Measure Preserving Transformations and Poincare Recurrence

Each $T^n$ is measurable and preserves $\mu$, so $W$ and every $T^{-n}W$ are measurable and $\mu (T^{-n}W)=\mu (W)$. Here $T^0$ is the identity.

For $0\le i<j$, membership of $x$ in both $T^{-i}W$ and $T^{-j}W$ would give $T^{i}x\in W\subseteq E$ and $T^{j-i}(T^{i}x)\in W\subseteq E$. This contradicts the absence of every positive return from $W$. Thus the tower sets are pairwise disjoint.

For every positive integer $N$, finite additivity and monotonicity give $N\mu (W)=\mu ({\bigcup }_{n=0}^{N-1}{T}^{-n}W)\le \mu (X)$. Since the right side is finite, a positive $\mu (W)$ would violate this bound for an integer $N>\mu (X)\mathrm{/}\mu (W)$. Hence $\mu (W)=0$, and step 1.1 makes every tower level null. This also covers $E=\mathrm{\varnothing }$ and $\mu (X)=0$.

Put $W=E\setminus {\bigcup }_{n\ge 1}{T}^{-n}E$ and $N={\bigcup }_{m\ge 0}{T}^{-m}W$. Every set in this union is measurable and null, whence $N$ is measurable and $\mu (N)=0$.

If $x\in E$ has only finitely many positive return times, the finite nonempty set $\{m\ge 0:T^{m}x\in E\}$ has a largest member $q$ (it contains zero). There is no positive return to $E$ from $T^{q}x$, so $T^{q}x\in W$ and $x\in N$. Thus every $x\in E\setminus N$ returns infinitely often. Conversely a point of $E\cap T^{-q}W$ has no visit after $q$, so the exceptional set is exactly $E\cap N$ and is measurable. The argument includes the case of no positive return by taking $q=0$.

For each basis member $B$ define its exceptional set explicitly as $N_B=B\cap {\bigcup }_{q\ge 0}{T}^{-q}(B\setminus {\bigcup }_{n\ge 1}{T}^{-n}B)$. The recurrence proof shows that $N_B$ is measurable and null. Thus $N={\bigcup }_{B\in \mathcal{B}}{N}_B$ is measurable and null. This is a prescribed family, not a choice of null covers. A finite basis is handled by a finite union, and the empty space has no points to check.

If $x\mathrm{\notin }N$ and $U$ is a neighborhood of $x$, choose an open set $V$ with $x\in V\subseteq U$ and a basis member $B$ with $x\in B\subseteq V$. Since $x\mathrm{\notin }{N}_B$, infinitely many positive iterates enter $B$, and hence $U$.

In the metric case set $n_0=0$ and let $n_j$ be the least integer exceeding $n_{j-1}$ for which $d(T^{n_j}x,x)<1\mathrm{/}j$, for $j\ge 1$. Infinitely many visits to the ball make this set nonempty. Least-element recursion supplies the sequence without countable choice; $n_j\ge j$ and the displayed bound proves convergence.

For $n\ge 1$, the fiber $H_n=\{x\in E:r_E(x)=n\}=E\cap T^{-n}E\cap {\bigcap }_{1\le j<n}{T}^{-j}(X\setminus E)$ is measurable. At $n=1$ the empty intersection is $X$. The fiber at infinity is $E\setminus {\bigcup }_{n\ge 1}{H}_n$. Every inverse image of a subset of the countable target is a countable union of these fibers, proving measurability of $r_E$.

For $B\in \mathcal{A}{\mathrm{\mid }}_{E_{\mathrm{\infty }}}$ the measurability of $E_{\mathrm{\infty }}$ implies $B\in \mathcal{A}$. The identity $T_E^{-1}B={\bigcup }_{n\ge 1}(E_{\mathrm{\infty }}\cap H_n\cap T^{-n}B)$ expresses its inverse image as a measurable subset of $E_{\mathrm{\infty }}$, hence a trace set. Empty B gives an empty union of pieces; full B gives all of the core.

