Rokhlin’s multiple-mixing problem for one transformation

Proof. Let k be the least failing order and choose sets witnessing failure along layouts whose least consecutive gap tends to infinity. Passing to a subsequence makes the discrepancy from the product of the means have a fixed sign and absolute value at least some \varepsilon>0. Mixing at every smaller order holds for bounded functions, by approximation with simple functions.

Let \alpha be the finite partition generated by the witnessing sets and put \alpha_i=T^{-i}\alpha. We work inside the completed factor generated by (\alpha_i)_{i\in\Z} and define its past tail by

All the factor \sigma-fields in this proof are understood with their probability completions. This \sigma-field is invariant under T and T^{-1}. Write g_i for the witnessing indicators and \bar g_i=\E[g_i\mid\mathcal C]. Replacing g_i\circ T^{d_i(b)} by \bar g_i\circ T^{d_i(b)}, successively from i=k down to i=1, changes the product integral by a quantity tending to zero. Indeed, after translating the time of the factor being replaced to zero, all earlier factors are measurable with respect to

and all later, already replaced factors are \mathcal C-measurable. Their product has absolute value at most one. The error at this step is therefore at most

which tends to zero by the reverse martingale theorem. The same argument applies to the first factor, whose remaining factors are all in the tail. Conditional expectation preserves the means, so the tail functions still witness failure along a subsequence.

We next verify that every finite \mathcal C-measurable partition \beta has entropy rate zero. For a finite partition \gamma, write \gamma_i=T^{-i}\gamma. Stationarity and the entropy chain rule give

For completeness, the increments in the chain rule are H(\gamma_0\mid\gamma_{-j},\ldots,\gamma_{-1}); they decrease to the right side, since the conditional probabilities of the finitely many atoms converge and entropy is a bounded continuous function on their probability simplex. Their Cesàro averages give the asserted limit.

For any \epsilon>0, there is M such that

Consequently the chain rule and monotonicity under conditioning give

After division by n, this proves h(\alpha\vee\beta)\leq h(\alpha)+\epsilon. The reverse inequality follows from refinement, and hence h(\alpha\vee\beta)=h(\alpha).

On the other hand, every translate of \beta belongs to the invariant tail \mathcal C and hence to the past of \alpha before each fixed time i. Conditioning on \beta_1,\ldots,\beta_n therefore does not add information beyond \sigma(\alpha_j:j<i) in the lower bound below:

Combining this with

and taking rates proves h(\beta)=0.

Choose, once and for all, finite-valued \mathcal C-measurable approximations h_i to the \bar g_i, with values in [0,1]. They can be made sufficiently close in L^1 to preserve a fixed positive discrepancy along the selected layouts: the change in either a product integral or the product of the means is bounded by the sum of the L^1 errors. Let \beta be their common finite partition. Coding the \beta-name gives a factor on \mathcal A^{\Z}, with its induced measure \mu_X. This is a standard probability space, its alphabet entropy rate is h(\beta)=0, and its shift inherits all mixing properties of the original transformation. The h_i become one-site functions.

Center these functions. Expanding their centered product, every proper subproduct tends to the product of its means, because its ordered sublayout still has all consecutive gaps tending to infinity and contains fewer than k times. Thus the centered product differs by o(1) from the original discrepancy. Each centered function has supremum norm at most one. A change of sign in the first function, followed by discarding finitely many layouts, gives (1).□

Proof. We first check closure under sequential inverse limits. For a sequence of extensions, take the local state and parameter spaces to be the spaces of compatible paths, with their consistent probability laws. They are standard Borel spaces. At every array level the consistent finite-stage laws determine a law on compatible array paths. Time acts coordinatewise. Projection to X, consistency under fixing any axis, and unconditional last-axis independence pass to the limit by testing finite-stage functions. Independence of the local X-state and the limiting parameter follows in the same way.

For the conditional requirement, fix a proper family J of last-axis planes and bounded functions F_j on those planes, all pulled from one finite stage. Let \mathcal P_n^{(s)} denote the full parameter array at stage s, and \mathcal P_{n-1,j}^{(s)} its restriction to plane j. For every sufficiently large s, the defining property gives

Increasing martingale convergence on both sides gives the same identity for the limiting parameter fields. The factors are uniformly bounded, so their products converge in L^2. This proves both independence and dependence of each plane law only on its own parameters, first for finite-stage tests and then, by density, for all bounded tests.

For a fixed test and data type, the norm of the indicated conditional expectation is nondecreasing under extension and bounded above by the norm of the test. At each constructed stage, choose countable dense families of bounded tests in every L^2(m_n), and list all pairs of a test and one of the finitely many data types at that level. Construct successive extensions as follows. At round q, process the first q listed pairs from each of the first q stages already constructed. For each pair choose an extension whose projection norm is within 2^{-q} of the supremum over extensions of the current stage. There is no requirement to attain the supremum. Append these finitely many extensions and continue. Every listed pair is processed infinitely often with error tending to zero. Taking the sequential inverse limit gives an object M_\infty in the class.

Let Z be any extension of M_\infty, and fix a test listed at some finite stage. At any sufficiently late processing step, Z is also an extension of the current stage. Its projection norm is consequently at most that step’s supremum, which is at most the chosen next-stage norm plus 2^{-q}. The next-stage norm is at most the limiting norm. Letting q\to\infty proves that Z cannot increase the norm for this test. Finite-stage functions are dense in the inverse-limit L^2 spaces, and conditional expectation is a contraction. The same conclusion follows for every F\in L^2(m_n) of M_\infty.

Finally, the pullback of \mathcal D is contained in \mathcal D'. The Pythagorean identity for these nested orthogonal projections is

The equality of norms therefore gives (4).□

Proof. The tower property shows that K_j(E_j,\cdot) is also the conditional law of O_j given E_j. Let M_j(x,\cdot) be its image under o\mapsto a_j(x,o). Disintegration gives, for almost every x with respect to the law of E_j,

Substituting x=e_j(E) preserves these almost-sure identities. The joint law of (E,O,A,B) thus factors as

The first three factors give the law of (E,A,B), proving (5).□

Proof. We prove the statement by induction on |R|, simultaneously for all array levels and all projection slots a\notin R. Saturation will be used only for extensions already supplied by smaller parent sets. The resulting screening identity and Lemma 3.3 will then verify the conditional product law for the larger parent set. Write an entry of the old level n+1 as y_{c,w}, where c\in I is the internal axis and w\in I^n comprises the remaining axes. The proposed projection retains y_{a,w}. It preserves every array law by axis consistency, preserves the X-state through \kappa(y_a), and recovers the old parameter from (6). Time invariance, consistency, and unconditional last-axis independence for the candidate follow directly from the old level n+1.

