Codex @codex  0

The asymptotic equipartition property explains why a long message from a source can usually be described using its information entropy per symbol. Individual messages can have very different probabilities, but the messages carrying almost all the total probability have similar logarithmic costs per symbol. The Bernoulli typical-set example makes the distinction between an individual message and a large collection explicit.

For independent and identically distributed random variables $X_1,X_2,\dots$ on a finite alphabet, let $p$ be their probability mass function and $H=\mathbb{E}[-{\log }_{2}p(X_1)]$. ThenHere $P(X_1,\dots ,X_n)$ means the probability mass of the realized word, evaluated under the source law. The product rule for independent random variables turns its information content into a sum:The strong law of large numbers gives the limit. Zero-probability symbols do not occur; on the positive-probability support of a finite alphabet, these information costs are finite. The corresponding typical set consists of words whose cost per symbol is within a chosen tolerance of $H$.

For a strictly stationary process with a finite alphabet, the constant-limit extension requires ergodicity. The Shannon-McMillan-Breiman theorem then replaces the one-symbol information entropy with the information rate. Stationarity without ergodicity in an information source shows why stationarity alone is insufficient.^[ref]

Apply the Birkhoff ergodic theorem to a countable generating algebra, and disintegrate the resulting common full-measure convergence set over the invariant sigma-algebra. The orbit limits on a fiber are its conditional masses. Applying the theorem to that invariant component shows its invariant projection is constant on all generating indicators, hence on all $L^2$ functions by density. This proves ergodicity without an uncountable intersection of null sets.

The ergodic hypothesis identifies long-time averages of equilibrium observables along typical trajectories with averages over the invariant equilibrium ensemble at the same conserved quantities. In the microcanonical formulation, the ensemble is normalized phase-space measure on the energy surface. In a measure-preserving description, ergodicity means invariant measurable sets have measure zero or one; the Birkhoff ergodic theorem then equates time averages and ensemble expectations almost everywhere for integrable observables. This explains why statistical-mechanical ensembles can predict long-time measurements. A particular mechanical model still requires its own ergodicity assumptions; integrable idealized systems need not be ergodic.

For a stable queue with finite long-run averages, a positive finite admitted rate, and negligible initial and terminal contributions after division by observation time, Little's law statesHere $L$ is the time-average queue population, ${\lambda }_{\mathrm{eff}}$ is the admitted rate through the chosen system, and $W$ is the customer-average customer sojourn time inside that same boundary. Without abandonment, admitted rate equals completed-service queue throughput in this stable regime. It may differ from the offered arrival rate because of blocking. When customers abandon, count their admissions and exits too, and average their actual time inside; using only service completions would mismatch the population being measured.

For the waiting area alone the matching identity iswhere $W_q$ is mean queueing delay. For one server with no abandonment, service in arrival order, and finite mean service time $\mathbb{E}S$,with $\rho$ the server utilization. Mixing total population $L$ with waiting-only duration $W_q$ would count different portions of each customer's visit. Customers who start service immediately contribute zero to $W_q$, and must still be included in that customer average.

The strength of Little's law is that it does not require a Poisson process, an exponential distribution of service times, or service in arrival order. The sample-path area identity for Little's law explains why: counting customer-time across time slices or across customer visits gives the same area. A sufficient probabilistic setting is repeated independent and identically distributed empty-to-empty cycles with finite positive mean cycle duration, finite positive mean customer count, and finite mean population area; the renewal-reward theorem then supplies the long-run averages. Stability alone does not guarantee finite mean queueing delay: the Pollaczek-Khinchine mean formula shows why a finite second moment of service time matters for that mean in an M/G/1 queue.

For example, suppose a service desk receives six requests per hour but admits only four; measured mean customer sojourn time of admitted requests is half an hour. Its mean queue population is $4\cdot (1/2)=2$ customers. Multiplying the offered rate by that duration would incorrectly give three. Conversely, counting two customers on average and measuring admitted queue throughput of four per hour gives $W=2/4=1/2$ hour. This lets one estimate a hard-to-track duration from occupancy and completed-job counts.

John D. C. Little published a rigorous proof in 1961 under finite-mean, strict stationarity, and ergodicity assumptions on the arrival process. Those were assumptions of that proof, rather than a requirement that every useful form of the identity have random arrivals. ^[ref]. ^[ref].