Fix a trace set $B\subseteq E_{\mathrm{\infty }}$; it is ambient measurable. Write $H_n=E\cap T^{-n}B\cap {\bigcap }_{j=1}^{n-1}{T}^{-j}{E}^c$ and $R_N=T^{-N}B\cap {\bigcap }_{j=0}^{N-1}{T}^{-j}{E}^c$. Pulling B back once and splitting at E gives $\mu (B)=\mu (H_1)+\mu (R_1)$. Pulling $R_N$ back and splitting at E gives $\mu (R_N)=\mu (H_{N+1})+\mu (R_{N+1})$. Thus for each $N\ge 1$, $\mu (B)={\sum }_{n=1}^N\mu (H_n)+\mu (R_N)$.

The H_n are disjoint first-return pieces. Each differs from $H_n\cap E_{\mathrm{\infty }}$ by a subset of the measurable null set $E\setminus E_{\mathrm{\infty }}$; these differences are themselves measurable. Consequently the finite-sum identity implies $\mu ({\bigcup }_{n=1}^N(H_n\cap E_{\mathrm{\infty }}))\le \mu (B)$. Passing to the increasing union gives $\mu (T_E^{-1}B)\le \mu (B)$.

Apply the same inequality to $C=E_{\mathrm{\infty }}\setminus B$. Since $T_E$ maps its entire domain to itself, $T_E^{-1}C=E_{\mathrm{\infty }}\setminus T_E^{-1}B$. All measures here are finite, so $\mu (E_{\mathrm{\infty }})-\mu (T_E^{-1}B)\le \mu (E_{\mathrm{\infty }})-\mu (B)$ gives the reverse inequality. Equality follows for every B; division by $0<\mu (E)<\mathrm{\infty }$ proves invariance of ${\mu }_E$.

Let $U={\bigcup }_{n\ge 0}{T}^{-n}{E}_{\mathrm{\infty }}$. It is measurable and conull by positive-set sweep-out. It is strictly invariant: if Tx eventually enters the core then x does; if x eventually enters at a positive time then Tx does, while if x is already in the core, its next positive return belongs to the core. Thus $T^{-1}U=U$. On U let $q(x)=\min \{n\ge 0:T^{n}x\in E_{\mathrm{\infty }}\}$. The fibers $\{q=n\}=T^{-n}{E}_{\mathrm{\infty }}\setminus {\bigcup }_{j<n}{T}^{-j}{E}_{\mathrm{\infty }}$ are measurable.

Take any finite real measurable f on the core with $f\circ T_E=f$ everywhere. Define $F(x)=f(T^{q(x)}x)$ for $x\in U$ and $F(x)=0$ otherwise. The measurable q-fibers make F measurable. If $x\in U\setminus E_{\mathrm{\infty }}$, then $q(Tx)=q(x)-1$, hence F(Tx)=F(x). If $x\in E_{\mathrm{\infty }}$, then $q(Tx)=r_E(x)-1$: before the first return there is no E visit, and the first return is in the core. Therefore $F(Tx)=f(T_{E}x)=f(x)=F(x)$. On $X\setminus U$, both x and Tx are outside U and F is zero.

Ergodicity of T and the everywhere invariant-function criterion give a constant c with F=c almost everywhere. Since F=f on the core, f=c for ${\mu }_E$-almost every point there. The same criterion applied to the probability-preserving T_E now proves its ergodicity. Working with everywhere invariant functions avoids choosing representatives for almost-everywhere invariance.

Put $C_0=X$ and $C_j={\bigcap }_{i=0}^{j-1}{T}^{-i}{E}^c$ for $j\ge 1$. Since $T^{-1}{C}_j={\bigcap }_{i=1}^j{T}^{-i}{E}^c$, its disjoint split at E consists of $E\cap \{r_E>j\}$ and $C_{j+1}$. Indicator invariance and finite additivity therefore give $\mu (E\cap \{r_E>j\})=\mu (C_j)-\mu (C_{j+1})$ for $j\ge 0$. In particular at j=0 this is $\mu (E)=1-\mu (E^c)$.

For $N\ge 1$, the simple function ${\sum }_{j=0}^{N-1}{\mathbf{1}}_{E\cap \{r_E>j\}}$ equals $\min (r_E,N)$ on E and zero elsewhere. Its integral telescopes to $1-\mu (C_N)$. The sets C_N decrease to points avoiding E at all nonnegative times, a null set by sweep-out. Since $\mu (C_0)=1$, continuity from above yields $\mu (C_N)\to 0$.