At the local level, condition on (u(y_c))_{c\in I} in m_1. The set R\cup\{a\} is proper. Its states are conditionally independent, with the law at a conditioned only on u(y_a). Since the X-image of that conditional law is \mu_X, the selected X-state still has law \mu_X after conditioning further on (y_c)_{c\in R}. This verifies the required independence from the enlarged parameter. If R=\varnothing, the remaining conditional last-axis requirement is exactly the old condition at level n+1, so the base case is proved.

Now suppose R\ne\varnothing and that the lemma holds for smaller parent sets. Fix n\geq1 and a missing last-axis slot j_0. In the old law m_{n+1} let

For j\ne j_0, let O_j be the full plane at last slot j, E_j its old parameter array, and A_j its states at internal indices c\in R. Finally, let

be the parent states in the missing plane. Conditional on E, the O_j are independent, with each plane law depending only on E_j. We claim that

Enumerate R and reveal its missing-plane slices one internal index c at a time. Let R' be the set of indices already revealed. It has smaller cardinality than R. The induction hypothesis supplies a valid extension with parent set R' and projection slot c. Apply saturation to a bounded test F of the missing-plane slice (y_{c,w}:w_n=j_0), viewed as a function on the projected old level n. For the data type \mathcal G_{n,j_0}, the old data are the parameters of the row at c and its states outside the missing plane; the enlarged data are E, all of O, and the already revealed missing parent slices. More explicitly, if B_{R'} denotes those already revealed missing slices, write A_j^{R'} for the states of O_j at internal indices in R'. This smaller-parent extension has \mathcal P'_n=\sigma(E,(A_j^{R'})_{j\ne j_0},B_{R'}), \mathcal O'_{n,j_0}=\sigma(O), and \mathcal G'_{n,j_0}=\sigma(E,O,B_{R'}). Hence (4) says exactly that

The right side is the old m_n conditional kernel at the fixed internal index c. It involves only E and the part of O belonging to that same c, and it does not depend on previously revealed missing states. Multiplication of these successive kernels proves (7).

Apply Lemma 3.3 to the proper family of planes O_j. Conditioning on E, all the A_j, and B preserves their independence and conditions each plane only on (E_j,A_j). These are precisely the complete enlarged parameters and each plane’s own enlarged parameters, respectively. The law of such a plane is the old m_n, now viewed as the candidate level n-1; its conditional law given (E_j,A_j) is therefore exactly the one required by Definition 3.1. We have proved the requirement for k-1 planes. Marginalization proves it for every proper family.

There are only countably many levels and finitely many choices of parents and slots at each induction step. Testing on countable determining classes makes all the kernel equalities simultaneous. The induction is complete.□

Proof. Column independence in (1) is the defining conditional independence at level two. To obtain rows, use the extension in Lemma 3.4 with parent set R and projection slot a\notin R. For a bounded test F of the entire projected row, saturation for the full parameter data \mathcal P_1 gives

Reveal the members of any proper row family in turn. The previous members number at most k-2, so (14) applies at every step. Their joint conditional law is the product of the indicated row laws.

Assertion (2) is the conditional last-axis requirement at level one of the extension with parents R. Its local law is m_1 on each column, and its own parameter there is (q_j,Y_{R,j}), giving precisely the stated marginal kernel. For (3), use the same extension and saturation for \mathcal G_{1,j_0}. The projected old data are p_a and the other entries of row a, while the extension data are E, the parents, and all columns except j_0. This gives (12).

We prove the grouped assertions without interchanging array axes. Each row group is a proper set of rows, so its full law is the product of its individual row laws. Within each individual row, any proper set of column coordinates is independent. Consequently any two grouped cells in that grouped row are independent. The identical conclusion for a grouped column follows by applying the corresponding column assertions. Since |P|\leq k-2, (2) applies to the parents P. The union of any two column groups is proper, which gives the claimed conditional independence of grouped columns.

It remains to prove (13). We first handle a single missing column j_0. Given E, Y_{P,I}, and all other columns, reveal y_{a,j_0} successively for a\in A. At each step the parents consist of P and the previously revealed rows in A. Their number is at most

All their remaining entries are already known from the observed columns. Equation (12) therefore gives the kernel of the individual row at that step. These kernels do not depend on the previously revealed entries in column j_0, and hence

Now choose j_0\in t, put J=t\setminus\{j_0\} and C=r\cup s, and retain W=Y_{A,J} as part of the output. The set C\cup J consists of exactly k-1 columns. By (2), conditional on E and Y_{P,I}, those columns are independent. In each of them, P\cup A is a proper row set. Thus the A-entries, even after conditioning on its P-entries, have the plain product marginal \bigotimes_{a\in A}\eta_{u_{a,j}}. It follows that

Use (15) and then integrate away the additional states in columns J, retaining W. Its right side depends on those columns only through W, so the resulting joint kernel for (W,Y_{A,j_0}) is the measure in (16), followed by the product of the individual row kernels in (15).

Under the row-group law \Lambda_A, the coordinates in C\cup J are also independent with exactly those marginals: the rows in A are independent and each row has independent proper coordinates. Consequently the just obtained kernel is exactly the conditional law of Y_{A,t} under \Lambda_A given Y_{A,C}. This proves (13), including the case J=\varnothing. Only cardinalities of the groups were used, so every ordering of the groups is allowed. All choices are finite; the earlier convention on kernel versions makes the assertions simultaneous.□

Proof. It is enough to prove (18) for J=I\setminus\{j_0\}, and then take marginals. Write

Use the extension with no parents and projection slot a. Saturation for the data \mathcal O_{1,j_0} says, for every bounded row test F,

Since S is a function of both Z and O, this is conditional independence of Z and O given S. One direct verification is to multiply (19) by a bounded function of O and condition on S; the resulting factorization is the conditional-independence identity. It also gives the reversed screening statement \Law(O\mid Z)=\Law(O\mid S).

Unconditionally the columns in O are independent, each with law m_1. Conditioning on their separate shared states therefore gives

We identify the parameter image of each factor. In a single column, the joint law of its parameter tuple q and its state y_a at slot a is

Disintegrating \pi at q[a] rewrites this as

Since u(y_a)=u almost surely under \eta_u, the conditional law of q given the shared state y_a is exactly K_a(u(y_a),\dd q). Thus

The right side is a function only of p_a, which itself is a function of Z. Conditioning down to p_a proves (18).

For the last assertion, take a=i and retain just j=i. The tuple p_i has marginal \pi, and its coordinate at i is u_{i,i}. Given p_i, the law of q_i is K_i(p_i[i],\cdot). Disintegrating once more over that coordinate shows that

as required.□

Proof. For an exact tuple put a_i=\abs{\int\chi_i\dd\mu_i}. Solving the product relation for \chi_i and using independence of the remaining coordinates gives

If some a_i=0, at least one other a_j is zero, and the same equations then force all a_j to be zero. Otherwise put b_i=-\log a_i. The equations give 2b_i=\sum_j b_j for every i; since m\ge3, all b_i vanish. Thus either all means have modulus one, in which case all the functions are constant almost surely, or all means are zero.

