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A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner-product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

For the Random-scan Gibbs sampler, the correct transition kernel is a measure, since a one-coordinate update is singular with respect to full-dimensional Lebesgue measure when $p>1$. Write $x_{-i}$ for all coordinates except $i$. ThenEach Dirac measure fixes an unupdated coordinate. Equivalently, $P(x,A)=p^{-1}{\sum }_i\int 1_A(x_{-i},z){\pi }_i(z\mid x_{-i})dz$, with $z$ placed in coordinate $i$. Replacing this kernel by an ordinary density on the whole product space would omit those fixed-coordinate constraints.

For the globally invertible Hamiltonian flow from (a), the general characteristic formula isEvery $\Phi (s,t)$ preserves Lebesgue measure, by (b). Thus composition with it is an isometry of each Lp space. The Minkowski integral inequality givesApplying the Minkowski inequality also in time yields the finite-time Lp bound for Hamiltonian transport:The smoothness assumptions allow the characteristic construction; the norm estimate itself only uses the Lp space data and volume preservation.

For a concrete failure on infinite time, choose $\omega =1$, $f_0=0$, andThis is smooth, time independent and in every finite Lp space; it is invariant under the isotropic harmonic oscillator flow. Therefore the solution grows linearly:This invariant-source secular growth in Hamiltonian transport supplies the counterexample even with zero initial data.

For each $x$, $g(x{)}^n\to 0$ if $g(x)<1$, and remains one if $g(x)=1$. Continuity gives pointwise convergence to $f(0)$ off $E=\{x:g(x)=1\}$ and to $f(1)$ on $E$. The functions are measurable and bounded by ${\max }_{[0,1]}|f|$, an integrable constant on a unit-length interval. The dominated convergence theorem therefore givesHere $|E|$ is Lebesgue measure. This convex combination belongs to $[f(0),f(1)]$, without requiring $f$ to be monotone.

For a Borel set $A\subset [0,1)$, the two inverse branches of the doubling map give the disjoint setsEach has half the Lebesgue measure of $A$, by translation and scaling invariance. Hence $\boxed{\mu (T^{-1}A)=\mu (A)}$. The piecewise affine map is Borel measurable, so it is a measure-preserving transformation. The printed one-dimensional “baker's map” is this doubling map.

The Szemerédi theorem says that for every integer $k\ge 2$ and every $\delta >0$, there is $N_0(k,\delta )$ such that, for $N\ge N_0$, every $A\subseteq \{1,\dots ,N\}$ with $|A|\ge \delta N$ contains a nonconstant arithmetic progressionEquivalently, every subset of the positive integers with positive upper asymptotic density contains arithmetic progressions of every finite length. Length one is immediate.

The Furstenberg multiple recurrence theorem states that for every probability measure-preserving system, every measurable $B$ with $\mu (B)>0$, and every integer $k\ge 2$, there is $n\ge 1$ withHere $T^{-jn}B$ means a preimage, so invertibility is unnecessary. The Furstenberg multiple recurrence theorem also does not assume an ergodic transformation.

We prove the implication to the finite Szemerédi theorem by constructing the Furstenberg correspondence principle explicitly. Suppose, to the contrary, that for fixed $k$ and $\delta >0$ there are $N_{\ell }\to \infty$ and $A_{\ell }\subseteq \{1,\dots ,N_{\ell }\}$ with $|A_{\ell }|\ge \delta N_{\ell }$, but with no length-$k$ arithmetic progression of positive common difference. In the binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$, encode $A_{\ell }$ as $x_j^{(\ell )}=1_{A_{\ell }}(j)$, taking all coordinates outside $A_{\ell }$ to be zero. Let $S$ be the left shift, $(S\omega {)}_j={\omega }_{j+1}$, and setThe cylinder set $C$ is a clopen set, and ${\nu }_{\ell }(C)=|A_{\ell }|/N_{\ell }\ge \delta$. The binary full shift is a compact metric space, so compactness of probability measures on a compact metric space gives a subsequence of these empirical measures with weak convergence of probability measures to a Borel probability measure $\nu$.