The tail sums increase pointwise to $r_E{\mathbf{1}}_E$, including infinite return times. Monotone convergence gives ${\int }_E{r}_{E}d\mu =1$. The complement of the infinitely returning core in E is measurable null, so its nonnegative integral is zero, even where $r_E=\mathrm{\infty }$. Restriction and normalization therefore give ${\int }_{E_{\mathrm{\infty }}}{r}_{E}d{\mu }_E=1\mathrm{/}\mu (E)$; multiplying by the positive denominator reverses the equivalence.

First let $0\le f\le M<\mathrm{\infty }$. Put $A_N={\bigcap }_{j=1}^N{T}^{-j}{E}^c$. Splitting the integral at E and pulling the complement part back gives $\int f={\int }_{E}f+\int f\circ T{\mathbf{1}}_{A_1}$. More generally pulling back the part of $\int f\circ T^N{\mathbf{1}}_{A_N}$ outside E gives the next remainder, while its part on E is ${\int }_{E}f\circ T^N{\mathbf{1}}_{\{r_E>N\}}$. Thus for N>=1 the exact finite identity is $\int f={\int }_E{\sum }_{j=0}^{N-1}f\circ T^j{\mathbf{1}}_{\{r_E>j\}}+\int f\circ T^N{\mathbf{1}}_{A_N}$. All products are defined piecewise as zero off their indicated sets.

The sets $A_N$ decrease, and their intersection is the complement of ${\bigcup }_{j\ge 1}{T}^{-j}E$. This intersection is null by [F1]. Since the ambient probability measure is finite, [F6] gives $\mu (A_N)\to 0$. The remainder is bounded between zero and $M\mu (A_N)$ and hence vanishes. The excursion partial sums on E increase pointwise to S_Ef; monotone convergence proves the identity for bounded nonnegative f.

For arbitrary nonnegative f use $f_m=\min (f,m)$. These increase pointwise to f, and $S_E{f}_m\uparrow S_{E}f$: one inequality is termwise, and the other follows by first restricting to any finite number of excursion terms and then increasing m. Monotone convergence on X and E proves the equality, including infinite values.

For integrable real or complex f, apply step 3.1 to $\mathrm{\mid }f\mathrm{\mid }$: ${\int }_E{S}_E\mathrm{\mid }f\mathrm{\mid }=\int \mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$. Thus $S_E\mathrm{\mid }f\mathrm{\mid }$ is finite a.e. (otherwise its integral is at least $K$ times the positive measure of its infinity set for every K). On this conull set the excursion is absolutely summable. Apply step 3.1 to the positive and negative parts of the real and imaginary components, each bounded by $\mathrm{\mid }f\mathrm{\mid }$. Their excursion integrals are finite, so linearity yields the asserted identity without subtracting infinities. Null-set values have no effect.

Let $a=\{\alpha \}$. For any Borel $B\subseteq [0,1)$, $R_{\alpha }^{-1}B=((B\cap [a,1))-a)\cup ((B\cap [0,a))+(1-a))$. The two pieces lie respectively in $[0,1-a)$ and $[1-a,1)$ and are disjoint. Translation invariance and additivity give $\lambda (R_{\alpha }^{-1}B)=\lambda (B\cap [a,1))+\lambda (B\cap [0,a))=\lambda (B)$. For a=0 the second piece is empty.

In particular this proves the inverse-image identity on the pi-system of half-open intervals and the whole circle. These generate the circle Borel sigma-algebra: ordinary open intervals are countable unions of half-open subintervals, and the Borel sigma-algebras agree by the circle definition. The whole-space measure is one, so the generating criterion applies (or directly use the identity for every Borel B in step 1.1). The completion clause extends preservation to all completed sets. The countable-choice use is exactly the earlier Lebesgue and completion construction.

Fractional-part arithmetic gives $R_{-\alpha }{R}_{\alpha }x=x=R_{\alpha }{R}_{-\alpha }x$. The same argument applies to $-\alpha$, so the inverse is measurable for either sigma-algebra. Thus invertibility here includes measurability of the inverse.