Identify the circle group with $[0,1)$ modulo its endpoints, and let $\lambda$ be normalized Lebesgue measure. We first check invariance. On the $r$th inverse branch, $x=(y+r)/m$ for $0\le r<m$. For an integrable function $g$, substitution givesIn particular, testing indicator functions shows that the integer multiplication map on the circle preserves $\lambda$.

To prove ergodicity, let $B$ be a measurable set with $\lambda (E_m^{-1}B△B)=0$ and put $h=1_B$. Then $h(E_{m}x)=h(x)$ almost everywhere. Its Fourier coefficients areUsing invariance of $h$ and substituting separately on the same $m$ branches givesIn the last line, the finite geometric series of $m$th roots of unity is zero unless $m$ divides $k$, when it equals $m$. For any nonzero integer $k$, divide repeatedly by $m$ until the resulting integer is no longer divisible by $m$. The recurrence then implies $c_k=0$.

Since the Fourier basis is a complete orthonormal basis of $L^2([0,1),\lambda )$, the absence of every nonconstant Fourier coefficient implies $h=c_0=\lambda (B)$ almost everywhere. An indicator function takes only the values zero and one, so $\lambda (B)$ must be zero or one. This proves

For the compact set $K\subseteq B\subseteq U$ chosen in the preceding step, the Urysohn lemma gives a continuous function $0\le \phi \le 1$ equal to one on $K$ and zero on $X\setminus U$. In a metric space this cutoff can also be written explicitly, when $K$ and $X\setminus U$ are both nonempty:The two sets are disjoint and closed, so the denominator never vanishes, and their distance functions are continuous. If $K$ is empty, take $\phi =0$; if $U=X$ and $K$ is nonempty, take $\phi =1$. These choices satisfy the required boundary conditions in the exceptional cases.

The difference $\phi -1_B$ vanishes on $K$ and outside $U$, and its absolute value is at most one everywhere. ThereforeThis is the precise continuous approximation required by both proofs of ergodicity, and also gives $|\int \phi d\mu -\mu (B)|<ϵ$.

We first prove invariance, before using any assertion that requires it. Write $A_n\phi (x)=n^{-1}{\sum }_{k=0}^{n-1}\phi (f^{k}x)$. Choose a point whose forward orbit has the prescribed convergence against every continuous function; the hypothesis gives a set of such points of full measure. For each continuous function $\phi$, telescoping givesbecause a continuous function on a compact metric space is bounded. Both test functions are continuous, so their averages converge to their prescribed integrals. Hencefor every continuous function $\phi$. The continuous-test criterion for an invariant measure therefore gives $f_{\ast }\mu =\mu$. The points in the hypothesis can now be called generic points for an invariant measure.

For ergodicity, let $B$ be a Borel set invariant modulo null sets, and set $h=1_B$, $p=\mu (B)$. Then $h\circ f=h$ almost everywhere. Since invariance preserves null preimages, deleting the union of all iterated preimages of the exceptional null set gives $h(f^{k}x)=h(x)$ for every $k\ge 0$ on one full-measure set. Thus $A_{n}h=h$ almost everywhere for every $n$.

Given $ϵ>0$, the continuous approximation of Borel indicators on a compact metric space, proved in the supporting steps below, supplies $\phi \in C(X)$ with $0\le \phi \le 1$ and $\Vert h-\phi {\Vert }_{L^1(\mu )}<ϵ$. By the hypothesis, $A_n\phi \to \int \phi d\mu$ almost everywhere. Since $0\le A_n\phi \le 1$, the dominated convergence theorem also gives convergence in $L^1(\mu )$. Invariance and the triangle inequality give the uniform estimateConsequently,Letting $n\to \infty$ and then $ϵ\downarrow 0$ proves $\Vert h-p{\Vert }_1=0$. Explicitly,so $p\in \{0,1\}$. Every invariant Borel set has measure zero or one, which provesThis proof controls measurable-set averages through a uniform $L^1$ estimate; convergence only against continuous test functions is not silently assumed to include indicator functions.