For every continuous $h$ on the full shift,Passing to the weak limit shows that $\nu$ is an invariant measure for the continuous left shift. Thus $(\Omega ,\mathcal{B},\nu ,S)$ is a probability measure-preserving system. Since the indicator function of $C$ is continuous, $\nu (C)\ge \delta$.

Apply the Furstenberg multiple recurrence theorem to $C$. For some $n\ge 1$, the clopen sethas $\nu (D)>0$. Its continuous indicator function gives ${\nu }_{\ell }(D)\to \nu (D)>0$. For large $\ell$, at least one $S^a{x}^{(\ell )}$ in the defining empirical measure therefore belongs to $D$. The choice of left shift impliesso $a,a+n,\dots ,a+(k-1)n$ lies in $A_{\ell }$. This contradicts the assumed absence of arithmetic progressions and proves the finite Szemerédi theorem. Applying the finite Szemerédi theorem on intervals where an infinite set has density bounded below proves the stated upper asymptotic density formulation.

For multiple recurrence for circle rotations, write $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ and $R_{\alpha }x=x+\alpha$, with normalized Lebesgue measure $m$, which is the Haar measure of the circle group. Fix a measurable $B\subseteq \mathbb{T}$ with $m(B)>0$ and $k\ge 2$. If $\alpha$ is rational, there is $q\ge 1$ with $R_{\alpha }^q$ the identity, and $n=q$ gives an intersection of measure $m(B)$.

For irrational $\alpha$, the pigeonhole principle applied to $0,\alpha ,\dots ,Q\alpha$ in $Q$ equal arcs gives $1\le n\le Q$ with $\Vert n\alpha {\Vert }_{\mathbb{T}}\le 1/Q$. In particular there are arbitrarily small nonzero returns to zero for the irrational rotation of the circle.

We also need translation continuity in L1 on the circle. Given a measurable $B$, approximate its indicator function in the L1 norm by a continuous $g$ on the circle group. Such approximation follows from regularity of Lebesgue measure, or approximation by finite unions of intervals. Translation invariance and uniform continuity givewhere the approximation error is first made arbitrarily small. Equivalently, $m(B△(B-t))\to 0$.

Choose a return time $n\ge 1$ so small modulo one that for each $1\le j<k$,This is possible because multiplication by each fixed $j$ is continuous on the circle group and translation continuity in L1 on the circle applies to the finitely many translates. The union bound now givesThis proves the Furstenberg multiple recurrence theorem for every circle rotation with its normalized Lebesgue measure, including both rational and irrational angles.

A continuous map $F$ of a compact metric space is uniquely ergodic if it has exactly one invariant measure that is a Borel probability measure. The uniqueness requirement ranges over all invariant Borel probability measures.

For an irrational rotation of the circle, let $R_{\alpha }x=x+\alpha$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\nu$ be any invariant Borel probability measure. For each integer $r$, set ${\chi }_r(x)=e^{2\pi irx}$. Invariance givesFor $r\ne 0$, irrationality forces $e^{2\pi ir\alpha }\ne 1$, so the corresponding Fourier coefficient is zero. For $r=0$ it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure $m$. By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence $\nu$ and $m$ integrate every continuous function equally and are the same Borel probability measure. Since $m$ is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure $m$.

For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with $T^{-1}(x,y)=(x-\alpha ,y-x+\alpha )$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.

Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the Fourier basis is $F={\sum }_{r,s}{c}_{r,s}{e}_{r,s}$ in $L^2$. Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor $s\ne 0$, the magnitudes of the Fourier coefficients along all distinct indices $(r+js,s)$, $j\in \mathbb{Z}$, are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha }{c}_{r,0}$, which forces $c_{r,0}=0$ for $r\ne 0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and

To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yieldsIn particular, for a Fourier basis element,If $s=0$ and $r\ne 0$, this is a geometric series in $n$, anduniformly in $(x,y)$.

For $s\ne 0$, we give the finite Van der Corput inequality for finite scalar sequences and its proof. For $|u_n|\le 1$, extend $u_n$ by zero outside $0\le n<N$. Fix an integer $1\le H\le N$ and set $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identityThe Cauchy-Schwarz inequality, followed by expansion of the squared window sums, givesConsequently the Van der Corput inequality for finite scalar sequences isThe finite prefactor is important; the order of limits will be $N\to \infty$ with $H$ fixed, followed by $H\to \infty$.