The quotient of two exact tuples is again exact. If their lines differ at one coordinate, its quotient function is nonconstant, so all quotient means vanish. In particular the two representatives are orthogonal at that coordinate. If their lines agree at one coordinate, the quotient there is constant, forcing constant quotients everywhere. This proves uniqueness of partners. Products and conjugates preserve exact tuples; uniqueness makes multiplication of lines well defined and makes each coordinate projection bijective. A family of nonzero mutually orthogonal lines in a separable Hilbert space is countable. Comparing any exact tuple with the constant tuple gives the final assertion.□

Proof. Condition on z,c,d,x,y. The marginal law of the two remaining entries in column s is the column law given c, and similarly column t has its law given d. Indeed, given row P, each of these columns is independent of column r. Denote the corresponding two joint L^2 spaces by \mathcal H_s and \mathcal H_t. Let Q:\mathcal H_t\to\mathcal H_s be the conditional Markov operator between them under this five-cell conditioning, and let \iota_{A,s},\iota_{B,s},\iota_{A,t},\iota_{B,t} be the isometric embeddings of the individual cell spaces into the appropriate joint spaces.

The own-row kernel assertion gives

The same assertion with s and t reversed gives

Also J_s=\iota_{A,s}^*\iota_{B,s} and J_t=\iota_{A,t}^*\iota_{B,t}. Consequently

Countable dense sets of tests make these operator equalities simultaneous on a conull set. The grouped locality assertions hold for all the indicated orderings, so the same proof applies to each of them.□

Proof. First, the essential closed range span of L_A(x) is V_{A,s}. Conditional expectations of product tests onto Y_{A,s} have the form

where w,h are bounded. Such tests span the active space by density of product functions in the L^2 space of the other two coordinates. A vector orthogonal to every integral in (24) is orthogonal to L_A(x)h almost everywhere, by varying w. A countable dense set of h then makes it orthogonal to the whole range for almost every x. The converse is immediate. This proves the assertion, and the same argument applies to every incident operator in either direction. In particular each such operator has its range in its own-direction active space and annihilates the orthogonal complement of the source active space.

By the first rectangle identity, J_s(c)L_B(y) has range in V_{A,s}. The pair (c,y) is independent. Fubini and the essential range span of L_B therefore give J_s(c)V_{B,s}\subseteq V_{A,s} for almost every c. The adjoint of the second identity is J_s(c)^*L_A(x)=L_B(y)J_t(d)^*. Independence of (c,x) gives the reverse inclusion for the adjoint. Together these say

Likewise, the first rectangle identity and independence of (x,d) show that L_A(x)W_{A,t}\subseteq W_{A,s}; the second identity and independence of (x,c) give the adjoint assertion. Thus

The same reasoning works at every edge.

Equation (25) shows that P^V_{A,s} preserves the essential range span of J_s, which is W_{A,s}. As an orthogonal projection it also preserves W_{A,s}^{\perp}, and therefore commutes with P^W_{A,s}. This proves commutation in every cell. Combining (26) with the own-row active-space reduction gives P^D_{A,s}L_A=L_AP^D_{A,t}, and similarly for all other edges. Finally, projecting the essential range span V_{A,s} onto D_{A,s} shows that the restricted ranges of L_A span D_{A,s}: P^D_{A,s}\Ran L_A=\Ran(L_AP^D_{A,t}). The corresponding argument handles each edge and its adjoint.□

Proof. The two rectangle equations and their adjoints imply

The parameter triples in these four equations are, respectively, (c,x,d), (c,y,d), (x,d,y) and (x,c,y). Each has product law by the independence observations preceding Lemma 4.3.

For the first equation, compare two independent values c,c' using the same x,d. Fubini gives

for almost every (c,c',x). The essential range span of F is its whole target D space, so the two target squares agree. The adjoint equation and the essential range span of F^* similarly make NN^* fixed. The second equation in (28) makes K^*K and N^*N fixed. The two equations in (29), using the range spans of N,K and their adjoints, make F^*F,G^*G,FF^*,GG^* fixed. These comparisons are operator comparisons: countable dense vector tests first give a common conull set, after which boundedness gives equality on every vector.

If v is in the kernel of the fixed square F^*F, then F(x)v=0 for almost every x. Hence v is orthogonal to the ranges of almost every F(x)^*, whose essential span is the whole source D. Thus v=0. The same reasoning proves injectivity of all source and target squares. Each operator consequently has zero kernel and dense range, so its polar part has full initial and final spaces and is unitary.

To preserve the rectangle identity, put A=|F| and B=|G|. From F^*FN=NG^*G we obtain A^2N=NB^2. Polynomial approximation to the square-root function on [0,1] gives AN=NB. Write F=U_FA and G=U_GB. The equation FN=KG becomes

The range of the injective self-adjoint B is dense, so the bounded operator in parentheses vanishes. If N=U_N|N| and K=U_K|K|, then

are polar decompositions of the same operator. Uniqueness yields U_FU_N=U_KU_G. This argument uses no inverse bound for the positive parts. Finally, polar parts are strongly measurable on vectors, for example by the bounded approximations F(F^*F+\varepsilon\id)^{-1/2} as \varepsilon\downarrow0.□

Proof. Give the unitary spaces the strong operator topology. On unitaries, inversion and multiplication are continuous, and these spaces are separable metrizable. Conditional on a typical fixed value z_0, the pair (a,b) and the pair (c_0,d_0) are independent, while each individual marginal equals its unconditional marginal. The continuous identity (30) therefore holds throughout the product of the two conditional pair supports. Choose one point (a_*,b_*) from the first support and one point (c_*,d_*) from the second. They satisfy a_*d_*=c_*b_*. Identify the four vertex spaces by these reference unitaries, starting at the lower right vertex. The two possible identifications of the upper left vertex agree by this equation, so all four reference edge values become identity. Use these fixed identifications from now on.

The conditional pair supports at z_0 now contain (\id,\id). Fixing the column pair at this point in (30) forces a=b throughout the other pair support. Fixing the row pair similarly forces c_0=d_0. The two row marginals therefore have a common law \rho and the two column marginals a common law \tau; these are also their unconditional laws. The supports of \rho and \tau both contain identity. Substitution of diagonal pair values into (30) shows that these two supports commute elementwise.

