does; it does not infer pointwise convergence from an average.

F1F2F3F4step 1.1∎

5 · Examples, counterexamples and false statements

None yet.