Kronecker’s theorem

1 Equivalent statements of Kronecker’s theorem

We shall now give two statements of Kronecker’s theorem, and prove that they are equivalent before proving that they are true.

Theorem 1.

If ${\theta }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\theta }_k\mathrm{,}\mathrm{1}$ are real numbers that are linearly independent over $\mathrm{Z}$, ${\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_k$ are real numbers, and $N$ and $ϵ$ are positive real numbers, then there are integers $n\mathrm{>}N$ and $p_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{p}_k$ such that for $m\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}k$,

Theorem 2.

If ${\theta }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\theta }_k$ are real numbers that are linearly independent over $\mathrm{Z}$, ${\alpha }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\alpha }_k$ are real numbers, and $T$ and $ϵ$ are positive real numbers, then there is a real number $t\mathrm{>}T$ and integers $p_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{p}_k$ such that for $m\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}k$,

We now prove that the above two statements are equivalent.

Lemma 3.

Theorem 1 is true if and only if Theorem 2 is true.

Proof.

Assume that Theorem 2 is true and let ${\theta }_1^{\prime },\mathrm{\dots },{\theta }_k^{\prime },1$ be real numbers that are linearly independent over $\mathbb{Z}$, let ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ be real numbers, let $N>0$ and let . Let ${\theta }_m={\theta }_m^{\prime }-q_m$ with . Because ${\theta }_1^{\prime },\mathrm{\dots },{\theta }_k^{\prime },1$ are linearly independent over $\mathbb{Z}$, so are ${\theta }_1,\mathrm{\dots },{\theta }_k,1$. Using Theorem 2 with $k+1$ instead of $k$, $N+1$ instead of $T$, $\frac{1}{2}ϵ$ instead of $ϵ$, applied with

$${\theta }_1,\mathrm{\dots },{\theta }_k,1,{\alpha }_1,\mathrm{\dots },{\alpha }_k,0,$$

there is a real number $t>N+1$ and integers $p_1,\mathrm{\dots },p_k,p_{k+1}$ such that for $m=1,\mathrm{\dots },k$,

and

Then $p_{k+1}>t-\frac{1}{2}ϵ>t-\frac{1}{2}>N$, and for $m=1,\mathrm{\dots },k$, because ,

$|p_{k+1}{\theta }_m-p_m-{\alpha }_m|$ $=|p_{k+1}{\theta }_m-p_m+t{\theta }_m-t{\theta }_m-{\alpha }_m|$

$\le |t{\theta }_m-p_m-{\alpha }_m|+|(p_{k+1}-t){\theta }_m|$

$\le |t{\theta }_m-p_m-{\alpha }_m|+|p_{k+1}-t|$

Thus for $n=p_{k+1}$, we have $n>N$, and for $m=1,\mathrm{\dots },k$,

proving Theorem 1.

Assume that Theorem 1 is true. The claim of Theorem 2 is immediate when $k=1$. For $k>1$, let ${\theta }_1^{\prime },\mathrm{\dots },{\theta }_k^{\prime }$ be linearly independent over $\mathbb{Z}$, let ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ be real numbers, and let $T$ and $ϵ$ be positive real numbers. Let ${\theta }_m=|{\theta }_m^{\prime }|>0$, and because ${\theta }_1^{\prime },\mathrm{\dots },{\theta }_k^{\prime }$ are linearly independent over $\mathbb{Z}$, so are ${\theta }_1,\mathrm{\dots },{\theta }_k$, and then

$$\frac{{\theta }_1}{{\theta }_k},\frac{{\theta }_2}{{\theta }_k},\mathrm{\dots },\frac{{\theta }_{k-1}}{{\theta }_k},1$$

are linearly independent over $\mathbb{Z}$. Applying Theorem 1 with $N=T{\theta }_k$ and

$$\frac{{\theta }_1}{{\theta }_k},\frac{{\theta }_2}{{\theta }_k},\mathrm{\dots },\frac{{\theta }_{k-1}}{{\theta }_k},\mathrm{sgn}{\theta }_1^{\prime }\cdot {\alpha }_1,\mathrm{\dots },\mathrm{sgn}{\theta }_{k-1}^{\prime }\cdot {\alpha }_{k-1},$$

we get that there are integers $n>T{\theta }_k$ and $p_1,\mathrm{\dots },p_{k-1}$ such that for $m=1,\mathrm{\dots },k-1$,