For general z, conditional independence of the two pairs and their fixed individual marginals imply

Indeed, after fixing z,c_0,d_0, the equation says a=c_0bd_0^{-1}, and a,b each have law \rho. Unconditionally, c_0,d_0 are independent, so their pair law is \tau\otimes\tau. The map (u,v)\mapsto u\rho v^{-1} is continuous for weak convergence of probability measures: apply strong-topology continuity and dominated convergence to each bounded continuous test. Thus (32) holds for every u,v\in\supp\tau. The symmetric argument gives \tau=u\tau v^{-1} for every u,v\in\supp\rho.

Taking v=\id shows that each support is preserved by left translation by elements of the other. Because identity belongs to both, they contain one another. Write their common closed support as \mathcal G. For g\in\mathcal G we have g\mathcal G=\mathcal G. This proves closure under products, and \id\in g\mathcal G supplies g^{-1}\in\mathcal G. Thus \mathcal G is a group. It is abelian by the earlier commutation, and (32) gives left invariance of \rho. Equation (30) now gives equality of the two ratios in (31). Given z, the ratios are independent and equal almost surely, hence have a common point-mass conditional law. A countable determining family in the standard unitary group makes its point measurable in z, giving h.

Choose a positive injective trace-class operator Q on \mathcal H and form the trace-norm integral

The integrand is trace-norm continuous under strong-unitary convergence, as follows first for finite-rank Q and then by trace-norm approximation. The integral is positive and trace-class, and left invariance of \rho makes it commute with every element of \mathcal G. For v\ne0 each quadratic form \ip{v}{UQU^*v} is positive; hence so is \ip{v}{Q_0v}. Thus Q_0 is injective. Its finite-dimensional positive eigenspaces span \mathcal H. Each is invariant under the commuting unitaries, so simultaneous diagonalization in each gives a full orthonormal eigenbasis (e_n).

Write Ue_n=\gamma_n(U)e_n, with \gamma_n(U)\in\T. These coordinates are continuous characters. On diagonal unitaries, coordinatewise convergence is equivalent to strong convergence, by approximation of any vector by its finite basis truncations. The coordinates therefore generate the Borel information of \mathcal G. Trigonometric polynomials on every finite coordinate torus are dense in L^2 of its actual probability law. Increasing conditional expectations onto the finite-coordinate fields then show that the functions

span L^2(\sigma(h(z))). No identification of the law of h as Haar measure is needed for this density statement. Finally, (31) gives

Each factor is unimodular. Multiplying by the conjugate factors shows that \gamma(h(z)) is exact both in column r and in row P.□

Proof. Since (Y_s,Y_t) has product marginal, the bilinear contraction is well defined on H_s\times H_t and satisfies

For bounded inputs let L_{r\leftarrow t}(v) be the row Markov operator from t to r given Y_s=v. Pairwise independence gives the vector-valued integral formula

Lemma 4.4 says P_rL_{r\leftarrow t}(v)=L_{r\leftarrow t}(v)P_t. Moving the projection through the integral proves P_rC_r(g,h)=C_r(g,P_th). Using instead the edge from s to r conditioned on Y_t proves P_rC_r(g,h)=C_r(P_sg,h). The bound (35) extends both identities to all L^2 inputs, including unbounded projected tests. Idempotence then gives

proving (33). Restricted contractions are fixed by P_r and hence lie in D_r. Unrestricted product contractions span the row-active space; projecting that space onto D_r proves the restricted spanning assertion. The column proof is identical.

Apply this result first in row A. For g\in D_{A,s} and h\in D_{A,t}, its contraction onto x=Y_{A,r} is

Conjugation preserves D_{A,s}. On the restricted spaces, L_A(x)=a(x)A_0 for a fixed positive operator A_0 by Lemma 4.5. Thus every displayed coefficient is a function of a(x) alone. The space L^2(\sigma(a(x))) is closed, being the range of a conditional-expectation projection, so the spanning assertion gives

Apply the column version of the spanning assertion in column r. Given z, we have a=h(z)b and b has its fixed marginal law \rho. For bounded measurable f,g the contraction of f(a) and g(b) onto z is therefore

a function of h(z). The product marginal of (x,y), the bound (35), and bounded approximation extend this conclusion to the two D spaces and their contraction span. Closure therefore yields D_{P,r}\subseteq L^2(\sigma(h(z))). In the other direction, the monomials of Lemma 4.6 are exact for both laws, hence lie in D_{P,r}, and they span L^2(\sigma(h(z))). This proves (34). We have used no assumption that D was already closed under multiplication. Permuting row groups and column groups places any grouped cell at the corner and proves the last assertion.□

Proof. Isolate row a and column j as singleton groups, and split each of their complements into two nonempty groups. At the isolated cell, the row-active and column-active spaces do not depend on the chosen splits: the other two grouped cells still contain exactly all the other entries of that row or column. Lemma 4.4 gives commutation, and Lemma 4.7 gives a spanning family of simultaneous exact lines for the two split three-cell laws. We take this family to contain all such lines: each is in the intersection, and orthogonality leaves no additional line outside the spanning family.

Fix the row-group partition and vary the column-group partition. Every resulting family is a subset of the mutually orthogonal exact lines of the same fixed split column law, by Lemma 4.2. All have the same closed span V_{a,j}\cap W_{a,j}. Two subsets of one orthogonal family with the same closed span must coincide: a line belonging only to one would be orthogonal to the other span. Now fix the initial column grouping and vary the row grouping. The same argument, using the fixed split row law, shows that the initial family is exact for every split in both directions.

For k=3 the grouped laws already have individual coordinates. For k>3, fix a unimodular representative \chi_j of one of the lines and consider its row completions. For each split S of I\setminus\{j\}, there are unimodular functions a_S,b_S and c_S\in\T with

Absorb c_S^{-1} into a_S. All the resulting completion functions then equal \overline{\chi_j(Y_{a,j})} under the row law, so they define the same function \phi of the other coordinates. Those k-1 coordinates have product marginal. For each singleton split, \phi lies in a one-dimensional line in that coordinate tensored with the remaining L^2 spaces. The orthogonal projections to these individual line factors commute and all fix \phi. Their product has the full one-dimensional tensor product as range, so \phi is a pure product of its unimodular individual factors, up to one scalar. This proves exactness for the full row law. The column argument is the same. Conversely, every line exact for both full laws is active for both and belongs to their intersection. This proves part (1).

For part (2), make the chosen row a singleton group, the chosen columns s,t singleton groups, and put all remaining columns in the third group. Split the remaining rows into two nonempty groups. The intertwining in Lemma 4.4 is exactly the stated identity for this grouping. To obtain the column assertion, isolate the two chosen rows instead and apply the already proved column intertwining; there is again no interchange of axes.

