Measure-Preserving Systems and Mixing Criteria

Set $\nu (E)=\mu (T^{-1}E)$. Inverse images take the empty set to the empty set and disjoint countable unions to disjoint countable unions. Thus $\nu (\mathrm{\varnothing })=0$ and $\nu ({\bigcup }_j{E}_j)={\sum }_j\nu (E_j)$ for disjoint measurable $E_j$, so $\nu$ is a measure.

The measures $\nu ,\mu$ agree on $\mathcal{P}$, and $\nu (P_n)=\mu (P_n)<\mathrm{\infty }$ on the stated increasing exhaustion. Uniqueness therefore gives $\nu (E)=\mu (E)$ for all $E\in \mathcal{A}$, which is measure preservation.

If $\mu (X)<\mathrm{\infty }$, adjoining $X$ keeps the family a generating pi-system; its preservation identity holds since $T^{-1}X=X$. The constant exhaustion $P_n=X$ then satisfies every required hypothesis.

For preserving $S,T$ and measurable $E$, $(S\circ T{)}^{-1}E=T^{-1}(S^{-1}E)$ is measurable and has measure $\mu (S^{-1}E)=\mu (E)$. The identity preserves measure; applying this composition calculation successively gives preservation for every $T^n$, $n\ge 0$.

Assume countable choice. The completion theorem supplies the complete measure extending $\mu$. For $E\in \bar{\mathcal{A}}$ choose $A,N\in \mathcal{A}$ with $\mu (N)=0$ and $E\mathrm{△}A\subseteq N$. Then $T^{-1}E\mathrm{△}{T}^{-1}A\subseteq T^{-1}N$, where $T^{-1}N$ is measurable and null. Thus $T^{-1}E$ is completion measurable and $\bar{\mu }(T^{-1}E)=\mu (T^{-1}A)=\mu (A)=\bar{\mu }(E)$.

For $E\in \mathcal{A}$, ${\mathbf{1}}_E\circ T={\mathbf{1}}_{T^{-1}E}$, hence its integral is $\mu (T^{-1}E)=\mu (E)$. A nonnegative simple function written over disjoint fibers has integral equal to the sum of coefficient times fiber measure, so the identity holds for every such function, also when the sum is infinite in value.

For nonnegative measurable $f$, take $s_n\uparrow f$ as supplied by simple approximation. Then $s_n\circ T\uparrow f\circ T$ measurably. Integral monotone convergence on both sides gives $\int f\circ T={\lim }_n\int s_n\circ T={\lim }_n\int s_n=\int f$.

For integrable $f$, applying step 2.1 to $\mathrm{\mid }f\mathrm{\mid }$ proves $\int \mathrm{\mid }f\circ T\mathrm{\mid }=\int \mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$. Apply that step to the positive and negative parts of each real component. Subtract their finite integrals and combine the real and imaginary parts by linearity to obtain the asserted equality.

Conversely the indicator identity is exactly $\mu (T^{-1}E)=\mu (E)$ for each measurable $E$. Thus it is measure preservation.

Composition with measurable $T$ is measurable. If $f=g$ outside a measurable null set $N$, then $f\circ T=g\circ T$ outside $T^{-1}N$, which is measurable and null. Thus pullback respects the a.e. equivalence relation used to define $U_T$.

For $p<\mathrm{\infty }$, integral invariance applied to the nonnegative function $\mathrm{\mid }f{\mathrm{\mid }}^p$ gives $\mathrm{\parallel }f\circ T{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }f{\mathrm{\mid }}^p\circ T=\int \mathrm{\mid }f{\mathrm{\mid }}^p=\mathrm{\parallel }f{\mathrm{\parallel }}_p^p$. Thus the pullback belongs to $L^p$ and preserves the norm, including at $p=1$.

For every $M\ge 0$, $\{\mathrm{\mid }f\circ T\mathrm{\mid }>M\}=T^{-1}\{\mathrm{\mid }f\mathrm{\mid }>M\}$ has the same measure as $\{\mathrm{\mid }f\mathrm{\mid }>M\}$. The sets of finite essential bounds therefore coincide, so their infima coincide. The least-essential-bound result applies to the real modulus, proving membership and equality of the infinity norms.

Pointwise, $(\alpha f+\beta g)\circ T=\alpha (f\circ T)+\beta (g\circ T)$. The real and complex quotient norm theorems make these the quotient vector operations. Together with steps 1.1, 2.1 and 2.2 this proves linear isometry.

