Measure Preserving Transformations and Poincare Recurrence — Examples

Write $E_j=[j\mathrm{/}q,j\mathrm{/}q+1\mathrm{/}(2q))$. For $x=j\mathrm{/}q+t$ with $0\le t<1\mathrm{/}(2q)$, one has $R_{p\mathrm{/}q}x=((j+p)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q)\mathrm{/}q+t$. The residue map $j\mapsto j+p<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q$ is a permutation, with inverse subtraction of $p$. Thus rotation maps the collection of $E_j$ onto itself and, being bijective, satisfies $R_{p\mathrm{/}q}^{-1}E=E$. The half-open convention makes the formula exact at every included left endpoint and excludes every right endpoint.

The $q$ intervals $E_j$ are pairwise disjoint and each has length $1\mathrm{/}(2q)$. Hence $\lambda (E)={\sum }_{j=0}^{q-1}1\mathrm{/}(2q)=1\mathrm{/}2$. It is Borel, and its complement also has measure $1\mathrm{/}2$. By [F1] the ambient system preserves the probability, so [F2] proves nonergodicity. For $q=1$, this is simply the identity rotation and $E=[0,1\mathrm{/}2)$. Countable choice is used only for the Lebesgue probability supplied by [F1].

On $[0,1\mathrm{/}2)$, $D(x)=2x$, and $0\le 2x<1\mathrm{/}2$ is equivalent to $0\le x<1\mathrm{/}4$. On $[1\mathrm{/}2,1)$, $D(x)=2x-1$, and $0\le 2x-1<1\mathrm{/}2$ is equivalent to $1\mathrm{/}2\le x<3\mathrm{/}4$. These two branches exhaust the circle and their solution intervals are disjoint. Each interval has length $1\mathrm{/}4$, so their union has measure $1\mathrm{/}2$, agreeing with the preservation in [F1].

For $x\in [0,1\mathrm{/}2)$, the image $2x$ ranges through every $y\in [0,1)$, with inverse $x=y\mathrm{/}2$. Thus the forward image has measure one although the source has measure $1\mathrm{/}2$. In particular inverse-image preservation does not assert equality of forward-image measures. The included points 0 and 1/2 map to 0, and the excluded right endpoints 1/4 and 3/4 map to 1/2. Countable choice is inherited from the Lebesgue probability in [F1].

For $x\in [a\mathrm{/}3,(a+1)\mathrm{/}3)$ with $a=0,1,2$, [F1] gives $D_{3}x=3x-a$. The condition $0\le 3x-a<1\mathrm{/}3$ is equivalent to $a\mathrm{/}3\le x<a\mathrm{/}3+1\mathrm{/}9$. Substitution of the three values of $a$ gives exactly the three stated disjoint intervals. Their lengths add to $3(1\mathrm{/}9)=1\mathrm{/}3$, as required by [F2].

The interval $I_{2,3a+c}=[a\mathrm{/}3+c\mathrm{/}9,a\mathrm{/}3+(c+1)\mathrm{/}9)$ lies in branch $a$. Its affine image under $3x-a$ is $[c\mathrm{/}3,(c+1)\mathrm{/}3)=I_{1,c}$. Surjectivity onto that interval is explicit: for $y\in I_{1,c}$ use $x=(y+a)\mathrm{/}3$, which lies in $I_{2,3a+c}$. The included left and excluded right endpoints are preserved by this increasing affine map, including $a=c=2$ where the right endpoint is the excluded point 1. No ambiguous choice of ternary expansion is required. Countable choice enters only in the Lebesgue measure statement of [F2].

The prefix $[0]$ prescribes the single value $x_0=0$, so [F1] gives $2^{-1}=1\mathrm{/}2$. The prefix $[01]$ prescribes $x_0=0,x_1=1$, so its mass is $2^{-2}=1\mathrm{/}4$. The prefix $[101]$ prescribes $x_0=1,x_1=0,x_2=1$, giving $2^{-3}=1\mathrm{/}8$. Repeated symbols do not reduce the number of distinct prescribed coordinates.

