Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 108 (Topics in Ergodic Theory)

Use the probability-system convention $\mu (X)=1$. The Birkhoff ergodic theorem, also called the pointwise ergodic theorem, states that for a measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The limit is integrable and has the same integral as $f$. On a probability space the convergence also holds in $L^1$, as in the allowed mean ergodic theorem. If $T$ is an ergodic transformation, $\mathcal{I}$ is trivial modulo null sets, giving

The integer multiplication map on the circle preserves Lebesgue measure: for any integrable $g$ on $[0,1)$,To prove the ergodicity of integer multiplication on the circle, suppose $u\in L^2(m)$ satisfies $u\circ T_K=u$. Let $c_j$ be its Fourier coefficients in the Fourier basis $e_j(x)=e^{2\pi ijx}$. Since $e_j\circ T_K=e_{Kj}$, the Fourier coefficients of $u\circ T_K$ at index $j$ are zero if $K$ does not divide $j$, and are $c_{j/K}$ otherwise. This identity holds for all $L^2$ functions by approximation with trigonometric polynomials and the isometry $u\mapsto u\circ T_K$. Invariance givesEvery nonzero integer $j$ can be divided by $K$ only finitely often. Thus $c_j=0$ for every $j\ne 0$, and completeness of the Fourier basis makes $u$ constant almost everywhere. Applying this to the indicator function of an invariant set gives measure zero or one, so

A normal number in base $K$ has every word $w=(w_1,\dots ,w_r)$ of $r$ base-$K$ digits occurring with limiting overlapping frequency $K^{-r}$. Use the expansion that is not eventually equal to $K-1$ when there are two expansions. The word $w$ corresponds to the half-open intervalA word starting at position $n+1$ occurs exactly when $T_K^{n}x\in I_w$. The Birkhoff ergodic theorem, applied to $1_{I_w}$, gives frequency $m(I_w)=K^{-r}$ almost everywhere. There are countably many pairs $(K,w)$, so their full-measure sets have a full-measure intersection. In particular,These are absolutely normal numbers, so existence follows as well. This interval description also proves normality and equidistribution under integer multiplication: the base-$K$ intervals form arbitrarily fine grids, so their frequencies imply the correct frequency for every interval by approximation from inside and outside.

For the growth assertion, put $g=|f|$. For every $\epsilon >0$, the Tonelli theorem gives the useful summability boundSince $T$ is a measure-preserving transformation, $\mu (g\circ T^n>\epsilon n)=\mu (g>\epsilon n)$. The first Borel-Cantelli lemma shows that $g(T^{n}x)>\epsilon n$ occurs only finitely often almost everywhere. Intersecting the resulting full-measure sets for $\epsilon =1,1/2,1/3,\dots$ proves the linear growth bound for integrable observables, $f(T^{n}x)/n\to 0$. Multiplication by $n^{1-a}\le 1$ then gives

The threshold is sharp. For $0<a<1$, choose $p$ with $1<p<1/a$, and take the Bernoulli shift on $X=(0,1{)}^{\mathbb{Z}}$ with the product measure of independent uniform coordinates. The left shift preserves that measure because it preserves the probability of every finite-coordinate event. Define $f(x)=x_0^{-1/p}$; it is integrable becauseThe variables $f(T^{n}x)=x_n^{-1/p}$ are independent. For any fixed $M\ge 1$,The probability sum diverges, so the second Borel-Cantelli lemma makes these events occur infinitely often almost surely. Intersecting over positive integer $M$ even yields ${\mathrm{limsup}}_{n}f(T^{n}x)/n^a=\infty$. For $a\le 0$, the constant observable $f=1$ already fails to give limit zero. Thus the sharpness of the linear growth bound for integrable observables gives

For an increasing sequence $(n_k)$ of nonnegative integers, full-density sequence means that its range $S$ has natural density one:For a sequence $(z_n)$ in a metric space, convergence in density of a sequence to $z$ means that, for every $\epsilon >0$,In particular, a bounded nonnegative scalar sequence converging to zero in density has averages tending to zero: the average is at most $\epsilon$ plus its bound times the proportion of indices above $\epsilon$.

Here is a Hilbert space form of the Van der Corput lemma, including the density version. Suppose $\Vert u_n\Vert \le M$, and putThenSince $0\le c_h\le M^2$, it suffices for $(c_h)$ to converge to zero in density. A frequently used special case is that the correlation average tends to zero for every fixed $h\ge 1$.

To prove the lemma, extend the finite list $u_0,\dots ,u_{N-1}$ by zero outside this range, and write $S_N={\sum }_{n=0}^{N-1}{u}_n$. For $1\le H\le N$,The Cauchy-Schwarz inequality for this sum, followed by expansion of the squared Hilbert space norm, givesThe coefficient $H-h$ counts the pairs of shifts separated by $h$. Dividing by $H^2{N}^2$, bounding real parts by absolute values, and first taking the limit superior in $N$ for fixed $H$ yieldsThe right side tends to zero as $H\to \infty$ under the stated hypothesis. This proves both the lemma and its density consequence.