For an integer $N\ge 2$, place $\{j\alpha \}$, $0\le j\le N$, in the N half-open intervals $[k\mathrm{/}N,(k+1)\mathrm{/}N)$. Two indices i<j lie in one interval. Hence for q=j-i, $1\le q\le N$ and $0<d(\{q\alpha \},0)<1\mathrm{/}N$; positivity follows from irrationality. Change q to -q if needed to obtain a subgroup point $\beta \in (0,1\mathrm{/}N)$.

The subgroup contains $0,\beta ,2\beta ,\dots ,m\beta$, where $m=\lfloor 1\mathrm{/}\beta \rfloor$; the last term is interpreted modulo one if necessary. Consecutive gaps are beta and the remaining gap to 1 is at most beta. For every x in [0,1), taking $k=\lfloor x\mathrm{/}\beta \rfloor$ gives $0\le x-k\beta <\beta$. Since N can be arbitrarily large, every circle neighborhood meets the subgroup.

For fixed M>=1 let $\delta ={\min }_{1\le q\le M}d(\{q\alpha \},0)>0$. Take N with $1\mathrm{/}N<\min (\delta ,\epsilon )$ and use the positive q from step 1.1 before changing its sign. That q exceeds M and has distance less than epsilon. For M=0 any q supplied there works after taking $1\mathrm{/}N<\epsilon$. Only finite minima and finite pigeonhole choices occur.

If $\alpha =p\mathrm{/}q$ with integers p and q>=1, the set $A={\bigcup }_{k=0}^{q-1}[k\mathrm{/}q,k\mathrm{/}q+1\mathrm{/}(2q))$ is Borel, has measure $1\mathrm{/}2$, and is permuted by R_alpha. Thus $R_{\alpha }^{-1}A=A$ and the rotation is not ergodic for either measure. This includes integral alpha with q=1.

Now let alpha be irrational and let A be a strictly invariant measurable set. Suppose both A and its complement have positive measure. Choose density-one points a in A and c in its complement, different from the cut point zero. For sufficiently small $0<r<1\mathrm{/}10$, the circle balls I=B(a,r), J=B(c,r) have length 2r and satisfy $\lambda (I\setminus A)<r\mathrm{/}5$ and $\lambda (J\cap A)<r\mathrm{/}5$. The ordinary density theorem applies because these small balls at the chosen points do not cross the cut.

Choose an integer m with $d(R_{\alpha }^{m}a,c)<r\mathrm{/}10$. Strict invariance and invertibility give $R_{\alpha }^{m}A=A$, and the rotated interval I has a portion outside A of measure less than r/5. Its intersection with J has length greater than $2r-r\mathrm{/}10=19r\mathrm{/}10$: both are radius-r arcs whose centers are less than r/10 apart. Subtracting the two exceptional portions, this intersection would contain points in both A and its complement on a set of measure at least $19r\mathrm{/}10-2r\mathrm{/}5=3r\mathrm{/}2>0$, impossible. Thus every strict invariant A has measure zero or one, proving ergodicity. Together with step 1.1 this proves both directions.

For $0\le a<c\le 1$, $D_b^{-1}[a,c)={\bigcup }_{j=0}^{b-1}[(a+j)\mathrm{/}b,(c+j)\mathrm{/}b)$, with disjoint pieces of length $(c-a)\mathrm{/}b$. Thus their total measure is c-a. The empty interval has empty inverse image. The map is Borel measurable by the Lipschitz bound from its definition.

The half-open intervals together with the empty set form a pi-system containing [0,1) and generating the circle Borel sets, as in the circle definition. Its whole-space mass is one, so the finite generator theorem gives Borel preservation; the completion theorem gives completed preservation. Countable choice is inherited by the Lebesgue, dilation and completion suppliers and is assumed here.

For every y in [0,1), the explicit preimage y/b lies in [0,1) and maps to y, proving surjectivity. The distinct points 0 and 1/b both map to 0, proving non-injectivity. Continuity is the already established inequality $d(D_{b}x,D_{b}y)\le bd(x,y)$.