Let $t=\frac{n}{{\theta }_k}$. Then $t>T$ and for $m=1,\mathrm{\dots },k-1$,

and

On the other hand, applying Theorem 1 with $N=T$ and

$${\theta }_1,\mathrm{\dots },{\theta }_k,0,\mathrm{\dots },0,\mathrm{sgn}{\theta }_k^{\prime }\cdot {\alpha }_k,$$

we get that there are integers $\nu >T$ and $q_1,\mathrm{\dots },q_k$ such that for $m=1,\mathrm{\dots },k-1$,

and

For $m=1,\mathrm{\dots },k-1$,

$|(t+\nu ){\theta }_m-(p_m+q_m)-\mathrm{sgn}{\theta }_m^{\prime }\cdot {\alpha }_m|$ $\le |t{\theta }_m-p_m-\mathrm{sgn}{\theta }_m^{\prime }\cdot {\alpha }_m|+|\nu {\theta }_m-q_m|$

and

$|(t+\nu ){\theta }_k-(p_k+q_k)-\mathrm{sgn}{\theta }_k^{\prime }\cdot {\alpha }_k|$ $\le |t{\theta }_k-p_k|+|\nu {\theta }_k-q_k-\mathrm{sgn}{\theta }_k^{\prime }\cdot {\alpha }_k|$

Therefore for $m=1,\mathrm{\dots },k$,

which proves Theorem 2. ∎

2 Proof of Kronecker’s theorem

We now prove Theorem 2.

Proof of Theorem 2.

Let ${\theta }_1,\mathrm{\dots },{\theta }_k$ be real numbers that are linearly independent over $\mathbb{Z}$, let ${\alpha }_1,\mathrm{\dots },{\alpha }_k$ be real numbers, and let $T$ and $ϵ$ be positive real numbers.

For real $c$ and $\tau >0$,

For $c_1,\mathrm{\dots },c_r\in ℝ$ with $c_m\ne c_n$ for $m\ne n$, and for $b_{\nu }\in ℂ$, let

$$\chi (t)=\sum _{\nu =1}^r{b}_{\nu }{e}^{i{c}_{\nu }t}.$$

Then for $1\le \mu \le r$,

$$\underset{\tau \to \mathrm{\infty }}{lim}\frac{1}{\tau }{\int }_0^{\tau }\chi (t)e^{-i{c}_{\mu }t}𝑑t=\sum _{\nu =1}^r{b}_{\nu }\underset{\tau \to \mathrm{\infty }}{lim}\frac{1}{\tau }{\int }_0^{\tau }{e}^{i(c_{\nu }-c_{\mu })t}𝑑t=b_{\mu }.$$

Let

$$F(t)=1+\sum _{m=1}^k{e}^{2\pi i(t{\theta }_m-{\alpha }_m)}=1+\sum _{m=1}^k{e}^{-2\pi i{\alpha }_m}{e}^{2\pi it{\theta }_m}$$

and let

$$\varphi (t)=|F(t)|,$$

which satisfies $0\le \varphi (t)\le k+1$.

Define $\varphi :{ℝ}^k\to ℝ$ by

$$\psi (x_1,\mathrm{\dots },x_k)=1+x_1+\mathrm{\cdots }+x_k$$

and let $p$ be a positive integer. By the multinomial theorem,

${\psi }^p$ $={(1+x_1+\mathrm{\cdots }+x_k)}^p$

$={\displaystyle \sum _{{\nu }_0+{\nu }_1+\mathrm{\cdots }+{\nu }_k=p}}\left({\displaystyle \genfrac{}{}{0pt}{}{p}{{\nu }_0,{\nu }_1,\mathrm{\dots },{\nu }_k}}\right)x_1^{{\nu }_1}\mathrm{\cdots }{x}_k^{{\nu }_k}$

$={\displaystyle \sum _{\nu }}{a}_{{\nu }_1,\mathrm{\dots },{\nu }_k}{x}_1^{{\nu }_1}\mathrm{\cdots }{x}_k^{{\nu }_k},$

for which

$$\sum _{\nu }{a}_{{\nu }_1,\mathrm{\dots },{\nu }_k}={(k+1)}^p$$

and the number of terms in the above sum is $\left(\genfrac{}{}{0pt}{}{p+k}{k}\right)$. We can write $F(t)$ as

$$F(t)=\psi (e^{2\pi i(t{\theta }_1-{\alpha }_1)},\mathrm{\dots },e^{2\pi i(t{\theta }_k-{\alpha }_k)}).$$