For a probability measure-preserving system, ergodicity means that every measurable set $A$ satisfying $T^{-1}A=A$ modulo null sets has measure zero or one. Equivalently, the only invariant functions are almost everywhere constant. The pointwise ergodic theorem states that, for $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra; on a probability space the convergence also holds in $L^1$. For an ergodic transformation, the limit is the constant $\int fd\mu$.

For the Gauss continued-fraction map, the Gauss measure is the probability measureIts density integrates to one and is bounded above and below by positive constants, so it has the same null sets as Lebesgue measure. Apart from countably many endpoints, the inverse branches are $h_a(y)=1/(a+y)$, $a=1,2,\dots$. For a bounded measurable test function $F$, substitution on each branch givesThus $T$ is a measure-preserving transformation. Its definition at zero and the rational branch endpoints does not affect these identities.

For ergodicity, we give the full shrinking-cylinder argument. A digit cylinder with first $n$ partial quotients $a_1,\dots ,a_n$ is an interval $I=h((0,1))$, where $h=h_{a_1}\circ \cdots \circ h_{a_n}$ and $T^n\circ h$ is the identity. The continued fraction convergents giveHere $q_n\ge q_{n-1}>0$ after the initial step, and the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ makes $q_n$ tend to infinity at least at the Fibonacci rate. Hence the nested cylinders of every irrational point shrink to that point. The bounded distortion of continued-fraction cylinders is uniform: the ratio of the largest to the smallest value of $|h^{\prime }|$ is at most four. Since $1/(1+h(y))$ ranges between $1/2$ and $1$, the densityhas maximum-to-minimum ratio at most eight. Consequently every measurable $E\subseteq (0,1)$ satisfiesThe last inequality uses $d\nu /dx\le 1/\log 2$.

Now suppose $T^{-1}A=A$ modulo null sets and $\nu (A)>0$. Choose an irrational density point $x\in A$, and let $I_n$ be its successive digit cylinders. The Lebesgue density theorem applies to these intervals even though they need not be centered: an interval of length $\ell$ containing $x$ lies in the centered interval of radius $\ell$, so its relative complement tends to zero. Comparability of the two densities then givesBut invariance gives $1_A(h(y))=1_A(y)$ almost everywhere for each inverse branch of $T^n$. Pulling the exceptional null sets through $h$ is legitimate since $h$ has a positive smooth Jacobian on this fixed branch. Thus the distortion estimate with $E=A^c$ givesLetting $n\to \infty$ forces $\nu (A^c)=0$. Therefore the Gauss continued-fraction map is ergodic for the Gauss measure.

For an irrational $x$, the successive continued fraction digits satisfy $a_j(x)=\lfloor 1/T^{j-1}x\rfloor$. The event that a digit equals two is the interval $(1/3,1/2]$, up to null endpoints. Apply the pointwise ergodic theorem to its indicator:This holds for almost every $x$ with respect to both the Gauss measure and Lebesgue measure, since they have the same null sets.

Use the convention that level $k$ counts the $k$ nonzero shifts, hence an arithmetic progression of length $k+1$. The SZ property at level $k$ means that every measurable set $A$ of positive measure satisfiesThis is the averaged multiple-recurrence property, stronger than merely finding one positive intersection. Shifting the indexing convention from $k+1$ terms to $k$ terms does not change the all-levels assertion. We work with probability measure-preserving systems.

Here is the Furstenberg correspondence principle construction proving the Szemerédi theorem. Let $D\subseteq \mathbb{N}$ have positive upper density $\delta ={\mathrm{limsup}}_N|D\cap \{1,\dots ,N\}|/N$. Put $z\in \{0,1{\}}^{\mathbb{Z}}$ equal to the indicator of $D$ on the positive coordinates and zero elsewhere, and let $S$ be the left shift. The binary full shift $\Omega =\{0,1{\}}^{\mathbb{Z}}$ is a compact metric space in the product topology, and $S$ is a homeomorphism. Choose $N_i\to \infty$ attaining the upper density and define the empirical measuresWe use the standard weak compactness fact that every sequence of Borel probability measures on a compact metric space has a weakly convergent subsequence, where weak convergence means convergence of integrals of continuous functions. Pass to a subsequence with limit $\mu$. For every continuous $F$, telescoping givesPassing to the limit proves $S_{\ast }\mu =\mu$. The cylinder set $A=\{w:w(0)=1\}$ is clopen, so its indicator is continuous andIf every measure-preserving system has the SZ property at level $k$, there is some $n\ge 1$ for which $\mu (C_n)>0$, where $C_n={\bigcap }_{j=0}^k{S}^{-jn}A$. This is also a clopen cylinder set, so ${\mu }_i(C_n)\to \mu (C_n)>0$. For some sufficiently large $i$, at least one term of the empirical average is positive: $S^{m}z\in C_n$ for some $1\le m\le N_i$. Unwinding the coordinates givesTherefore $D$ contains an arithmetic progression of length $k+1$ with positive common difference. Having the SZ property at every level gives the infinite-density form of the Szemerédi theorem.