The b-adic intervals at all depths, together with the empty set, form a pi-system: two such intervals are nested or disjoint, and depth zero is X. They generate the circle Borel sets. Indeed every ordinary open interval in (0,1) is the union of the b-adic cells whose closures lie within it; the cell containing any specified interior point has arbitrarily small diameter. The endpoint zero is the intersection of $[0,b^{-r})$ over r, so relative interval-open sets are generated as well. Conversely each cell is Borel. The countable family of cells permits these unions in the generated sigma-algebra.

Let $I=I_{r,k}$ and $J=I_{s,l}$. For $n\ge r$, the interval I contains exactly $b^{n-r}$ depth-n cells. On each such cell, the inverse image under $D_b^n$ of J is a half-open interval of length $b^{-(n+s)}$. Hence $\lambda (I\cap D_b^{-n}J)=b^{n-r}{b}^{-(n+s)}=b^{-r}{b}^{-s}=\lambda (I)\lambda (J)$. The same equality holds when either set is empty. The generator theorem proves Borel strong mixing.

For completed A,B take Borel A_0,B_0 and Borel null sets Z_A,Z_B containing their respective symmetric differences. For every n, $(A\cap D_b^{-n}B)\mathrm{△}(A_0\cap D_b^{-n}{B}_0)\subseteq Z_A\cup D_b^{-n}{Z}_B$, a completed null set by preservation. Thus the correlation and both marginal measures agree with those for A_0,B_0. The Borel limit proves the completed limit. Countable choice is inherited from the measure and completion suppliers; only finitely many representatives are selected here.

By the definitions, $D=D_2$. Since 2 is an allowed integer base, the base-map theorem proves preservation on both sigma-algebras and continuity. Its countable-choice measure and completion hypothesis is the assumption here.

Explicitly $D^{-1}[a,c)=[a\mathrm{/}2,c\mathrm{/}2)\cup [(a+1)\mathrm{/}2,(c+1)\mathrm{/}2)$ for $0\le a<c\le 1$; the two pieces have total length c-a. For each y, y/2 is a preimage, while $D(0)=D(1\mathrm{/}2)=0$ exhibits failure of injectivity. These branch identities also show why preservation concerns inverse images.

Let A be strictly invariant: $D^{-1}A=A$. Then $(D^n{)}^{-1}A=A$ for every n. On $I_{n,k}=[k\mathrm{/}{2}^n,(k+1)\mathrm{/}{2}^n)$ the map $D^n$ is $x\mapsto 2^{n}x-k$, so $A\cap I_{n,k}=(A+k)\mathrm{/}{2}^n$. Translation and dilation give $\lambda (A\cap I_{n,k})=2^{-n}\lambda (A)=\lambda (I_{n,k})\lambda (A)$.

If $\lambda (A)>0$, choose a density-one point $x\in A\cap (0,1)$. Let I_n be its unique half-open depth-n cell and put h_n=2^{-n}. This cell lies in the centered interval $(x-h_n,x+h_n)$ apart from possibly an endpoint of measure zero. Thus $\lambda (I_n\setminus A)\mathrm{/}{h}_n\le 2\lambda ((x-h_n,x+h_n)\setminus A)\mathrm{/}(2h_n)\to 0$. But step 1.1 makes the left side identically $1-\lambda (A)$, so $\lambda (A)=1$. Every strict invariant set is therefore null or conull, and the invariant-set criterion proves ergodicity. Countable choice is used in the measure, scaling and density suppliers.

The definition gives D=D_2. Taking b=2 in the integer-base theorem, with exactly its countable-choice assumption, yields the stated correlation limit for every Borel pair and every completed pair.

For clarity, if A and B are dyadic intervals of depths r and s, the proof gives the exact value $2^{n-r}{2}^{-(n+s)}=2^{-r-s}=\lambda (A)\lambda (B)$ for every n>=r. This includes the depth-zero whole circle; an empty test set gives zero. Thus the specialization retains the explicit dyadic correlation calculation as well as the general measurable-set conclusion.