For the polynomial application, fix a nonzero integer $m$ and let $u_n=e^{2\pi imP(n)}\in \mathbb{C}$. For each fixed $h\ge 1$, the correlation phase is linear:Thus, with $q_h=e^{2\pi i(2mah)}$,The coefficient $2mah$ is irrational, so $q_h\ne 1$. This geometric series is bounded in modulus by $2/(N|1-q_h|)$ and tends to zero. All $c_h$ vanish, so the Van der Corput lemma gives

The Weyl criterion now shows that the fractional parts form an equidistributed sequence. For completeness, its sufficiency here follows directly: the displayed limits give the correct average for every trigonometric polynomial, including the constant term. The Stone-Weierstrass theorem extends this to every continuous function by uniform approximation. Approximating an interval's indicator function above and below by continuous functions gives its length as the limiting frequency. HenceThis proves the equidistribution of a quadratic polynomial with irrational leading coefficient using the required correlation argument.

Let ${\xi }_0^{n-1}={\bigvee }_{j=0}^{n-1}{T}^{-j}\xi$ denote the join of measurable partitions recording the first $n$ observations, and set $H_n=H_{\mu }({\xi }_0^{n-1})$, with $H_0=0$. The entropy of a countable measurable partition is $H_{\mu }(\xi )=-{\sum }_{A\in \xi }\mu (A)\log \mu (A)$, using natural logarithms and $0\log 0=0$. Null atoms can be discarded. Throughout, $\mu$ is a probability measure.

The stronger property needed for monotonicity of normalized block entropy is that its increments decrease. DefineSince conditioning reduces entropy, $c_{k+1}\le c_k$. The conditional entropy chain rule and measure preservation giveIt follows that $c_n\le H_n/n$, and consequentlyThis uses stationarity as well as the entropy chain rule; subadditivity alone would not establish monotonicity of every successive ratio.

The entropy rate of a measurable partition and the Kolmogorov-Sinai entropy are, respectively,One may equivalently take the supremum over countable finite-entropy measurable partitions.

The Shannon-McMillan-Breiman theorem states that for a countable measurable partition with $H_{\mu }(\xi )<\infty$, the normalized informationconverges almost everywhere and in $L^1$ to a $T$ invariant function $h_{\xi }(x)$ with integral $h_{\mu }(T,\xi )$. Here ${\xi }_0^{n-1}(x)$ is the atom containing $x$. If $T$ is an ergodic transformation, then $h_{\xi }(x)=h_{\mu }(T,\xi )$ almost everywhere. We prove the general form, including an explicit formula for its limit.

Let ${\mathcal{F}}_k$ be the sigma-algebra generated by ${\bigvee }_{j=1}^k{T}^{-j}\xi$, let ${\mathcal{F}}_0$ be trivial, and put ${\mathcal{F}}_{\infty }={\bigvee }_{j=1}^{\infty }{T}^{-j}\sigma (\xi )$. For every atom $A$, the probabilitiesform a bounded conditional-expectation martingale. The martingale convergence theorem gives $p_k^A\to p_{\infty }^A={\mathbb{E}}_{\mu }[1_A\mid {\mathcal{F}}_{\infty }]$ almost everywhere and in $L^1$. Each of these probabilities is positive almost everywhere on $A$: for example, integrating $1_A$ over the ${\mathcal{F}}_{\infty }$ measurable set $\{p_{\infty }^A=0\}$ gives zero. Countability of $\xi$ lets us choose a common full-measure set for every atom.

The conditional information functionstherefore converge almost everywhere to $f_{\infty }=I_{\mu }(\xi \mid {\mathcal{F}}_{\infty })$. We need an integrable bound on this sequence; convergence of the probabilities alone would not supply one after taking logarithms. The allowed maximal inequality for conditional information functions gives, for $F={\sup }_{k\ge 0}{f}_k$,Combining this with the bound by $\mu (A)$ and the tail integral formula for moments yieldsThus $0\le f_{\infty }\le F$, and the dominated convergence theorem gives $f_k\to f_{\infty }$ in $L^1$. In particular,because $H_n/n$ is the average of the decreasing sequence $(c_k)$.

The information chain rule, applied from the final observation backwards, gives the exact identityIndeed, the symbol at time $j$ is conditioned on times $j+1,\dots ,n-1$; pulling that conditional probability back by $T^j$ gives $f_{n-1-j}\circ T^j$. Measure preservation is sufficient for this pullback identity.

