Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 14 (Topics in Ergodic Theory)

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.

Use a standard Borel probability space so that conditional measures can be realized by measures. The disintegration theorem for a probability measure states the following. For a measurable map $\pi :X\to Y$ between standard Borel spaces and $\nu ={\pi }_{\ast }\mu$, there is a measurable probability kernel $y\mapsto {\mu }_y$, unique for $\nu$-almost every $y$, such thatMeasurability means $y\mapsto {\mu }_y(A)$ is measurable for every Borel set $A$. More generallyfor integrable $f$, with equality of the last expression almost everywhere. The ${\mu }_y$ are the conditional measures of a factor. This also defines conditional measures for a countably generated sub-sigma-algebra, or a measurable partition represented by such a map. Sigma-algebras are considered modulo null sets where necessary.

For an invariant probability measure, disintegrate over the invariant sigma-algebra $\mathcal{I}$. On a standard probability space it can be represented, modulo null sets, by a countably generated measurable factor $\pi$. One justification for this countability is the separability of $L^2$: a countable dense family of $\mathcal{I}$-measurable functions generates $\mathcal{I}$ modulo null sets. The resulting conditional measures are the ergodic components:Their invariance is one of the permitted facts. It remains to prove that almost every component is ergodic, rather than simply invariant.

Choose a countable generating algebra $\mathcal{A}$ generating the Borel sigma-algebra of $X$. Apply the Birkhoff ergodic theorem simultaneously to its indicators. For $\mu$-almost every $z$ and all $A\in \mathcal{A}$,Disintegration transfers this common full-measure set to ${\mu }_y$-almost every $z$, for $\nu$-almost every $y$. On that fiber the conditional expectation is constant and equals ${\mu }_y(A)$. Thus, simultaneously for every $A\in \mathcal{A}$,Fix such a $y$, for which ${\mu }_y$ is also invariant. Applying the Birkhoff ergodic theorem to the system with measure ${\mu }_y$ identifies these limits withFinite linear combinations of the algebra's indicators are dense in $L^2({\mu }_y)$, by the Monotone class theorem. Since conditional expectation is an $L^2$ contraction, its projection onto invariant functions consequently sends every $L^2({\mu }_y)$ function to its constant integral. In particular, for any ${\mu }_y$-invariant set $D$,and so ${\mu }_y(D)\in \{0,1\}$. This proves almost every component is ergodic. The countable-test proof of ergodicity of conditional components avoids taking an invalid intersection of uncountably many full-measure sets.

For rotation by $1/5$, take the quotient map $\pi (x)=5x(\mathrm{mod}1)$. Its fibers are exactly the five-point orbits. The ergodic components of a rational circle rotation are thereforeEquivalently the component through $x$ is $\frac{1}{5}{\sum }_{j=0}^4{\delta }_{x+j/5}$. To check the disintegration, for bounded measurable $f$ use the change of variables on the five consecutive fifth-intervals:The rotation cyclically permutes the five atoms, preserving their equal masses. A set invariant modulo ${\mu }_y$ must contain all or none of this finite orbit, since every atom has positive mass. Thus each ${\mu }_y$ is ergodic. Finally, the invariant sigma-algebra is the pullback under $\pi$: a function invariant under translation by $1/5$ is constant on each orbit and is a measurable function of $5x(\mathrm{mod}1)$. This confirms that these are the conditional measures over $\mathcal{I}$, not merely some decomposition into invariant measures.

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.