Positive definiteness and symmetry follow immediately from the first disagreement. If x,y agree before index r and y,z agree before index s, then x,z agree before min(r,s). Thus $d(x,z)\le \max (d(x,y),d(y,z))\le d(x,y)+d(y,z)$, with equal pairs included. A prefix cylinder fixing indices 0 through m-1 is the open ball of radius $2^{-(m-1)}$ about any of its points when m>=1. Arbitrary finite-coordinate cylinders are finite unions of sufficiently long prefixes. Conversely prefixes of arbitrarily small diameter fit in every ball. Hence cylinders generate exactly the metric topology and are clopen.

Fix an open cover with no finite subcover. The empty prefix cylinder has no finite subcover. If a prefix cylinder has no finite subcover, at least one of its two children also lacks one; otherwise the union of the two finite subcovers would cover it. Choose the zero child if it lacks a finite subcover, and the one child otherwise. This is a uniquely specified recursion, producing a binary sequence x whose every prefix cylinder lacks a finite subcover.

Some member U of the cover contains x. Since U is open, it contains a prefix cylinder about x. That cylinder has the one-element subcover {U}, contradicting its construction. Therefore no open cover without a finite subcover exists, proving compactness. The recursion makes no arbitrary infinite choices; neither Tychonoff nor dependent choice is used.

Let $A={\bigcup }_{n\ge 0}{A}_n$ with disjoint $A_n\in \mathcal{C}$ and $A\in \mathcal{C}$. All these sets are clopen, being finite unions of clopen cylinders. The family consisting of $\mathrm{\Omega }\setminus A$ and all A_n is an open cover of Omega. Compactness gives finitely many members covering Omega and hence finitely many A_n covering A. Using their least indices if the same member repeats gives a finite index set F with $A={\bigcup }_{n\in F}{A}_n$. This uses ambient compactness only of Omega itself, not an unstated compact-subset criterion.

For every n outside F, disjointness gives $A_n\subseteq A\setminus {\bigcup }_{j\in F}{A}_j=\mathrm{\varnothing }$. Thus all other terms have zero content, and finite additivity gives $p_0(A)={\sum }_{n\in F}{p}_0(A_n)={\sum }_{n\ge 0}{p}_0(A_n)$. If A is empty every A_n is empty and the same identity is zero equals zero. Along with $p_0(\mathrm{\varnothing })=0$ and $p_0(\mathrm{\Omega })=1$, this is precisely a finite premeasure.

The prefix cylinders form a countable base: length m has exactly $2^m$ possible prefixes, enumerated by the binary integers from 0 to $2^m-1$, and the pairs of length and index admit a diagonal enumeration. Every cylinder is open, and every open set is the union of the subfamily of prefix cylinders contained in it. It follows that the sigma-algebra generated by the cylinder algebra is exactly the metric Borel sigma-algebra.

Apply the extension theorem to the premeasure p_0. It gives a Borel measure p agreeing with every cylinder mass and with $p(\mathrm{\Omega })=p_0(\mathrm{\Omega })=1$. Any other measure with these cylinder masses agrees on finite disjoint unions by finite additivity, hence on the entire algebra; finite-premeasure uniqueness makes it equal to p.

The completion theorem gives a complete probability extending p, since the whole-space mass remains one. Countable choice is used by the cited extension and completion constructions; the preceding finite-algebra and compactness arguments themselves were choice-free. This constructs this particular binary measure, not a general infinite product measure.

For a cylinder C prescribed on F, ${\sigma }^{-1}C$ imposes the same values on $F+1$. It is a cylinder with the same number of fixed coordinates and the same mass. Inverse images of prefix cylinders are open, so sigma is continuous and hence Borel measurable. Cylinders together with the empty set form a generating pi-system containing Omega. Since its mass is one, the preservation criterion applies; the completion clause then gives completed preservation.

If C and H prescribe finite coordinate sets F and G, then ${\sigma }^{-n}H$ prescribes G+n. For all sufficiently large n these are disjoint from F, so their intersection is a cylinder prescribing $\mathrm{\mid }F\mathrm{\mid }+\mathrm{\mid }G\mathrm{\mid }$ coordinates. Its mass is $2^{-\mathrm{\mid }F\mathrm{\mid }-\mathrm{\mid }G\mathrm{\mid }}=p(C)p(H)$. Empty sets give zero correlations. The mixing criterion now proves mixing for arbitrary Borel pairs.