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For a probability measure-preserving system, weak mixing is the vanishing of averaged absolute correlation discrepancies:By approximation with simple functions and the Cauchy-Schwarz inequality, this is equivalent toHere $⟨f,g⟩=\int f\overline{g}d\mu$, and $U=U_T$. Absolute values are part of the definition: signed Cesaro convergence of a sequence of these correlations alone expresses ergodicity, not weak mixing.

Suppose the product system is ergodic. Fix a mean-zero $f\in L^2(\mu )$ and any $g\in L^2(\mu )$. On the product takeThese belong to $L^2(\mu \otimes \mu )$, and $\int F=|\int f{|}^2=0$. The mean ergodic theorem on the product, followed by pairing with $G$, givesThe Cauchy-Schwarz inequality in $n$ bounds the averaged absolute correlation by the square root of this quantity. Subtract the mean from a general $f$ to obtain the definition above. This proves product ergodicity implies weak mixing, through the square-correlation proof of weak mixing from product ergodicity.

For the multiple averages, write $m(f)=\int fd\mu$ and use the following Van der Corput lemma. If $(v_n)$ is bounded in a Hilbert space, every correlation averageexists, and $H^{-1}{\sum }_{h=1}^H|L_h|\to 0$, then $N^{-1}{\sum }_{n<N}{v}_n\to 0$ in norm. One way to see the estimate is to replace $v_n$ by $H^{-1}{\sum }_{j=0}^{H-1}{v}_{n+j}$; for fixed $H$ the change in its long average tends to zero. The Cauchy-Schwarz inequality and expansion of the squared block norm giveLet $H\to \infty$. This proves the auxiliary implication needed here.

A weak mixing system is ergodic: a mean-zero invariant $f$ would have the nonvanishing correlation $⟨U^{n}f,f⟩=\Vert f{\Vert }_2^2$. Every positive power $T^r$ is also weak mixing, since for nonnegative correlation discrepancies $b_n$,This part of the stability of weak mixing under powers and products will control the induction.

In fact the stronger arithmetic-progression multiple averages under weak mixing holds:For $k=1$ this is the mean ergodic theorem and ergodicity. Suppose it is known for $k-1$ and first assume $m(f_k)=0$. Put $v_n={\prod }_{j=1}^k{U}^{jn}{f}_j$. For fixed $h$, setUsing invariance of the integral to remove the common $U^n$ givesThis identity does not require an inverse of $T$. Apply the induction hypothesis to the $k-1$ factors and pair their $L^2$ limit with $g_{1,h}$. It follows thatThe average in $h$ of the right side tends to zero because $T^k$ is weak mixing and $m(f_k)=0$. The Van der Corput lemma proves the zero $L^2$ limit. For general $f_k$, split it into $f_k-m(f_k)$ and its constant mean; the first term has the zero limit just proved, and the other term is $m(f_k)$ times the induction average for $k-1$. This completes the induction.

Pair this result with the real bounded $f_0$. Forthe requested Cesaro limit is

To obtain convergence in density of a sequence, we also need the product system to be weak mixing. For simple tensors, its correlations are products of single-system correlations. If $c_n\to c$ and $d_n\to d$ in averaged absolute discrepancy and both sequences are bounded, thenso the product discrepancy has zero average. Finite sums of tensors are dense in $L^2(\mu \otimes \mu )$, and the Cauchy-Schwarz inequality extends the conclusion to arbitrary $L^2$ functions. Hence $T\times T$ is weak mixing.

Apply the multiple-average result to $F_j=f_j\otimes f_j$ on that product. Because the original $f_j$ are real,Together with the first-moment limit this givesThe mean-square criterion for convergence in density now yields, for every $\epsilon >0$,Therefore the density convergence of multiple weak-mixing correlations gives the same answer:The second-moment argument is essential; the signed Cesaro limit alone would not imply this conclusion.

For each positive integer $n$, consider the threshold set $A^{(n)}=\{a\in A:a\ge 1/n\}$. If it were infinite, one could choose recursively a sequence of distinct elements from it. Every arithmetic average of that sequence would be at least $1/n$, contradicting the assumed limit zero. Hence every $A^{(n)}$ is a finite set.

Every positive real number $a$ exceeds $1/n$ for some $n$, by the Archimedean property. ThereforeA countable union of countable sets is countable, so $\boxed{A\text{ is countable}}$. In fact, under the stipulated infinitude, $A$ is a countably infinite set. Positivity matters: it ensures that these thresholds cover $A$. This is countability from vanishing averages of distinct sequences; ordinary Cesaro convergence of a sequence of one chosen enumeration would not suffice to control the whole set.

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.

On a probability measure-preserving system, for a finite measurable partition $\xi =\{A_1,\dots ,A_r\}$, the entropy of a finite measurable partition isUse natural logarithms, so information entropy is measured in nats; another fixed logarithm base rescales all answers. The join of measurable partitions is their common refinement, and write ${\xi }_a^b={\bigvee }_{j=a}^b{T}^{-j}\xi$, with an empty join the trivial partition. The entropy rate of a measurable partition and Kolmogorov-Sinai entropy are respectivelyThe block entropies form a subadditive sequence, so the first limit exists by the Fekete lemma. For finite partitions the conditional entropy of finite measurable partitions is $H(\eta \mid \zeta )=H(\eta \vee \zeta )-H(\zeta )$.