Then

$$F{(t)}^p=\sum a_{{\nu }_1,\mathrm{\dots },{\nu }_k}\exp \left(\sum _{m=1}^k{\nu }_m\cdot 2\pi i(t{\theta }_m-{\alpha }_m)\right).$$

Because ${\theta }_1,\mathrm{\dots },{\theta }_k$ are linearly independent over $\mathbb{Z}$, for $\nu \ne \mu$ it is the case that $2\pi {\sum }_{m=1}^k{\nu }_m{\theta }_m\ne 2\pi {\sum }_{m=1}^k{\mu }_m{\theta }_m$. Write $c_{\nu }=2\pi \nu \cdot \theta$ and

$$b_{\nu }=a_{{\nu }_1,\mathrm{\dots },{\nu }_k}\exp \left(-2\pi i\sum _{m=1}^k{\nu }_m{\alpha }_m\right),$$

with which

$$F{(t)}^p=\sum b_{\nu }{e}^{i{c}_{\nu }t}.$$

Then for each multi-index $\mu$,

$$\underset{\tau \to \mathrm{\infty }}{lim}\frac{1}{\tau }{\int }_0^{\tau }F{(t)}^p{e}^{-i{c}_{\mu }t}𝑑t=b_{\mu }.$$ (1)

Suppose by contradiction that

Then there is some and some $t_0$ such that when $t\ge t_0$,

$$|F(t)|=\varphi (t)\le \lambda .$$

Thus for $p$ a positive integer,

$\underset{\tau \to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{1}{\tau }}{\displaystyle {\int }_0^{\tau }}{|F(t)|}^p𝑑t$ $\le \underset{\tau \to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{1}{\tau }}{\displaystyle {\int }_0^{t_0}}{|F(t)|}^p𝑑t+\underset{\tau \to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{1}{\tau }}{\displaystyle {\int }_{t_0}^{\tau }}{|F(t)|}^p𝑑t$

$=\underset{\tau \to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{1}{\tau }}{\displaystyle {\int }_{t_0}^{\tau }}{|F(t)|}^p𝑑t$

$\le \underset{\tau \to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{1}{\tau }}{\lambda }^p(\tau -t_0)$

$={\lambda }^p.$

But then by (1),

$$|b_{\mu }|\le \underset{\tau \to \mathrm{\infty }}{lim\; sup}\frac{1}{\tau }{\int }_0^{\tau }{|F(t)|}^p𝑑t\le {\lambda }^p,$$

and then

${(k+1)}^p$ $={\displaystyle \sum _{\nu }}{a}_{{\nu }_1,\mathrm{\dots },{\nu }_k}$

$={\displaystyle \sum _{\nu }}|b_{\nu }|$

$\le {\displaystyle \sum _{\nu }}{\lambda }^p$

$\le {\lambda }^p\cdot \left({\displaystyle \genfrac{}{}{0pt}{}{p+k}{k}}\right).$

Let $r=\frac{\lambda }{k+1}$, for which , and so for each positive integer $p$ it holds that

$$1\le r^p\cdot \left(\genfrac{}{}{0pt}{}{p+k}{k}\right).$$ (2)

Now,

$$\left(\genfrac{}{}{0pt}{}{p+k}{k}\right)=\left(\genfrac{}{}{0pt}{}{p+k}{p}\right)=\frac{p^k}{\mathrm{\Gamma }(k+1)}\left(1+\frac{k(k+1)}{2p}+O(p^{-2})\right),p\to \mathrm{\infty }.$$

In particular,

$$r^p\cdot \left(\genfrac{}{}{0pt}{}{p+k}{k}\right)=O(r^p\cdot p^k),p\to \mathrm{\infty },$$

and because , $r^p\cdot p^k\to 0$ as $p\to \mathrm{\infty }$, contradicting (2) being true for all positive integers $p$. This contradiction shows that in fact

$$\underset{t\to \mathrm{\infty }}{lim\; sup}\varphi (t)\ge k+1,$$

and because $\varphi (t)\le k+1$,

$$\underset{t\to \mathrm{\infty }}{lim\; sup}\varphi (t)=k+1.$$ (3)