Finally, solving the exact product relation for \chi_t and conditioning gives (36). Its scalar has modulus one. The projections P^W commute with conjugation. Thus the intertwining in part (2) carries membership of \C\chi_t in W_{a,t} to membership of \C\chi_s in W_{a,s}. The reverse row Markov operator gives the converse. The same proof applies to column partners.□

Proof. Fix i. By Lemma 3.6, the row and column tuples meeting at cell (i,i) are conditionally independent with law

given their common entry u. Proposition 4.8 therefore gives, for almost every such pair (p,q),

Here the almost-everywhere assertion does not presuppose a measurable description of C_i. Indeed, choose a measurable conull set of parameter squares on which the conditional kernel hypotheses of Proposition 4.8 hold on countable determining classes. That Proposition gives its pointwise conclusion on every square in this set. Disintegrate this set over the two tuples meeting at (i,i). On a measurable conull set of pairs the conditional measure assigns probability one to admissible squares, and is concentrated on the prescribed pair. Each such pair has an admissible completion. The conclusions in (38) depend only on the two tuple laws, so they hold for that pair. No measurable choice of a completion or of an exact line is required. Disintegration over u then gives conull sections for the product law \pi_i^u\otimes\pi_i^u.

For a typical u define the positive contraction on H_u

The weak operator integral is well defined, and its matrix entries are measurable in u. Since 0\le B_{i,u}\le\id, Q_{i,u} is the strong limit of (\id-B_{i,u})^n and is measurable too. In the fixed ambient \ell^2 coordinates, use the projection of each standard basis vector e_j into H_u and extend the operators by zero on H_u^\perp. Then

After a countable intersection, \Pi_i(p)Q_{i,u}=0 for almost every p. Taking adjoints gives

This range containment is an operator assertion on a measurable conull set: it is tested by Q_{i,u}\Pi_i(p)e_j=0 for all j.

Fix p satisfying (40) and having a conull q-section in (38). If v\in\Ran\Pi_i(p) is orthogonal to C_i(p), commutation implies that \Pi_i(q)v belongs to the intersection in (38). Consequently

Thus \ip{v}{B_{i,u}v}=0, and v\in\Ker B_{i,u}. Equation (40) forces v=0. The conull q-section used here gives whole-subspace statements, so it works for every such v; a measurable selection of v is unnecessary. Every exact line is active, by its defining product relation. This proves (37). There are finitely many slots i.□

Proof. We first enumerate the exact lines over tuple parameters, then show that these enumerations describe a family depending only on a local entry. The distinction is necessary: Lemma 5.1 alone gives no such independence from the other parameters.

Enumerating the tuple-dependent lines. Fix i and another slot j\ne i. Let

be conditional expectation onto all the other coordinates under \lambda_p. Their joint marginal is product, so the displayed target is their L^2 space. This is a measurable contraction field. It is the adjoint of conditional expectation in the other direction, and hence vanishes on the inactive complement. For a unit representative of an exact line, \Delta_i(p) gives, up to a scalar phase, the unit tensor of the conjugated matched representatives in the other slots.

Choose dense unit-ball test sections g_n(u) in each H_u. If \theta_g h=g\ip{g}{h}, set

All their matrix entries are measurable before any exact-line enumeration. Pointwise, Lemma 5.1 and Lemma 4.2 give an orthogonal exact-line basis of the active space. Write w_{j,\ell} for the unit partner appearing at j in \Delta_i(p)\chi_\ell. The operators in (41) are diagonal on this basis, with eigenvalues

Indeed, for two distinct labels the partners are orthogonal in every other slot. Since k\ge3, at least one slot outside i,j is untouched by the insertion, and its inner product kills the off-diagonal entry. The partners at j are orthonormal, so Bessel’s inequality gives

Dense tests detect each line and distinguish any two different lines: the corresponding rank-one projections at j are different.

Here is a measurable spectral construction which does not assume that the basis just used pointwise is already enumerated measurably. Choose independent continuously distributed weights w_n\in(0,2^{-n}) and form

The series converges in operator norm and is jointly measurable. It is positive and has trace at most \sum_n w_n on the conull set where the preceding arguments apply. Trace finiteness is itself measurable, being the finiteness of the sum of nonnegative diagonal entries in the fixed coordinate space. Set the operator to zero on its null exceptional set, so that the ensuing recursion always acts on a positive compact operator.

For such an operator A, its norm is measurable from a countable dense set of unit vectors. If a=\norm A>0, the strong limit

is its top-eigenvalue projection. Its matrix entries and rank are measurable. Remove aQ and repeat, taking zero projections if the remainder is zero. Compactness ensures that this lists every distinct positive eigenvalue and its finite-dimensional eigenspace. Thus a repeated positive eigenvalue is detected by the measurable event that some nonzero projection in the recursion has rank at least two.

For a fixed p, two distinct exact lines have different coefficient sequences (a_n(\ell))_n. Conditioning on all but a weight where they differ shows that their weighted eigenvalues coincide with probability zero. There are only countably many lines in the separable Hilbert space. Fubini applied to the measurable collision event therefore chooses one deterministic sequence of weights for which the positive spectrum is simple for almost every p. Every active line has a strictly positive eigenvalue, because some test detects it and all weights are positive. The zero eigenspace is precisely the inactive complement; eigenvalues tending to zero add no kernel vectors. The positive spectral projections therefore enumerate exactly the exact lines. Zero padding of the Hilbert coordinates supplies no positive eigenvectors and hence no extra lines.

Choose a unit vector in each one-dimensional projection by projecting the first coordinate vector with nonzero image and normalizing. These measurable Hilbert sections have jointly measurable function representatives. One direct verification is to approximate a section in each fibre within 2^{-m} by the first suitable member of a countable family of rational combinations of simple sections. The choices are measurable. The successive differences are summable in the integrated L^2 norm, and hence in L^1, giving a measurable almost-everywhere limit that represents the section. Each unit vector on an exact line has modulus one, so its representative does also. For a listed tuple of such representatives, their product Z is constant under \lambda_p exactly when

This measurable scalar test enumerates the unique matched tuples and is unchanged by all choices of scalar phases.

Forcing all column memberships. Return to a typical parameter square. Fix a matched tuple along row a, with unit representatives v_j in its cells. At cell (a,j) let Q_j denote the column-active projection \Pi_a(q_j). Proposition 4.8 and orthogonality of distinct row exact lines show that

Here \varepsilon_j=1 precisely when this row line is also an exact line of the column law. The row Markov operator between two cells, given the remaining cells, intertwines their column-active projections. The exact relation maps one chosen representative to the conjugate partner times a modulus-one scalar depending on the remaining cells. Since the projections commute with conjugation, this intertwining and (44) imply \varepsilon_j=\varepsilon_t for every j,t.