Put $c_0=H(\xi )$ and $c_k=H(\xi \mid {\xi }_1^k)$ for $k\ge 1$. The chain rule for information entropy, applied from the last coordinate backwards, and measure preservation giveThe second equality uses invariance of the joint partition atom probabilities under the common pullback $T^{-j}$; invertibility is unnecessary. Since conditioning reduces entropy, $c_k$ decreases to a nonnegative limit $c$. The Cesaro convergence of a sequence of this convergent sequence has the same limit. ThereforeEquivalently $h_{\mu }(T,\xi )=H_{\mu }(\xi \mid {\mathcal{F}}_1)$, where ${\mathcal{F}}_1=\sigma ({\bigvee }_{j\ge 1}{T}^{-j}\xi )$. Here conditional entropy of a countable measurable partition conditioned on a sigma-algebra is computed using conditional partition atom probabilities; the martingale convergence theorem gives continuity under increasing conditioning sigma-algebras. This is the infinite-future formula for partition entropy rate.

The Kolmogorov-Sinai generator theorem states that if a finite or countable measurable partition $\eta$ has finite entropy of a countable measurable partition and its iterates generate the whole completed sigma-algebra modulo null sets, then $h_{\mu }(T)=h_{\mu }(T,\eta )$. For an invertible system, generating means $\sigma ({\bigvee }_{j\in \mathbb{Z}}{T}^{-j}\eta )=\mathcal{B}$ modulo null sets. For a noninvertible system a one-sided generator, using $j\ge 0$, suffices. The two-sided and one-sided versions must not be confused.

For a Bernoulli shift with discrete symbol probabilities $(p_i)$, the coordinate-zero measurable partition has independent coordinate iterates. ThusFor a finite alphabet the coordinate partition is a generator of finite entropy of a finite measurable partition; on the two-sided sequence space use all integer coordinate iterates, and on the one-sided space use the nonnegative ones. The Kolmogorov-Sinai generator theorem proves the displayed answer in both cases. The same calculation applies to countably many symbols when their Shannon entropy is finite, using the countable finite-entropy version of the theorem. If the Shannon entropy is infinite, merge all but the first $r$ symbols into one cell. These finite coordinate partitions have entropy rate $-{\sum }_{i\le r}{p}_i\log p_i-p_{>r}\log p_{>r}\to \infty$, so the system entropy is infinite. Zero-probability symbols contribute zero. In particular a fair $r$-symbol shift has entropy $\log r$.

For the final assertion, let ${\mathcal{F}}_0=\sigma ({\bigvee }_{j\ge 0}{T}^{-j}\xi )$ and complete it modulo null sets. The approximation property forces ${\mathcal{F}}_0=\mathcal{B}$ modulo null sets. Indeed for each $A\in \mathcal{B}$, choose approximating sets from finite blocks with error tending to zero. Their indicators approach $1_A$ in $L^2$, and $L^2({\mathcal{F}}_0)$ is a closed vector subspace, so $A$ is ${\mathcal{F}}_0$-measurable modulo a null set.

Invertibility now gives ${\mathcal{F}}_1=T^{-1}{\mathcal{F}}_0=T^{-1}\mathcal{B}=\mathcal{B}$ modulo null sets. In particular the present partition $\xi$ is measurable with respect to its entire future, so $H(\xi \mid {\mathcal{F}}_1)=0$. The infinite-future formula gives $h_{\mu }(T,\xi )=0$. Since the given one-sided generator is also a two-sided generator, the Kolmogorov-Sinai generator theorem finishes the proof:This is finite one-sided generator of an invertible system forces zero entropy. Invertibility is essential: a fair binary one-sided Bernoulli shift has a finite one-sided generator and Kolmogorov-Sinai entropy $\log 2$.

The statement $D-lim{a}_n=a$ means convergence in density of a sequence: for every $\epsilon >0$, the proportion of $n\le N$ with $|a_n-a|\ge \epsilon$ tends to zero. The statement $C-lim{a}_n=a$ means Cesaro convergence of a sequence, $N^{-1}{\sum }_{n=1}^N{a}_n\to a$.

Suppose $|a_n-a|\le M$. Splitting into the indices where $|a_n-a|<\epsilon$ and its complement givesThus density convergence implies Cesaro convergence of the absolute deviations to zero. Conversely, the Markov inequality givesproving the reverse implication.

The product characterization of weak mixing says that if $T$ is weakly mixing, then $T\times S$ is ergodic exactly when $S$ is ergodic; in particular, ergodicity of $T\times T$ characterizes weak mixing.

Suppose $U_{T}f=\lambda f$. Unitarity of the Koopman operator gives $|\lambda |=1$. Weak mixing implies ergodicity, so $|f|$ is almost everywhere constant. On the product,is invariant under $T\times T$. Since the square is ergodic, $F$ is constant, which forces $f$ to be constant almost everywhere. Thus there are no nonconstant Koopman eigenfunctions.

Finally use the density-one correlation characterization of a weakly mixing measure-preserving transformation. For each positive-measure pair $(A,B)$, the integers $n$ for which $\mu (T^{-n}A\cap B)>0$ form a density-one set after discarding finitely many terms; the corresponding sets for $(A,B)$ and $(A,C)$ therefore intersect, proving simultaneous hitting. Conversely, the simultaneous-hitting property is precisely the simultaneous hitting characterization of weak mixing, so it implies weak mixing.

A probability measure-preserving transformation $T$ is weakly mixing when, for every pair of measurable sets $A,B$, the average absolute correlation discrepancy tends to zero:Equivalently, the correlations converge to their products in convergence in density of a sequence, its Cartesian product with itself is an ergodic transformation, or its product with every ergodic transformation is ergodic. The absolute values are essential: signed Cesaro convergence of a sequence of the correlations to their products characterizes ergodic transformations and does not imply weak mixing. For example, the transformation interchanging two equally weighted points has discrepancy $(-1{)}^n/4$ for a singleton $A=B$: its signed average tends to zero while its average absolute discrepancy stays $1/4$.