For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed $h\ge 1$, $s\alpha h$ is irrational, so another geometric series estimate givesuniformly in $(x,y)$. With fixed $H$, the Van der Corput inequality for finite scalar sequences therefore impliesLetting $H\to \infty$ proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.

The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in $C({\mathbb{T}}^2)$. If $p$ approximates a continuous $g$ with $\Vert g-p{\Vert }_{\infty }<\epsilon$, thenTaking $N\to \infty$ and then $\epsilon \downarrow 0$ proves uniform convergence to $\int gd{m}_2$ for every continuous $g$.

Finally, if $\nu$ is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence giveThus $\nu =m_2$, since continuous functions determine Borel probability measures on a compact metric space. We concludeThe uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the pointwise ergodic theorem.

First take nonempty bounded open sets $A,B\subset {\mathbb{R}}^n$. Their Minkowski sum is open and hence a Lebesgue measurable set. Apply the Prékopa–Leindler inequality to their indicator functions and the indicator function of $(1-\theta )A+\theta B$. This gives the multiplicative formTo obtain the usual additive Brunn–Minkowski inequality, put $a={\lambda }_n(A{)}^{1/n}$, $b={\lambda }_n(B{)}^{1/n}$, $A_0=A/a$ and $B_0=B/b$. Both normalized open sets have Lebesgue measure one. For $\theta =b/(a+b)$,The multiplicative bound and the scaling rule for Lebesgue measure implyReplacing $A,B$ by $A\cap B(0,R),B\cap B(0,R)$ and using continuity from below extends this form of the Brunn–Minkowski inequality to all nonempty open sets, including those of infinite Lebesgue measure. This open-set form is sufficient here and avoids any measurability qualification for sums of arbitrary measurable sets.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

For each $t\in [0,1]$, evaluation ${\delta }_t:f\mapsto f(t)$ is a bounded linear functional of norm one. Thus weak convergence to zero gives $f_n(t)\to 0$ at every $t$. The set $\{f_n:n\ge 1\}$ is a weakly bounded set, since every scalar sequence $\ell (f_n)$ converges, and part (a) supplies a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$.

The constant $M$ is integrable for Lebesgue measure on $[0,1]$. Applying the dominated convergence theorem to $|f_n|$ gives the weakly null continuous functions converge in L1 conclusionPointwise convergence alone would not provide the needed uniform dominating function; it is the weak boundedness argument that supplies it.

A continuous transformation $T$ of a compact metric space $X$ is uniquely ergodic when there is exactly one $T$-invariant Borel probability measure on $X$. We use the usual compact-space convention for unique ergodicity; the compactness and continuity hypotheses matter in the assertion about all starting points.

Let $R_{\alpha }(x)=x+\alpha (\mathrm{mod}1)$ on the circle group, with $\alpha$ irrational. Normalized Lebesgue measure $m$ is invariant under this irrational rotation of the circle. If $\nu$ is any invariant Borel probability measure, define its Fourier coefficients by $c_k=\int e^{2\pi ikx}d\nu (x)$. Invariance givesFor $k\ne 0$, irrationality makes $e^{2\pi ik\alpha }\ne 1$, so $c_k=0$; also $c_0=1$. These are the Fourier coefficients of $m$. Hence $\nu$ and $m$ have the same integrals against every trigonometric polynomial. Such polynomials are uniformly dense in the continuous functions by the Stone-Weierstrass theorem, so the measures agree on every continuous test function and therefore agree as Borel measures. Thus