Choose the row tuple measurably from the enumeration just obtained. Given p_a, each \varepsilon_j depends only on q_j and this fixed tuple. Lemma 3.6 makes these indicators pairwise conditionally independent. Their common conditional probability r(p_a) therefore satisfies r(p_a)=r(p_a)^2: indeed, writing r_j(p_a)=\E[\varepsilon_j\mid p_a], the almost-sure equality of the indicators makes all r_j equal, while conditional independence gives r_j=r_jr_t for j\ne t. At the matched column j=a, however,

The strict inequality follows from (40), since v_a is a nonzero active vector. Thus r(p_a)=1. Countably many row labels and finitely many cells make the resulting inclusions simultaneous: every row exact line is a column exact line at its cell, almost surely in the parameter square.

Descent to the local parameter. At (i,i) two conditionally iid families of exact lines over the shared parameter u satisfy inclusion almost surely. Swapping the two copies gives reverse inclusion, so the families agree. To descend this agreement measurably, for each local dense test g_n(u) encode a family \mathcal S of orthogonal unit lines by

The sum is independent of phases and order, has trace at most one, and is measurable for our tuple-dependent enumeration. For every matrix entry, the two conditional iid encodings are equal almost surely. Their conditional squared difference is twice the conditional variance, so that variance is zero. Each entry equals its conditional expectation given u. The encodings consequently give measurable operator fields depending only on u. The family of encodings detects and distinguishes its orthogonal lines, so the same positive-weight spectral construction recovers a measurable family \mathcal L_i(u) agreeing with the slot-i exact lines for almost every tuple with that entry.

At a nonmatched cell (a,j) the already proved inclusion now reads \mathcal L_j(u_{a,j})\subseteq\mathcal L_a(u_{a,j}). The marginal of u_{a,j} is \nu. Applying this to every ordered pair a,j yields one common family \mathcal L(u) almost everywhere. We have now removed the dependence on the other entries of a tuple: the remaining construction can be made on the local parameter space.

By Lemma 4.2, multiplication of representatives modulo scalars makes each family an abelian group. Orthogonality and the zero-mean assertion for nonidentity labels follow from the same Lemma. Equality of unit lines is tested by \abs{\ip v w}=1. Testing the product of two representatives against the countable list therefore makes multiplication measurable; conjugation gives the inverse. The constant line is measurably identifiable and we choose its representative to be 1. Equation (43) and uniqueness give the measurable group isomorphisms \Phi_{st}(p).

The equivariant versions of \eta_u and \lambda_p show that composition with T_Y^{-1} carries exact tuples to exact tuples. It consequently induces \tau_u and respects all matchings. Equality tests against the measurable line lists show that this map and its inverse are measurable. Intersecting local conull sets over all integer translates makes the local assertions time invariant. Tuple and square assertions are imposed separately on their own conull sets, also intersected over integer time translates; nothing asserts validity for every tuple over the local conull set.

Consistency around the square. Start with a matched label tuple along a row and extend every one of its labels along its column by exact matching. Choose representatives for the resulting labels at all cells. The product down each column is constant under the conditional square law, as is the product along the initial row. Dividing the product of the column relations by that row relation shows that the product of the remaining k-1 row products is constant. Those rows are independent as wholes by Proposition 3.5. Each row product has modulus one. The absolute mean of their product is one, so each has absolute mean one and is itself constant. Thus every row is matched. Starting from a single cell, its row and then all columns are uniquely determined; the preceding argument supplies consistency along the other rows. Countability of the label lists makes this simultaneous for every initial label. This proves the final assertion.□

Proof. For a nonnegative measurable function F on Z^I, define

Unique exact matching makes this independent of the chosen source slot s. Define the square measure in the same way, counting its unique matched arrays from one chosen cell. The last assertion of Proposition 5.2 makes this independent of the chosen cell. Counting from a coordinate, row, or column gives the claimed marginals. The preimages of the first N labels of any one coordinate have measure at most N and exhaust the lifted spaces. Thus the measures are sigma-finite.

Disintegrating (48) first over p_s=u gives (46). For each prescribed source label there is exactly one completion over every compatible tuple; fixing that label introduces no weight into the parameter kernel. This proves directly that the conditional kernels are probability kernels even when the total lifted measure is infinite. Similarly, given a labelled row of the square, its label has exactly one extension over every compatible parameter square. Its conditional parameter-square law is therefore the original law given that row’s parameter tuple. Lemma 3.6 gives independent proper columns; their labels are then determined separately by their shared labels and their own parameter tuples. This proves the last assertion of the Lemma.

Write \widetilde\pi_{st} for the (s,t) marginal. Its two marginals are \xi, and its kernels in the two directions are K_{st} and K_{ts}. Jensen’s inequality and the marginal identities give L^2 contraction and M_{st}^*=M_{ts}. Time equivariance of exact matching preserves both lifted measures and gives VM_{st}=M_{st}V.

Finally, test the (s,t) pair in row b of the lifted square with f,g\in L^2(\xi). Conditional on labelled row a, its two column transitions are independent applications of M_{ab}, since two columns form a proper family when k\ge3. Hence

These integrals are absolutely integrable by Cauchy–Schwarz with the pair marginals \xi. Alternatively one may first use bounded tests of finite-measure support and then pass in L^2. The calculation proves (47) without normalizing the sigma-finite measures.□

Proof. Fix s\ne t and write M=M_{st}. Taking (a,b)=(s,t) and (t,s) in (47) gives

Let \mathcal H_1=\overline{\Ran M} and \mathcal H_2=\overline{\Ran M^*}. The first identity puts \mathcal H_1 in the fixed space of M^*M, which lies in \mathcal H_2. The adjoint of the second identity puts \mathcal H_2 in the fixed space of MM^*, which lies in \mathcal H_1. They are therefore equal to a closed space \mathcal H. On it M^*M=MM^*=\id, while M vanishes on \mathcal H^\perp=\Ker M. Thus M is unitary on \mathcal H, with inverse M^* there. For every F\in\mathcal H, equality in the conditional-expectation norm gives

Indeed, the integral of the squared difference is \norm F_2^2-\norm{MF}_2^2=0.

Choose a measurable enumeration \ell_n(u) without repetition, and set h(u,\ell_n(u))=2^{-n} on its domain, starting at n=1. Then h>0 and \norm h_2^2\le\sum_{n\ge1}4^{-n}<\infty. The probability kernel sends h to a strictly positive function v=Mh almost everywhere: the kernel avoids every \xi-null set for almost every source point, by its marginal identity, and the integral of a positive function under a probability law is positive. Also v\in\mathcal H.

The time operator V is unitary and commutes with M in both directions, so it preserves \mathcal H. Define the measurable countable vector

Negative powers of M are taken on \mathcal H. All its coordinates have finite versions outside one null set, and its (0,0) coordinate is positive there. On \C^{\Z^2} define

These are commuting invertible coordinate reindexings. Using (51) for the countably many coordinates gives

almost everywhere in the respective measures.