By [F2], membership in ${\sigma }^{-3}[10]$ prescribes $x_3=1,x_4=0$. Intersecting with $[01]$ therefore prescribes exactly coordinates $0,1,3,4$, with respective values $0,1,1,0$. Coordinate 2 and all later coordinates remain free. The cylinder mass is $2^{-4}=1\mathrm{/}16$ by [F1]. Equivalently this intersection is the disjoint union $[01010]\cup [01110]$, whose masses sum to $2(1\mathrm{/}32)=1\mathrm{/}16$. Countable choice is inherited from the measure construction in [F1], not from the finite coordinate count.

The set $E$ is Borel, with $\lambda (E)=1\mathrm{/}4$, and [F2] gives a measure-preserving system of total mass one. Applying [F1] with exactly this $E$ proves the stated almost-everywhere infinitely-many-positive-returns conclusion. Countable choice enters through [F2]; the recurrence theorem itself needs no choice axiom.

The two inverse branches give $D^{-1}E=[0,1\mathrm{/}8)\cup [1\mathrm{/}2,5\mathrm{/}8)$. Intersecting with $E$ leaves $[0,1\mathrm{/}8)$, whose measure is $1\mathrm{/}8$. This is the first-return-one set because there is no smaller positive time. Return times need not all be one: $1\mathrm{/}5\in E$ has successive images $2\mathrm{/}5,4\mathrm{/}5,3\mathrm{/}5,1\mathrm{/}5$, so its first positive return time is four. Zero is fixed and returns at every positive time. These calculations are compatible with recurrence; recurrence alone supplies no return-frequency value or every-point assertion.

The set $E$ is Borel and is the disjoint union of $[0,1\mathrm{/}2)$ and $\{1\mathrm{/}2\}$. The first has length $1\mathrm{/}2$ and the second is null by [F3], so $\lambda (E)=1\mathrm{/}2>0$. By [F2] irrationality gives ergodicity of the Lebesgue probability system. All hypotheses of [F1] hold; applying its first conclusion yields ${\int }_E{r}_{E}d\lambda =1$. This uses the return time to the closed set $E$ as stated, so no unproved comparison between return times for different endpoint conventions is involved.

The recurrent core has restricted mass $\lambda (E_{\mathrm{\infty }})=\lambda (E)=1\mathrm{/}2$ in [F1], and normalization divides restricted measure by $1\mathrm{/}2$. The second conclusion of [F1] therefore gives ${\int }_{E_{\mathrm{\infty }}}{r}_{E}d{\lambda }_E=1\mathrm{/}(1\mathrm{/}2)=2$. This is an average, not an assertion that all return times equal two. The unnormalized mean is one, and multiplying it by the normalization factor two gives the same answer. Countable choice is propagated from the Lebesgue ergodicity and endpoint-null suppliers [F2] and [F3].

For $0\le x<1$, the equation $D(x)=0$ means $2x$ is an integer. The only possibilities are $2x=0$ or $2x=1$, giving $D^{-1}\{0\}=\{0,1\mathrm{/}2\}$. Its symmetric difference with $E=\{0\}$ is exactly $\{1\mathrm{/}2\}$, a null measurable set by [F3]. Thus $E$ is invariant modulo null sets in the probability system of [F1], by [F2].

The point $1\mathrm{/}2$ belongs to $D^{-1}E$ and not to $E$, so the sets are not equal and $E$ is not strictly invariant by [F2]. Both sets are finite Borel sets; this is an exact failure for the given representative, despite their equality modulo null sets. Countable choice is used only through the Lebesgue measure and null-set suppliers [F1] and [F3].

The circle map $D(x)=\{2x\}$ fixes zero and exchanges $1\mathrm{/}3$ with $2\mathrm{/}3$. Thus ${\delta }_0(D^{-1}E)={\mathbf{1}}_E(D0)={\mathbf{1}}_E(0)={\delta }_0(E)$ for every Borel $E$. Every Borel set has ${\delta }_0$-measure either zero or one, so in particular every strictly invariant Borel set does; [F2] proves ergodicity for ${\delta }_0$. Ergodicity for $\lambda$ is [F1].