For the general uniquely ergodic system, fix $x\in X$ and form the empirical measuresOn a compact metric space, the Borel probability measures are compact for weak convergence of probability measures. Any subsequential limit $\nu$ is invariant: for every continuous $g$,Here $g$ is bounded and $g\circ T$ is continuous, so the identity passes to the limit. Uniqueness of the invariant Borel probability measure gives $\nu =\mu$. Every subsequential limit is therefore $\mu$, and compactness implies convergence of the entire sequence. Testing against $f$ provesIn fact this proves uniform ergodic convergence for uniquely ergodic systems: if convergence were not uniform in $x$, choose $N_j\to \infty$ and $x_j$ where the discrepancy stays above a fixed positive number. The same compactness and telescoping argument applied to ${\nu }_{N_j,x_j}$ forces a subsequence to converge to $\mu$, a contradiction. This also makes clear why an almost-everywhere Birkhoff ergodic theorem alone would not establish the requested everywhere assertion. Without the compact-space hypothesis the assertion need not hold: on the discrete space $\{p\}\bigsqcup {\mathbb{Z}}_{\ge 0}$, set $T(p)=p$ and $T(n)=n+1$. The only invariant probability is ${\delta }_p$, but the continuous bounded function which is zero at $p$ and one on the integer orbit has orbit average one there.

For the decimal application put $\alpha ={\log }_{10}2$. This is irrational: if $\alpha =p/q$ with positive integers $p,q$, then $2^q=10^p=2^p{5}^p$, contradicting unique prime factorization. Writing $n\alpha =m_n+t_n$ with $m_n=\lfloor n\alpha \rfloor$ and $t_n\in [0,1)$ gives $2^n=10^{m_n}10^{t_n}$. Its leading decimal digit is seven exactly whenThe half-open upper endpoint correctly excludes powers whose leading digit is eight, including $2^3=8$.

The indicator of $I$ is not continuous, so an extra step is needed. For any $\epsilon >0$, choose continuous functions $a_{\epsilon },b_{\epsilon }$ on the circle with $0\le a_{\epsilon }\le 1_I\le b_{\epsilon }\le 1$ and $\int (b_{\epsilon }-a_{\epsilon })dm<\epsilon$, by tapering in small neighbourhoods of the two endpoints. Applying the everywhere averaging result to these functions traps the lower and upper limits of the interval frequency between their integrals. Letting $\epsilon \downarrow 0$ proves everywhere interval frequency under an irrational rotation. ConsequentlyThis is the leading-seven case of Benford frequencies for powers of an integer.

For a probability measure-preserving system, strong mixing means that for every pair of measurable sets $A,B$,A sequence $a_n$ has convergence in density of a sequence to $a$ when for each $\epsilon >0$ the exceptional set $\{n:|a_n-a|\ge \epsilon \}$ has natural density zero. A weakly mixing measure-preserving transformation has convergence in density of $\mu (T^{-n}A\cap B)$ to $\mu (A)\mu (B)$ for every $A,B$. Equivalently, because these correlations are bounded,Indeed the mean of the absolute discrepancy is at least $\epsilon$ times the exceptional frequency, while it is at most $\epsilon$ plus a uniform bound times that frequency. A signed Cesaro convergence of a sequence without absolute values is insufficient to define weak mixing.

The standard three-cut Chacon map is obtained by cutting and stacking. Start on $[0,1)$ with normalized Lebesgue measure, an initial tower consisting of $[0,2/3)$, and a reservoir $[2/3,1)$ for spacers. At stage $j$, cut every level of the current tower into three equal subintervals, producing three subcolumns. Add one new interval of the same width above the middle subcolumn. Stack the first subcolumn at the bottom, then the middle subcolumn with its spacer, then the third at the top. Define the partial transformation by translation from each level to the next, leaving the current top unmapped. These assignments extend the earlier partial transformation. Repeat indefinitely.

If $h_j$ is the number of levels and $w_j$ their width, thenHence $h_j=(3^{j+1}-1)/2$, $w_j=2/3^{j+1}$, and the stage-$j$ tower has measure $1-3^{-j-1}$. The spacers consume total measure ${\sum }_{j\ge 0}{w}_{j+1}=1/3$, precisely the reservoir. The increasing partial maps give the Chacon map modulo null sets. The tower tops and unused reservoir have measures tending to zero, and the construction yields an invertible measure-preserving transformation. With $0$ marking an original level and $1$ a spacer, the tower words satisfy $W_0=0$ and $W_{j+1}=W_j{W}_{j}1W_j$, so the first new word is $0010$. This describes the spacer placement without needing a proof of well-definedness. The three-cut convention agrees with the classical constant spacer vector $(0,1,0)$ described in Ryzhikov's construction.