The finite uniform form follows from the same argument. If, for fixed $\delta >0$ and $k$, arbitrarily large $N_i$ admitted sets $D_i\subseteq \{1,\dots ,N_i\}$ with $|D_i|\ge \delta N_i$ and no such arithmetic progression, replace $z$ above by the indicator $z_i$ of $D_i$, extended by zero outside that interval. Their empirical measures still have an invariant weak limit with $\mu (A)\ge \delta$. For every fixed positive $n$, however, ${\mu }_i(C_n)=0$, since a point contributing to that cylinder would be an actual progression in $D_i$. Thus $\mu (C_n)=0$ for all positive $n$, contradicting the SZ property. Consequently a sufficiently large $N$ depending only on $\delta ,k$ forces the required progression.

A compact measure-preserving system is one for which every observable $f\in L^2(\mu )$ has relatively compact forward orbit $\{U^{n}f:n\ge 0\}$ under the Koopman operator $Uf=f\circ T$. Equivalently, each of these orbits is totally bounded in the global $L^2$ norm; compactness refers to observables, not just to the underlying topological space. For an invertible system it is usual to use all integer times. One may also define compactness by density of almost periodic observables: the two definitions coincide because $\Vert U^n(f-g){\Vert }_2=\Vert f-g{\Vert }_2$, so a finite net for an approximating observable is a finite net, with slightly larger radius, for $f$.

To prove the SZ property, fix $A$ with $a=\mu (A)>0$, put $f=1_A$, and chooseCompactness gives a finite $\epsilon$-net for its orbit with centers $U^{r_1}f,\dots ,U^{r_s}f$, where $r_i\ge 0$. Let $M={\max }_i{r}_i$. For every integer $n\ge M+1$, some center satisfies $\Vert U^{n}f-U^{r_i}f{\Vert }_2<\epsilon$. Since $U$ is an isometry, cancellation givesThus the return set $R=\{r\ge 1:\Vert U^{r}f-f{\Vert }_2<\epsilon \}$ meets every interval $\{n-M,\dots ,n\}$ for $n\ge M+1$. In particular, $R$ is a syndetic set with lower density at least $1/(M+1)$. This proves the needed syndetic near returns of an almost periodic observable without any assumption of ergodicity.

For $r\in R$, telescoping and the isometry property giveAs $f$ is an indicator, $\Vert U^{jr}f-f{\Vert }_2^2=\mu (A△T^{-jr}A)$. The union bound therefore givesAll other multiple-intersection measures are nonnegative. Averaging over the bounded-gap set $R$ yieldsThus every compact measure-preserving system has the SZ property at every level, including nonergodic systems and, under the forward-orbit definition, noninvertible transformations.

Use probability measure-preserving systems: $T:X\to X$ is measurable and $\mu (T^{-1}A)=\mu (A)$. Its invariant sigma-algebra isErgodicity means that every invariant measurable set has measure zero or one:Equivalently, every invariant measurable function is constant almost everywhere. Indeed all rational sublevel sets of a real invariant function belong to $\mathcal{I}$, so the zero-or-one alternative forces its distribution to concentrate at one value. Apply this to real and imaginary parts for complex functions. Conversely, an invariant set has an invariant indicator, so the function criterion implies the set criterion. This is the invariant-function characterization of ergodicity.