We claim that \psi has countable essential image. Given a source point z=(u,\ell), the parameter of its K_{st} target has law \nu, since p_s,p_t are independent under \pi and fixing a source label does not tilt that law. Consider the occurrence relation

It is measurable: existence is a countable union and equality of vectors is a countable conjunction of coordinate equalities. The first identity in (53) implies \nu\{u':R(z,u')\}=1 for \xi-almost every z. Tonelli’s theorem for the sigma-finite product \xi\otimes\nu selects one u' for which R(z,u') holds for \xi-almost every z. That target fibre has only countably many \psi-values. Invertibility of D_0 proves the claim.

Let \mathcal B be the positive-mass values in this countable essential image. For b\in\mathcal B put

Every b has positive (0,0) coordinate b_{0,0}, and

Equal marginals in the first identity of (53) and time invariance in the second show that D_0 and S_0 permute \mathcal B, preserving each level’s mass along their orbits. After discarding a \nu-null set, all labels in every remaining fibre belong to these levels, and each B_b(u) is finite. Countability of \mathcal B makes these assertions simultaneous.

Counting the labels in (53) shows that, for \pi-almost every p, its matching isomorphism \Phi_{st}(p) carries every block b onto block D_0b. Because each fibre’s label list and \mathcal B are countable, intersecting the corresponding conull sets upgrades the transported label identity to every label in every block for such a tuple. Disintegrate the measurable conull set of such tuples over (p_s,p_t). Since its marginal is \nu\otimes\nu, there is a measurable conull set of pairs (u,w) admitting a tuple with this property. Fubini chooses a common source u_* for which almost every target is in this set. For each such target u, choose one resulting group isomorphism I_u:\mathcal L(u_*)\to\mathcal L(u). Each I_u takes block b to block D_0b simultaneously for every b. The choices are used only pointwise.

Fix one typical target u_0, set \mathcal L=\mathcal L(u_0), and define J_u=I_uI_{u_0}^{-1}. These group isomorphisms preserve every block index. Thus all |B_b(u)| are constant almost everywhere. Equation (54) makes each constant finite, and its positive left side makes it nonzero. Let B_b=B_b(u_0) and m_b=|B_b|.

The second identity of (53) gives the asserted time permutation of blocks. Restrict the local parameter set to the intersection of all integer translates of the conull sets just used and of the time-equivariance set. Choose u in this invariant conull set. The automorphism

of \mathcal L maps every B_b to B_{S_0b}. Take G=A^{-1}. Iteration gives (49). This last step fixes one automorphism realizing the permutation of blocks; it makes no assertion about the maps inside a block at other parameters or later times.□

Proof. Condition the nonzero product integral at level 1 on p. For any chosen slot i, the conditional expectation of the product of the other bounded inputs onto that slot belongs to its active space. If f_i(x) were orthogonal to that space for almost every p, the product integral would vanish. In using complex inner products here one can conjugate the other product; the active space is conjugation invariant. Lemma 5.1 and Proposition 5.2 therefore give a choice f=f_i with a nonzero coefficient on some local label on a set of positive \nu-measure. The identity-label coefficient is zero:

Thus nonidentity labels supply the nonzero coefficient.

For a block b\in\mathcal B define the measurable function

The countably many blocks cover all labels, so some fixed block b has \int c_b\dd\nu>0. Its cardinality m_b is fixed and finite, and hence c_b\le m_b\norm f_\infty. Set B=B_b\subset\mathcal L using Lemma 5.4, and c=\int c_b\dd\nu.

For each typical u choose the block-preserving identification J_u from that Lemma and put \chi_\ell^u=\chi_{u,J_u\ell}. Proposition 5.2 gives (55). By (49), the sum in (56) equals the intrinsic measurable sum

Only the set of lines is relevant; changing representative phases or permuting labels inside this set changes no summand magnitude. This proves measurability without a measurable choice of J_u.

The conditional laws are equivariant: (T_Y^r)_*\eta_u=\eta_{T_U^r u}. Time carries B_{S_0^{-r}b}(u) bijectively onto B_b(T_U^r u), taking representatives to representatives up to scalar phase. The change of variables y'=T_Y^r y in (58) consequently gives

Invariance of \nu proves its integral is c for every r. Finally, G^r bijects B onto B_{S_0^{-r}b}, so every moving block has the same cardinality |B|=m_b. Each coefficient is at most \norm f_\infty. This proves the bound in (57) and completes the Proposition.□

Proof. For deterministic complex numbers w_m with |w_m|\le1 and 0<\theta\le1, the elementary inequality e^t\le1+t+2t^2 for |t|\le1 gives

The right side is nonnegative. Multiply these inequalities over m<n and integrate. On expanding the product, a nonconstant term contains one factor \chi_{l_m} or \overline{\chi_{l_m}} at each of a nonempty set of distinct indices. By (60), it is a constant multiple of the representative of a signed sum of those labels. Dissociation makes that label nonzero, so its integral vanishes. Hence

For 0<\varepsilon\le1, choose \theta=\varepsilon/4 and apply Markov’s inequality. Uniformly over the deterministic coefficients,

We next allow the coefficients to be the random word (F_m). Put

Call a word likely if its probability is at least e^{-a_n n}. There are at most e^{a_n n} likely words. For the random word, the mean of minus the logarithm of its probability is H_n, so the total probability of the other words is at most H_n/(a_n n). Both a_n and this last bound tend to zero.

Fix arbitrary deterministic phases \omega_m\in\T. On the event that (F_m)_{m<n} equals a particular likely word (v_m), the sum

is the sum in (63) with w_m=\omega_m v_m. We do not condition the character law on that word: the event in question is simply a subset of the corresponding event for deterministic coefficients. A union bound therefore gives

This explains why arbitrary coupling is allowed.

Finally choose each \omega_m to make \omega_m\int F_m\overline{\chi_{l_m}}\dd\eta nonnegative real, choosing any phase if that integral is zero. These choices are deterministic. Since Y_n\le1, we have

First let n tend to infinity and then let \varepsilon tend to zero.□

Proof. Suppose the first alternative fails. Remove any initial zero coefficients from a nonempty signed relation and apply a power of G^{-1}. We obtain

If q=0, then l=0 and the third alternative holds with p=d=1. For q\ge1, the unit coefficient at the upper end expresses G^ql as an integral combination of the preceding q labels. The coefficient at the lower end does the same in the other direction. Applying the relation to all translates shows that K is generated by l,Gl,\ldots,G^{q-1}l. In particular, its torsion subgroup K_{\rm tor} is finite and its quotient \overline K is a finite-rank lattice.