The finite weighted sum $\mu$ is a Borel probability by [F3] and the finite-sum interchange with nonnegative series. Its inverse-image mass is $\frac{1}{2}{\mathbf{1}}_E(0)+\frac{1}{4}{\mathbf{1}}_E(2\mathrm{/}3)+\frac{1}{4}{\mathbf{1}}_E(1\mathrm{/}3)=\mu (E)$, so it too is invariant. Let $Z={\bigcup }_{n\ge 0}(D^n{)}^{-1}\{0\}$. Continuity of the iterates makes this Borel. Since zero is fixed, $x$ eventually reaches zero exactly when $Dx$ does; hence $D^{-1}Z=Z$. Zero belongs to $Z$ and neither point of the two-cycle does, so $\mu (Z)=1\mathrm{/}2$. This contradicts the ergodicity criterion in [F2]. The atomic calculations are choice-free; countable choice is used only to include the Lebesgue system of [F1].

By [F3] and [F4], $e^{2\pi i}=(e^{\pi i}{)}^2=1$, and the same holds for every integer multiple of $2\pi i$ by multiplication and inversion. Thus $f(x)=e^{2\pi ix}$ is well defined across the circle cut. By the Cartesian formula [F4], it equals $\cos (2\pi x)+i\sin (2\pi x)$ and is continuous by [F5] (the limit from below at 1 equals its value at 0), has $\mathrm{\mid }f\mathrm{\mid }=1$ by [F4], and belongs to both $L^1(\lambda )$ and $L^2(\lambda )$ because the circle has mass one. The addition law gives $f\circ R_{1\mathrm{/}2}=-f$. By [F6] and [F7], if $m=\int fd\lambda$ then $m=\int f\circ R_{1\mathrm{/}2}d\lambda =-m$, so $m=0$.

For every $n\ge 0$, the addition law and integer-period identity give $f(R_{\alpha }^{n}x)=e^{2\pi in\alpha }f(x)$. Hence $\int (f\circ R_{\alpha }^n)\bar{f}d\lambda =e^{2\pi in\alpha }\int \mathrm{\mid }f{\mathrm{\mid }}^{2}d\lambda =e^{2\pi in\alpha }$. The centered correlation of [F2] is the same because the mean is zero. Its modulus is one by [F4], so for every $N\ge 1$, $N^{-1}{\sum }_{n=0}^{N-1}\mathrm{\mid }{C}_n(f,f)\mathrm{\mid }=1$. It cannot tend to zero; the necessary implication in [F2] disproves weak mixing. Irrationality supplies ergodicity by [F1]. Countable choice is inherited from [F1] and [F6]; only the implication from weak mixing to vanishing absolute $L^2$ correlation averages in [F2] is needed. No Fourier-series completeness or spectral theorem is used.

By [F3] the restricted Lebesgue measure has $\lambda ([0,1])=1$ and $\lambda (E)=1\mathrm{/}2$. The identity is measurable and satisfies $T^{-1}A=A$ for every Borel $A$, so it preserves this probability. In particular $E$ is strictly invariant of measure $1\mathrm{/}2$, and [F4] shows that the system is not ergodic. Countable choice is inherited from the Lebesgue measure in [F3].

For each $x\in E$ and each positive integer $n$, $T^{n}x=x\in E$. Thus the least positive return time in [F1] is 1 and the infinitely-returning core is all of $E$. The integral of the constant one over $E$ is $\lambda (E)=1\mathrm{/}2$. The normalized restricted probability has total mass one, so the same constant return time has mean one there. Both values differ from the respective ergodic conclusions of [F2], namely one before normalization and $1\mathrm{/}(1\mathrm{/}2)=2$ afterwards. All endpoints return as well, so no exceptional-point convention is involved.

By [F3], $\log 2>\log 1=0$, so $h$ is positive, bounded by $1\mathrm{/}\log 2$, and continuous on $[0,1]$. By [F4] it is Riemann integrable on every closed subinterval. The derivative of $\log (1+x)$ is $1\mathrm{/}(1+x)$ by [F2], using the translated difference quotient, also one-sided at subinterval endpoints. Thus [F5] and [F6] yield ${\int }_{[a,b]}hd\lambda =[\log (1+b)-\log (1+a)]\mathrm{/}\log 2$ for $0\le a<b\le 1$. Singletons and countable sets have zero density integral because they are null by [F7] and $h$ is bounded. Consequently endpoints do not change this interval value. By [F1] the density defines a Borel measure, and the value with $a=0,b=1$ is one; removing the endpoint 1 does not change it.