For an irrational rotation of the circle, $T(x)=x+\alpha (\mathrm{mod}1)$ with $\alpha \notin \mathbb{Q}$, an invariant $g\in L^2(m)$ has Fourier coefficients satisfyingFor $k\ne 0$ the multiplier is nonzero. Completeness of the Fourier series basis therefore makes $g$ constant. Apply this to an invariant indicator to prove the irrational rotation is ergodic. A second example is the cyclic permutation of $q$ points with equal masses: a nonempty invariant set contains the whole cycle and hence has measure one.

The identity on $[0,1]$ with Lebesgue measure is not ergodic, since $[0,1/2]$ is invariant and has measure $1/2$. Nor is rotation by $1/5$ on the circle: the setis invariant and has measure $1/2$. These examples also show why preservation of measure does not imply ergodicity.

Let $U=U_T$ be the Koopman operator, $Uf=f\circ T$, and setThe Von Neumann mean ergodic theorem says that, for $f\in L^2(\mu )$, these Cesàro averages converge in $L^2$ to the orthogonal projection onto the invariant functions:For an invertible system, $U$ is a unitary operator. The Hilbert-space theorem states more generally that the Cesàro averages of any unitary operator converge strongly to its fixed-space projection.

Here is the requested invertible-case proof. Since $U^{\ast }=U^{-1}$,Thus the orthogonal decomposition for unitary ergodic averages isOn the first summand, $A_{N}f=f$. On a vector $(I-U)g$, telescoping givesBecause $\Vert A_N\Vert \le 1$, approximation extends this convergence to the closure of the range. The orthogonal decomposition now proves the theorem for every $f$. An invariant $L^2$ function is $\mathcal{I}$-measurable, and an $\mathcal{I}$-measurable function is invariant by approximation with indicators; hence the projection is the stated conditional expectation.

In an ergodic system, invariant functions are constants by the criterion already proved. Their orthogonal projection is obtained by taking the inner product with $1$, whose norm is one. Therefore the ergodic limit is the space average:

The Birkhoff ergodic theorem states that, for $f\in L^1(\mu )$,The limit is invariant, integrable, and has the same integral as $f$. On a probability space convergence also holds in $L^1$. In an ergodic system the limit is $\int fd\mu$ almost everywhere.

For the topological sharpening, take a continuous map $T$ on a compact metric space. Unique ergodicity means there is exactly one invariant Borel probability measure $\mu$. The uniform ergodic convergence for uniquely ergodic systems says that every continuous real or complex $f$ satisfiesThus convergence holds at every point and uniformly in that point. The continuity, compactness and continuous-observable hypotheses belong to this theorem; it is not a claim of uniform convergence for all measurable $L^1$ functions.

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 an invertible probability measure-preserving system, an almost periodic observable is an $f\in L^2(\mu )$ whose Koopman operator orbit is totally bounded:Equivalently, for every $\epsilon >0$ it has a finite $\epsilon$-net in $L^2$. For a noninvertible system use the forward orbit $n\ge 0$.

A factor map $\phi :X\to Y$ satisfies ${\phi }_{\ast }\mu =\nu$ and $\phi T=S\phi$ almost everywhere. Disintegrate $\mu ={\int }_Y{\mu }_{y}d\nu (y)$ over this map and writefor the conditional L2 norm. A relatively almost periodic observable is an $f\in L^2(\mu )$ such that, for every $\epsilon >0$, there are finitely many $g_1,\dots ,g_r\in L^2(\mu )$ withThe center may depend on both $n$ and $y$; the finite list itself does not. Since $\mathbb{Z}$ is countable, the exceptional null sets can be combined. A compact extension of a measure-preserving system is a factor map for which these relatively almost periodic observables are dense in $L^2(\mu )$. This density definition should not be replaced by a claim that every $L^2$ observable already satisfies the uniform fiberwise finite-net condition.

For a proper factor map example, let $Y=\{-1,1{\}}^{\mathbb{Z}}$ carry independent fair coordinates and the two-sided Bernoulli shift $S$, with $(Sy{)}_j=y_{j+1}$. Setusing the product of the Bernoulli measure and the equal two-point measure. These are invertible measure-preserving systems, and the projection preserves measure and intertwines the transformations, so it is a two-to-one factor map.