For an actual measurable inverse $S=T^{-1}$, $TE=S^{-1}E$ is measurable and $\mu (TE)=\mu (T^{-1}(TE))=\mu (E)$. Thus $S$ preserves measure. Both compositions $U_T{U}_S$ and $U_S{U}_T$ are the identity, so $U_T$ is onto.

In the modulo-null case restrict to the measurable conull invariant $X_0$ of the definition. For measurable $E\subseteq X_0$, $T^{-1}E$ differs from its restricted inverse image only within $X\setminus X_0$, so the restricted map preserves restricted measure. Step 4.1 applies there. For any measurable $f$ on $X$, compose $f{\mathrm{\mid }}_{X_0}$ with the restricted inverse and extend by zero on $X\setminus X_0$. This extension is measurable, has the same Lp norm as $f$, and pulls back to $f$ on $X_0$. It supplies a preimage class.

$T^{-1}X=X$ and $T^{-1}\mathrm{\varnothing }=\mathrm{\varnothing }$. Also $T^{-1}(X\setminus E)=X\setminus T^{-1}E$ and $T^{-1}({\bigcup }_j{E}_j)={\bigcup }_j{T}^{-1}{E}_j$. Thus exact invariance is preserved under complements and countable unions, proving that $\mathcal{I}$ is a sigma-algebra.

The symmetric difference of the two complements is $T^{-1}E\mathrm{△}E$. Further, $T^{-1}({\bigcup }_j{E}_j)\mathrm{△}{\bigcup }_j{E}_j\subseteq {\bigcup }_j(T^{-1}{E}_j\mathrm{△}{E}_j)$. If all component differences are null, countable subadditivity makes the union null. Therefore ${\mathcal{I}}^{\mathrm{\prime }}$ is a sigma-algebra as well. Exact invariance gives empty symmetric difference, proving $\mathcal{I}\subseteq {\mathcal{I}}^{\mathrm{\prime }}$.

Write $D=E\mathrm{△}{T}^{-1}E$, which is measurable and null. For $n\ge 1$, $E\mathrm{△}{T}^{-n}E\subseteq {\bigcup }_{k=0}^{n-1}{T}^{-k}D$: whenever membership at times zero and n differs, it differs at a consecutive pair. Iterates preserve measure, so every set on the right is null; finite subadditivity makes the left null. For n=0 the difference is empty.

The displayed countable intersection and unions make F measurable. Outside the measurable null set $N={\bigcup }_{n\ge 0}(E\mathrm{△}{T}^{-n}E)$ all these indicators equal ${\mathbf{1}}_E$, so membership in their limsup equals membership in E. Thus $E\mathrm{△}F\subseteq N$ and its measure is zero.

Pulling back the displayed formula gives $T^{-1}F={\bigcap }_{m\ge 0}{\bigcup }_{n\ge m}{T}^{-(n+1)}E=F$: removing the initial term does not change membership infinitely often. Hence F is strictly invariant and is the required representative.

If the system is ergodic, a set in ${\mathcal{I}}^{\mathrm{\prime }}$ has by F2 a strict invariant representative differing by a null set, hence has measure zero or one. Conversely (ii) applies to strictly invariant sets. Thus (i) and (ii) are equivalent.

Assume (ii) and let $f:X\to \mathbb{R}$ be measurable and a.e. invariant. For $n\ge 1$, $k\in \mathbb{Z}$, let $E_{n,k}=f^{-1}([k\mathrm{/}n,(k+1)\mathrm{/}n))$. Measurability follows from F3. Its pullback differs from it only where $f\circ T\mathrm{\ne }f$, so it is in ${\mathcal{I}}^{\mathrm{\prime }}$. For each n these fibers partition X. Their measures are zero or one; countable subadditivity excludes all zero, and disjointness and total mass one exclude two fibers of measure one. There is therefore a unique k(n) with $\mu (E_{n,k(n)})=1$.

The set $Y={\bigcap }_{n\ge 1}{E}_{n,k(n)}$ is conull by countable subadditivity. It is nonempty since $\mu (Y)=1$. Fix one $x_0\in Y$. For any $x\in Y$, $\mathrm{\mid }f(x)-f(x_0)\mathrm{\mid }<1\mathrm{/}n$ for every n, so $f(x)=f(x_0)$ by the Archimedean property of the real numbers. Thus (ii) implies (iv). The uniquely determined k(n) require no countable choice.

For a complex a.e. invariant f, its real and imaginary parts are measurable and a.e. invariant by F5. Apply the preceding argument to both, and intersect the two conull sets; f is constant there. This proves both real and complex versions of (iv), and each implies the corresponding version of (iii).