The Borel sets $J_n=(1\mathrm{/}(n+1),1\mathrm{/}n]\cap X$, $n\ge 1$, partition $(0,1)$. On $J_n$, $G(x)=1\mathrm{/}x-n$, and on the singleton $\{0\}$ it is zero. Each branch is the restriction of a continuous real function and takes values in $[0,1)$, so for every open subset of $X$ its inverse image is a countable union of Borel branch inverse images and possibly $\{0\}$. This proves Borel measurability. For $0<s<1$, the exact inverse image is $G^{-1}[0,s]=\{0\}\cup {\bigcup }_{n\ge 1}([1\mathrm{/}(n+s),1\mathrm{/}n]\cap X)$. The displayed intervals are pairwise disjoint because $1\mathrm{/}(n+s)>1\mathrm{/}(n+1)$. The only endpoint outside $X$ is 1 when $n=1$.

Using the interval integral of step 1.1 and countable additivity, $\mu (G^{-1}[0,s])$ is $1\mathrm{/}\log 2$ times the sum over $n\ge 1$ of $\log (1+1\mathrm{/}n)-\log (1+1\mathrm{/}(n+s))$. The quotient law [F3] and the identity $(1+1\mathrm{/}n)\mathrm{/}(1+1\mathrm{/}(n+s))=(1+s\mathrm{/}n)\mathrm{/}(1+s\mathrm{/}(n+1))$ rewrite the partial sum through $M$ as $\log (1+s)-\log (1+s\mathrm{/}(M+1))$. All original summands are nonnegative. Continuity of $\log$ at 1 gives the limit $\log (1+s)$, so $\mu (G^{-1}[0,s])=\log (1+s)\mathrm{/}\log 2=\mu ([0,s])$. For $s=0$, the preimage is $\{0\}\cup \{1\mathrm{/}n:n\ge 2\}$, an explicitly enumerated countable null set, and both masses are zero. The full space $X$ also has equal inverse-image mass one.

The family consisting of $X$, the empty set, and all $[0,s]$ with $0\le s<1$ is a pi-system. It generates the Borel sets of $X$: complements give $(s,1)$, and increasing unions of initial closed intervals give $[0,b)$; intersections give ordinary open intervals, which form a countable rational-endpoint base for the interval topology. Conversely all generators are Borel. The measure is finite and $G$ is measurable, so [F8] applies to step 2.1 and proves preservation for every Borel set. By [F9] the map is measurable and preserving on the completion as well. Countable choice is inherited in the Lebesgue/Riemann comparison, null-set and completion suppliers; the branch sums and partial-sum telescoping make no choices. The rational-base assertion uses [F10]. The rational-base assertion uses [F11].

The corresponding inverse-branch density calculation can also be seen directly. On $0<y<1$, the inverse branch is ${\eta }_n(y)=1\mathrm{/}(n+y)$, with $\mathrm{\mid }{\eta }_n^{\mathrm{\prime }}(y)\mathrm{\mid }=1\mathrm{/}(n+y{)}^2$. Hence $h({\eta }_n(y))\mathrm{\mid }{\eta }_n^{\mathrm{\prime }}(y)\mathrm{\mid }=1\mathrm{/}[\log 2(n+y)(n+y+1)]$. The algebraic identity $1\mathrm{/}[(n+y)(n+y+1)]=1\mathrm{/}(n+y)-1\mathrm{/}(n+y+1)$ gives the partial sum $[1\mathrm{/}(1+y)-1\mathrm{/}(M+1+y)]\mathrm{/}\log 2$, tending to $h(y)$. This verifies the density balance numerically; the interval proof in steps 1.1–3.1 already establishes measure preservation without assuming a change-of-variables theorem. The separate endpoint computations in step 2.1 account for $y=0$.