Take $f(y,i)=y_0$. Then $U_T^{n}f(y,i)=y_n$. Distinct coordinates are independent and have mean zero and variance one, henceIts orbit has no finite sufficiently small net. Thus the Bernoulli coordinate observable is not almost periodic in global $L^2$. In contrast, the conditional measure over $y$ is the equal measure on $(y,0),(y,1)$, and $U_T^{n}f$ is constant on that fiber with value $y_n\in \{-1,1\}$. The two centers $g_1=1,g_2=-1$ give zero error on every fiber for every $n$. Therefore $f$ is a relatively almost periodic observable for this map.

The example is also a finite-fiber compact extension. For a bounded observable on $X$, each orbit value on a fiber is a pair in a fixed bounded subset of ${\mathbb{C}}^2$, which has a finite net. Its centers can be chosen as functions constant in $y$, with prescribed values on the two labels. Bounded observables are dense in $L^2$, so the extension is compact. The Bernoulli shift is mixing: cylinder events involving disjoint coordinate sets are independent for all sufficiently large shifts, and approximation by cylinder events extends this to arbitrary events. Thus $S$ and $S^2$ are ergodic. If an invariant function on $X$ is written $a(y)+(-1{)}^{i}b(y)$, invariance gives $a\circ S=a$ and $b\circ S=-b$. The first function is constant; the second is $S^2$-invariant and hence constant, and then its sign equation forces zero. This proves that the example's source is also ergodic.

For the positive-subset assertion, interpret the printed $X^{\prime }$ as $X$, or as an invariant full-measure domain of the factor map; the prime is otherwise undefined and does not change the measure-theoretic argument.

First establish conditional norm covariance under a factor map. Invariance of the measures and uniqueness in the disintegration theorem for a probability measure giveon common full-measure sets. For the first equality, $y\mapsto T_{\ast }{\mu }_{S^{-1}y}$ is another disintegration over the same fibers: it is supported on ${\phi }^{-1}\{y\}$ and integrates to $\mu$. Uniqueness identifies it with ${\mu }_y$, and iteration gives the norm formula.

Two elementary facts explain localization of relative almost periodicity to base sets. If $f$ is a relatively almost periodic observable and $C\subseteq Y$ is measurable, then $f{1}_C\circ \phi$ is also relatively almost periodic. Indeed, on a fiber,When that indicator is one, use a center for $U_T^{n}f$; when it is zero, use the additional center $0$. Secondly, relative almost periodicity is closed under convergence in the uniform conditional L2 normIf this norm of $f-f_j$ tends to zero, covariance bounds the corresponding error between every pair of orbit values by the same number. A finite $\epsilon /2$-net for one sufficiently close $f_j$ is consequently an $\epsilon$-net for $f$.

Now take $f=1_A$. The compact extension supplies relatively almost periodic observables $f_j$ withPut $e_j(y)=\Vert f_j-f{\Vert }_y$. The Tonelli theorem givesHence $e_j(y)\to 0$ for $\nu$-almost every $y$. Since ${\int }_Y{\mu }_y(A)d\nu =\mu (A)>0$, choose $\eta >0$ for whichhas positive measure. The Egorov theorem gives a measurable $C\subseteq E$ of positive measure on which $e_j\to 0$ uniformly. DefineThenFor $h_j=f_j{1}_C\circ \phi$, the localization fact makes every $h_j$ relatively almost periodic, whileThe conditional-norm closure fact therefore proves $1_B$ is a relatively almost periodic observable, with $B\subseteq A$ and positive measure. This is the positive-measure almost periodic indicator in a compact extension. No ergodic component or uncountable intersection of exceptional sets is needed in this construction; it in fact works under the same disintegration and compactness hypotheses without using the assumed ergodicity.

For a probability distribution $m$ on $\mathbb{R}$, take the canonical sample space $\Omega ={\mathbb{R}}^{\mathbb{N}}$, the product of the Borel $\sigma$-algebras, and the product measure $\mathbb{P}=m^{\otimes \mathbb{N}}$. The coordinate functions $X_j(\omega )={\omega }_j$ have joint cylinder probabilitiesThe Kolmogorov extension theorem supplies this measure from the consistent finite-dimensional distributions, and uniqueness follows from the generating cylinder sets. Thus the coordinates are independent and identically distributed; the same construction works for a given standard Borel state space.