If either version of (iii) holds and E is strictly invariant, then ${\mathbf{1}}_E$ is an everywhere invariant measurable function. A constant indicator on a conull nonempty set must have constant value zero or one, so E is null or conull. Thus (iii) implies (i). Also (iv) directly implies (ii) by the same argument applied to an indicator invariant a.e. All listed implications are now closed.

Assume ergodicity and put $U={\bigcup }_{n\ge 1}{T}^{-n}A$. Then $T^{-1}U\subseteq U$ and $\mu (T^{-1}U)=\mu (U)\le 1$. The finite-measure difference formula gives $\mu (U\setminus T^{-1}U)=0$. Since $\mu (U)\ge \mu (T^{-1}A)=\mu (A)>0$, the modulo-null invariant-set criterion yields $\mu (U)=1$.

If the sweep-out property holds and $\mu (B)>0$, then $\mu (B\cap U)=\mu (B)>0$. Were every $B\cap T^{-n}A$ null, their countable union $B\cap U$ would be null. Thus at least one intersection has positive measure.

If the positive-intersection property holds and $T^{-1}A=A$, then $T^{-n}A=A$ for all n. Taking $B=X\setminus A$ gives $B\cap T^{-n}A=\mathrm{\varnothing }$ for every n. The property excludes both A and its complement having positive measure, proving ergodicity.

For a fixed measurable pair let $d_n=\mu (A\cap T^{-n}B)-\mu (A)\mu (B)$. Strong mixing says $d_n\to 0$. Given $\epsilon >0$, choose m so that $\mathrm{\mid }{d}_n\mathrm{\mid }<\epsilon$ for $n\ge m$. For $N>m$, $N^{-1}{\sum }_{n<N}\mathrm{\mid }{d}_n\mathrm{\mid }\le N^{-1}{\sum }_{n<m}\mathrm{\mid }{d}_n\mathrm{\mid }+\epsilon$. The first term tends to zero because it is a fixed finite sum divided by N. As $\epsilon$ is arbitrary, weak mixing follows.

For a strictly invariant E, $T^{-n}E=E$ for every n. Thus its weak-mixing average against itself is exactly $\mathrm{\mid }\mu (E)-\mu (E{)}^2\mathrm{\mid }$. A constant sequence tends to zero only if that constant is zero. Since $0\le \mu (E)\le 1$, this gives $\mu (E)=0$ or $1$, which is ergodicity.

Let $\mathcal{D}=\{E\in \mathcal{A}:\text{for every }\epsilon >0\text{ some }C\in \mathcal{C}\text{ has }\mu (E\mathrm{△}C)<\epsilon \}$. Every $C\in \mathcal{C}$ lies in $\mathcal{D}$ by using itself. For complements, $(X\setminus E)\mathrm{△}(X\setminus C)=E\mathrm{△}C$, and $X\setminus C\in \mathcal{C}$. Thus $\mathcal{D}$ contains X and is closed under complements.

Let $E_j\in \mathcal{D}$ and $E={\bigcup }_{j\ge 1}{E}_j$. For $H_m={\bigcup }_{j=1}^m{E}_j$, continuity from below gives $\mu (H_m)\uparrow \mu (E)<\mathrm{\infty }$. By additivity on $E=H_m\bigsqcup (E\setminus H_m)$, some $m\ge 1$ satisfies $\mu (E\setminus H_m)<\epsilon \mathrm{/}2$. Choose finitely many $C_j\in \mathcal{C}$ with $\mu (E_j\mathrm{△}{C}_j)<\epsilon \mathrm{/}(2m)$. Then $C={\bigcup }_{j=1}^m{C}_j\in \mathcal{C}$ and $\mu (E\mathrm{△}C)\le \mu (E\setminus H_m)+{\sum }_{j=1}^m\mu (E_j\mathrm{△}{C}_j)<\epsilon$.

Thus $\mathcal{D}$ is a sigma-algebra containing $\mathcal{C}$. By the minimality of $\sigma (\mathcal{C})=\mathcal{A}$, $\mathcal{A}\subseteq \mathcal{D}$, and the reverse inclusion is built into its definition. This proves the approximation assertion.

Let $\mathcal{C}$ consist of finite Boolean combinations of members of $\mathcal{P}$. Each indicator of an atom of a finite Boolean partition is a product of factors ${\mathbf{1}}_P$ and $1-{\mathbf{1}}_P$. Expanding the product expresses it as a finite integer linear combination of indicators of intersections of members of $\mathcal{P}$. Empty intersections are X, which lies in $\mathcal{P}$, and all other intersections lie in $\mathcal{P}$ by the pi-system property. Finite sums of these atom indicators express every ${\mathbf{1}}_C$, $C\in \mathcal{C}$, in this way.