The induced map \overline G is a lattice automorphism. If it has an eigenvalue \lambda with |\lambda|>1, choose a nonzero complex left eigenfunctional \varphi on \overline K\otimes_\Z\C. The two-sided orbit of \overline l spans this vector space. Thus \varphi(\overline l)\ne0, since otherwise \varphi(\overline G^r\overline l)=\lambda^r \varphi(\overline l)=0 for every integer r. Choose p with R=|\lambda|^p>2. On each residue class j, the absolute values of the projected labels are |\lambda^j\varphi(\overline l)|R^m. In any finite signed relation on distinct indices, the term with largest index strictly dominates the sum of all earlier terms, because \sum_{a<m}R^a<R^m. There can be no such relation in K. This proves the second alternative.

It remains to consider the case with no expanding eigenvalue. If the lattice has positive rank, \det\overline G=\pm1 implies that every eigenvalue has modulus one. We use the bounded-coefficient argument in Kronecker’s root-of-unity criterion [13, Theorem I]; see also [5, Theorem 1 and its proof]. The characteristic polynomial of \overline G^n has integer coefficients, each bounded independently of n: its coefficients are elementary symmetric functions of a fixed number of numbers of modulus one. Only finitely many such polynomials exist. For each eigenvalue \lambda, all the powers \lambda^n are roots of these finitely many polynomials. Hence two powers agree and \lambda is a root of unity.

Choose p such that \overline G^p is unipotent and G^p is the identity on K_{\rm tor}. The latter is possible because the automorphism group of a finite group is finite. This argument includes the rank-zero case. For some d_0\ge1, the map N=G^p-\id satisfies N^{d_0}K\subseteq K_{\rm tor}, and NK_{\rm tor}=0. Consequently N^{d_0+1}K=0. The additional application of N accounts for possible maps from a free part into torsion; an invariant splitting of K has not been assumed. Taking d=d_0+1 and expanding (\id+N)^m proves (65).□

Proof. We induct on the degree. If P(m)=l_0 is constant, condition \overline{\chi_{l_0}} onto x. Its conditional expectation is a bounded measurable function g on X, and

by mixing and centering. Mixing for bounded functions follows from the set formulation by simple-function approximation. This proves the base case, including l_0=0.

For the inductive step, choose deterministic phases \omega_m aligning the integrals in the assertion. In the fixed Hilbert space L^2(\eta) put

Their norms are bounded by \norm F_\infty. At a fixed positive lag h, the absolute value of \ip{v_m}{v_{m+h}} equals

The constant phases from alignment and from (60) disappear inside this absolute value. Apply the induction hypothesis to the centered function F_h-\int F_h\dd\mu_X and the polynomial of lower degree. The remaining constant contribution has absolute value at most |\int F_h\dd\mu_X|. Therefore

The following averaging step is the Hilbert-space van der Corput argument; see [4, Theorem 4.1] for a real Hilbert-space formulation, using the real part of the inner product in the complex case. For completeness, fix H\ge1. Replacing each v_m in its mean by H^{-1}\sum_{a=0}^{H-1}v_{m+a} changes the mean by a vector of norm at most 2H\norm F_\infty/n. Convexity of the squared norm bounds the squared norm of this new mean by the mean of the squared norms of its summands. Expanding those squares, grouping terms by the lag, and using (66) gives

The harmless shift of the averaging range by at most H indices does not change any of these limiting bounds. Mixing and centering imply \int F_h\dd\mu_X\to0 as h\to\infty. Letting H\to\infty shows that the Hilbert-space means tend to zero. Testing them against the constant function 1 yields exactly the mean of the aligned absolute integrals, proving the induction.□

Proof of Proposition 6.1. Apply Lemma 6.4. In the first alternative, Lemma 6.3 applies directly. In the second, apply it on each of the finitely many residue classes modulo p, using the progression-word entropy estimate after that lemma. In the third, (65) gives a polynomial on every residue class. Apply Lemma 6.5 there with F=f\circ T_X^j. This function is centered by invariance of \mu_X. In each case the finitely many progression averages tend to zero and combine to give the full average in (61).□

Proof of Theorem 1.1. Suppose the conclusion fails, and let k\ge3 be the least failing order. Proposition 2.1 gives a mixing finite-alphabet process (X,\mu_X,T_X) of zero entropy, centered finite-valued one-site functions, and a sequence of layouts witnessing failure at order k. All proper sublayouts satisfy the lower-order mixing conclusions.

Construct the ordered array laws and their saturated extension by Section 3. The square locality and operator results of Sections 3 and 4 apply to this extension. Proposition 5.5 supplies a centered one-site witness f, a countable abelian group \mathcal L, an automorphism G, a finite nonempty set B\subset\mathcal L, and representatives \chi_l^u in each typical fibre. The measurable coefficient sums s_r(u) defined in (56) satisfy

Fix a typical u. The marginal of x under \eta_u is exactly \mu_X. The representatives obey (60) by (55). Apply Proposition 6.1 to each of the finitely many l\in B. It gives

This is a pointwise argument in the fibre; it uses neither invariance of \eta_u under time nor a measurable choice of the abstract identification. The original sums are already measurable, and are bounded by |B|\norm f_\infty for every r and almost every u. Bounded convergence therefore makes their integrals tend to zero. Every such integrated average equals c>0, a contradiction.

There is consequently no least failing finite order. Since ordinary mixing supplies order two, all the conclusions in Theorem 1.1 follow.□

Proof of Corollary 8.1. For the endomorphism assertion, fix the tested sets and let \alpha be the finite partition they generate, with alphabet D. The one-sided name map \pi records the atom of \alpha containing T^n\omega in coordinate n\ge0. It takes values in D^{\Z_{\ge0}} and intertwines T with the left shift V. The Borel measure \nu=\pi_*\mu is stationary and mixing, by pulling back any two Borel sets. Its completion is still mixing, by null representatives. Take the mixing invertible natural extension of this standard factor, (\widehat Y,\widehat\nu,\widehat V) with projection p [16, Section 9]. If C_i is the time-zero cylinder coding A_i, then for every nonnegative layout,

Theorem 1.1 gives the required product limit. Only Borel cylinders are pulled back to the original space; no natural extension of that possibly nonstandard space is asserted.

For the flow assertion, S_1 is mixing: integer positive times are a subsequence of the assumed limit, and invertibility gives negative times by exchanging the tested sets. Subtract t_1 from each layout and write t_i=n_i+r_i, where n_i=\lfloor t_i\rfloor and 0\le r_i<1. The successive integer gaps still diverge. Along any sequence of layouts, pass to a subsequence on which every r_i converges to r_i^*\in[0,1]. The product telescoping bound and the L^1 isometry of U_{n_i} give

Theorem 1.1 applied to S_1 and the fixed indicators U_{r_i^*}1_{A_i} gives the product of their unchanged means. Every hypothetical sequence violating the real-time limit has such a subsequence, proving the assertion.□