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A probability measure-preserving transformation is strong mixing exactly when $\mu (T^{-n}A\cap A)\to \mu (A{)}^2$ for every measurable $A$. The converse follows by applying decay of autocorrelation implies weak convergence of an observable to $f=1_A$ and then testing against $1_B$. Diagonal polarization alone controls a sum of the two directed cross-correlations and does not suffice as a proof.

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.