For the correlation assertion, take the L2 inner product to be $⟨u,v⟩=\int u\overline{v}d\mu$, linear in its first argument, and put $U=U_T$. The Koopman operator is an isometry on $L^2$, even when $T$ is not invertible. For $n\ge k$ the hint givesTo extend rigorously to all test functions, centre the observable: $v=f-a1$. Invariance of the integral gives $⟨U^{n}v,v⟩=⟨U^{n}f,f⟩-|a{|}^2\to 0$. LetFor each fixed $k$, $⟨U^{n}v,U^{k}v⟩\to 0$ by the same isometry identity. Therefore the limit is zero for every finite linear combination of these orbit vectors. By the Cauchy-Schwarz inequality and $\Vert U^{n}v{\Vert }_2=\Vert v{\Vert }_2$, approximation extends this conclusion to every $g\in M$: the approximation error in the correlation is bounded by $\Vert v{\Vert }_2\Vert g-g_0{\Vert }_2$ uniformly in $n$. For $g\in M^{\perp }$, the correlation is identically zero because $U^{n}v\in M$. The orthogonal decomposition by a closed subspace now gives the result for all $g\in L^2$. ThusThis is decay of autocorrelation implies weak convergence of an observable; no assumption that $T$ is an ergodic transformation, and no invertibility hypothesis was used, and $a=0$ is included.

Finally, strong mixing immediately implies the stated diagonal limit by taking $B=A$. Conversely, suppose that limit holds for every $A$. Apply the correlation result with $f=1_A$ and $g=1_B$. Its hypothesis is exactly $⟨U^n{1}_A,1_A⟩\to \mu (A{)}^2$, and its conclusion isHence strong mixing is equivalent to diagonal set-correlation criterion for mixing. Using the orbit-span proof avoids an invalid polarization argument that would recover only the sum of the two directed cross-correlations from diagonal correlations.

Write $D={\sum }_{i=1}^n{b}_i$ and let $\lambda$ denote Lebesgue measure. Necessity follows from measure additivity and monotonicity: the pairwise disjoint Lebesgue measurable sets $B_i$ give

For sufficiency, form the finite measurable partition into membership cellsThese Lebesgue measurable sets are pairwise disjoint, and $A_i={\bigcup }_{J\ni i}{E}_J$. Empty cells may be retained. Construct a flow network with a source, one vertex for each index $i$, one vertex for each cell $E_J$, and a sink. Give the source-to-$i$ edge capacity $b_i$, the $i$-to-$J$ edge capacity $D$ whenever $i\in J$, and the $J$-to-sink edge capacity $\lambda (E_J)$.

Consider any cut of a flow network, and let $I$ be its index vertices on the source side. If an index-to-cell edge crosses the cut of a flow network, its capacity alone is $D$. Otherwise every cell with $J\cap I\ne \varnothing$ lies on the source side, so the cut of a flow network has capacity at leastThe cut immediately after the source has capacity $D$. The max-flow min-cut theorem, valid for finite flow networks with real capacities, therefore supplies a flow of value $D$. Its cell allocations $f_{iJ}\ge 0$ satisfy

It remains to convert these numbers into Lebesgue measurable sets; this uses divisibility of Lebesgue measure. Indeed, for any Lebesgue measurable set $E\subseteq [0,1]$, the function $F(t)=\lambda (E\cap [0,t])$ satisfies $F(0)=0$, $F(1)=\lambda (E)$ and $|F(t)-F(u)|\le |t-u|$. Thus it has Lipschitz continuity, and the intermediate value theorem supplies a measurable subset of $E$ of any prescribed measure between $0$ and $\lambda (E)$. Successively apply this to the remaining portion of each $E_J$, choosing disjoint pieces $E_{iJ}$ of measure $f_{iJ}$; choose the empty set when $f_{iJ}=0$. ThenThis proves the measurable Hall theorem. The splitting step is essential: an arbitrary measure with an atom of a measure would not support the same conclusion. Here inclusion allows equality; even if strict inclusion is required, deleting one point of each nonempty $A_i$ from every $B_j$ preserves all measures and ensures $B_i\ne A_i$.