Integration and multiplication of finite sums now express $d_n(C,D)$ as a finite linear combination ${\sum }_{j,k}{a}_j{b}_k{d}_n(P_j,Q_k)$. In the strong case each summand tends to zero. In the weak case the average of the absolute value is at most ${\sum }_{j,k}\mathrm{\mid }{a}_j{b}_k\mathrm{\mid }{N}^{-1}{\sum }_{n<N}\mathrm{\mid }{d}_n(P_j,Q_k)\mathrm{\mid }$, which tends to zero. The respective test therefore holds on $\mathcal{C}$.

The family $\mathcal{C}$ is an algebra generating $\mathcal{A}$. For measurable A,B and $\delta >0$, choose C,D in it with $\mu (A\mathrm{△}C),\mu (B\mathrm{△}D)<\delta$. Measure preservation gives $\mu (T^{-n}(B\mathrm{△}D))=\mu (B\mathrm{△}D)$ by repeated pullback. Thus the difference of the intersection measures is bounded by the sum of these two errors. Also $\mathrm{\mid }\mu (A)\mu (B)-\mu (C)\mu (D)\mathrm{\mid }\le \mu (A\mathrm{△}C)+\mu (B\mathrm{△}D)$, since all masses are at most one. Consequently $\mathrm{\mid }{d}_n(A,B)-d_n(C,D)\mathrm{\mid }<4\delta$ uniformly in n.

In the strong case the limsup of $\mathrm{\mid }{d}_n(A,B)\mathrm{\mid }$ is at most $4\delta$. In the weak case the same bound holds for the limsup of its absolute Cesaro averages. Letting $\delta$ decrease to zero proves the full respective property. Conversely either full property restricts to the pairs in $\mathcal{P}$ by its definition.

The constant one has $L^2$ norm one. Complex Cauchy–Schwarz gives $\mathrm{\mid }\int f\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$ and $\mathrm{\mid }\int g\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_2$. Koopman isometry, iterated n times, gives $\mathrm{\parallel }f\circ T^n{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Applying the real-modulus Cauchy–Schwarz inequality to $\mathrm{\mid }f\circ T^n\mathrm{\mid }$ and $\mathrm{\mid }g\mathrm{\mid }$ proves absolute integrability of the product. Thus every term defining $C_n$ exists and is independent of representatives, with the indicated complex pairing.

For indicators $C_n({\mathbf{1}}_B,{\mathbf{1}}_A)=\mu (A\cap T^{-n}B)-\mu (A)\mu (B)$. For finite complex simple $a={\sum }_j{\alpha }_j{\mathbf{1}}_{B_j}$ and $b={\sum }_k{\beta }_k{\mathbf{1}}_{A_k}$, sesquilinearity gives $C_n(a,b)={\sum }_{j,k}{\alpha }_j\overline{{\beta }_k}{C}_n({\mathbf{1}}_{B_j},{\mathbf{1}}_{A_k})$. Therefore the set version of strong mixing gives convergence for simple pairs, and the weak version gives it for absolute Cesaro averages by the finite triangle inequality.

For any $L^2$ h,k, the two Cauchy–Schwarz estimates in step 1.1 give $\mathrm{\mid }{C}_n(h,k)\mathrm{\mid }\le 2\mathrm{\parallel }h{\mathrm{\parallel }}_2\mathrm{\parallel }k{\mathrm{\parallel }}_2$. Choose finite simple a,b with $\mathrm{\parallel }f-a{\mathrm{\parallel }}_2<\delta$ and $\mathrm{\parallel }g-b{\mathrm{\parallel }}_2<\delta$ using only the finite-simple clause of density. Then sesquilinearity gives $\mathrm{\mid }{C}_n(f,g)-C_n(a,b)\mathrm{\mid }\le 2(\mathrm{\parallel }f-a{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2+\mathrm{\parallel }a{\mathrm{\parallel }}_2\mathrm{\parallel }g-b{\mathrm{\parallel }}_2)\le 2\delta (\mathrm{\parallel }g{\mathrm{\parallel }}_2+\mathrm{\parallel }f{\mathrm{\parallel }}_2+\delta )\mathrm{.}$ The bound is independent of n.

Taking limsups of absolute values, or of their Cesaro averages, uses step 2.1 to remove the simple-pair term. Sending $\delta$ to zero proves the respective $L^2$ property. Conversely the $L^2$ property applies to indicators, which lie in $L^2$ because $\mu (X)=1$, and their identity in step 2.1 is precisely the set property.