We now prove the required triangular ergodic averaging lemma in this instance. Put $g_M={\sup }_{k\ge M}|f_k-f_{\infty }|$. Then $g_M\downarrow 0$ almost everywhere, $g_M\le 2F$, and $\Vert g_M{\Vert }_1\to 0$. Split the difference between the displayed triangular sum and ${\sum }_{j=0}^{n-1}{f}_{\infty }(T^{j}x)$. Terms whose index is at least $M$ are bounded by ${\sum }_{j=0}^{n-1}{g}_M(T^{j}x)$. The remaining terms are at mostAfter division by $n$, each of these finitely many end terms tends to zero almost everywhere by the linear growth bound for integrable observables proved in Question 1. Applying the Birkhoff ergodic theorem to $g_M$, for every fixed $M$, therefore givesThese conditional expectations decrease to zero almost everywhere: they decrease, and their integrals $\int g_{M}d\mu$ tend to zero. Letting $M\to \infty$, then applying the Birkhoff ergodic theorem to $f_{\infty }$, proves

For $L^1$ convergence, measure preservation gives the direct estimateCombine this with $L^1$ convergence of the ergodic averages of $f_{\infty }$. Finally, $\int h_{\xi }d\mu =\int f_{\infty }d\mu =h_{\mu }(T,\xi )$. In the ergodic case the invariant sigma-algebra is trivial, completing the familiar form of the theorem:

The Rudolph measure rigidity theorem has an essential ergodicity hypothesis. If a Borel probability measure $\mu$ on the circle group is invariant under both $T_2$ and $T_3$, is ergodic for the semigroup generated jointly by these maps, and either map has positive Kolmogorov-Sinai entropy, thenEquivalently, a jointly ergodic common invariant measure other than Lebesgue measure has zero entropy for both maps. Joint ergodicity means that every set invariant modulo $\mu$ under both maps has measure zero or one. Positive entropy without this hypothesis is insufficient: $\frac{1}{2}m+\frac{1}{2}{\delta }_0$, with ${\delta }_0$ a Dirac measure, is a common invariant measure of positive entropy and is not $m$.

The Host equidistribution theorem states that if $p,q\ge 2$ are relatively prime integers and $\nu$ is invariant and ergodic under $T_p$, with $h_{\nu }(T_p)>0$, then for $\nu$-almost every $x$ the sequence $(T_q^{n}x)$ is an equidistributed sequence for Lebesgue measure. Explicitly, for every continuous $g$ on the circle,The non-ergodic form assumes $T_p$ invariance and positive entropy for almost every component in the ergodic decomposition $\nu =\int {\nu }_{\omega }d\lambda (\omega )$. Applying the ergodic theorem of Host on each such component gives the same almost-everywhere conclusion for $\nu$. More generally, its conclusion holds on the part supported on positive-entropy components. A positive value of $h_{\nu }(T_p)$ alone does not eliminate zero-entropy components.

To deduce the joint version of the Rudolph measure rigidity theorem, suppose $h_{\mu }(T_2)>0$; if only $T_3$ has positive entropy, interchange the roles. Write the $T_2$ ergodic decomposition as $\mu =\int \nu d\tau (\nu )$. Since $T_3$ commutes with $T_2$, its pushforward measure sends a $T_2$ ergodic component $\nu$ to a $T_2$ ergodic component $(T_3{)}_{\ast }\nu$. On each component, $T_3$ is a factor of a measure-preserving system with fibres of size at most three. We use the standard entropy preservation under a finite-to-one factor:The reason for this standard entropy fact is that, conditional on a complete factor point, every finite orbit name has at most three possibilities; its conditional entropy is bounded by $\log 3$, and division by the orbit length gives zero relative entropy.

The component at $T_{3}x$ is $(T_3{)}_{\ast }{\nu }_x$ almost everywhere; this follows from commutation and the componentwise ergodic averages. The component entropy function $r(x)=h_{{\nu }_x}(T_2)$ is therefore invariant under both $T_2$ and $T_3$. Joint ergodicity makes it constant almost everywhere, and affinity of entropy under ergodic decomposition identifies the constant as $h_{\mu }(T_2)>0$. Thus almost every $T_2$ component has positive entropy, exactly the condition required in the non-ergodic Host equidistribution theorem. It follows that $\mu$-almost every point equidistributes for $m$ under $T_3$.

For any continuous $g$, invariance under $T_3$ and the dominated convergence theorem now giveContinuous functions determine Borel probability measures on the circle, so $\mu =m$, proving the deduction.

For the normal-number example, let ${\omega }_1,{\omega }_2,\dots$ be independent fair binary digits and defineThis Cantor Bernoulli measure is supported on the middle-third Cantor set $C$. If $S$ is the Bernoulli shift, then $T_3\circ \pi =\pi \circ S$ on the circle. Consequently $\nu$ is $T_3$ invariant and ergodic: a $T_3$ invariant event pulls back to an $S$ invariant event, which has probability zero or one.

Take the ternary digit measurable partition $\eta =\{[0,1/3),[1/3,2/3),[2/3,1)\}$. Its block partition of length $n$ has, up to null endpoints, $2^n$ positive-measure atoms under $\nu$, each of measure $2^{-n}$, and all other atoms have measure zero. HenceApply the Host equidistribution theorem with $p=3$, $q=2$. For $\nu$-almost every $x\in C$, the sequence $(T_2^{n}x)$ equidistributes for Lebesgue measure, so $x$ is a normal number in base $2$ by normality and equidistribution under integer multiplication.

The ternary expansion of $\nu$-almost every such $x$ contains only $0$ and $2$, so the frequency of digit $1$ is zero rather than $1/3$. The ambiguous ternary endpoints form a countable $\nu$ null set and can be removed. Therefore $x$ is not a normal number in base $3$. We have proved the stronger almost-everywhere existence statement