Now let . By (3) there is some $t\ge T$ for which $\varphi (t)\ge k+1-\eta$. For $1\le m\le k$, write

$$z_m=e^{2\pi i(t{\theta }_m-{\alpha }_m)}=x_m+i{y}_m.$$

It is straightforward from the definition of $\varphi (t)$ that

$$k+1-\eta \le \varphi (t)\le (k-1)+|1+e^{2\pi i(t{\theta }_m-{\alpha }_m)}|,$$

which yields

$$2\ge |1+e^{2\pi i(t{\theta }_m-{\alpha }_m)}|\ge 2-\eta .$$

Because $|z_m|=1$,

$${|1+z_m|}^2={(1+x_m)}^2+y_m^2={(1+x_m)}^2+(1-x_m^2)=2+2x_m,$$

hence

$$2+2x_m\ge {(2-\eta )}^2=4-4\eta +{\eta }^2>4-4\eta ,$$

so

Furthermore,

Therefore

hence

For $x\in ℝ$, denote by $\parallel x\parallel$ the distance from $x$ to the nearest integer. We check that

$$|\sin (\pi x)|=\sin (\pi \parallel x\parallel )\ge \frac{2}{\pi }\cdot \pi \parallel x\parallel =2\parallel x\parallel .$$

Thus, for each $m=1,\mathrm{\dots },k$,

We have taken $t\ge T$ .Take ${\eta }^{1/2}=ϵ$, i.e. $\eta ={ϵ}^2$, and take $p_m$ to be the nearest integer to $t{\theta }_m-{\alpha }_m$, for which , proving the claim. ∎

3 Uniform distribution modulo $1$

For $x\in ℝ$ let $[x]$ be the greatest integer $\le x$, and let $\{x\}=x-[x]$, called the fractional part of $x$. For $P=(x_1,\mathrm{\dots },x_d)\in {ℝ}^d$ let $\{P\}=(\{x_1\},\mathrm{\dots },\{x_d\})$, which belongs to the set $Q={[0,1)}^d$. Let $P_j=(x_{j,1},\mathrm{\dots },x_{j,d})$, $j\ge 1$, be a sequence in ${ℝ}^d$, and for $A\subset Q$ let

$${\varphi }_n(A)=\{k:1\le k\le n,\{P_j\}\in A\}.$$

We say that $(P_j)$ is uniformly distributed modulo $\mathrm{1}$ if for each closed rectangle $V$ contained in $Q$,

$$\underset{n\to \mathrm{\infty }}{lim}\frac{{\varphi }_n(V)}{n}=\lambda (V),$$

where $\lambda$ is Lebesgue measure on ${ℝ}^d$: for $V=[a_1,b_1]\times \mathrm{\cdots }[a_d,b_d]$, $\lambda (V)={\prod }_{j=1}^d(b_j-a_j)$.

We have proved that if ${\theta }_1,\mathrm{\dots },{\theta }_k,1$ are linearly independent over $\mathbb{Z}$, then the sequence $\{n\theta \}=(\{n{\theta }_1\},\mathrm{\dots },\{n{\theta }_k\})$ is dense in $Q$.a It can in fact be proved that $(n\theta )$ is uniformly distributed modulo $1$.

4 Unique ergodicity

Let $X$ be a compact metric space, let $C(X)$ be the Banach space of continuous functions $X\to ℝ$, and let $ℳ(X)$ be the space of Borel probability measures on $X$, with the subspace topology inherited from $C{(X)}^{*}$ with the weak-* topology. One proves that $\mu$ and $\nu$ in $ℳ(X)$ are equal if and only if ${\int }_{X}f𝑑\mu ={\int }_{X}f𝑑\nu$ for all $f\in C(X)$. $ℳ(X)$ is a closed set in $C{(X)}^{*}$ that is contained in the closed unit ball, and by the Banach-Alaoglu theorem that closed unit ball is compact, so $ℳ(X)$ is itself compact. $C{(X)}^{*}$, with the weak-* topology, is not metrizable, but it is the case that $ℳ(X)$ with the subspace topology inherited from $C{(X)}^{*}$ is metrizable.

For a continuous map $T:X\to X$, define $T_{*}:ℳ(X)\to ℳ(X)$ by

$$(T_{*}\mu )(A)=\mu (T^{-1}A)$$

for Borel sets $A$ in $X$. For ${\mu }_n\to \mu$ in $ℳ(X)$ and $f\in C(X)$, by the change of variables theorem we have

$${\int }_{X}fd(T_{*}{\mu }_n)={\int }_{X}f\circ T𝑑{\mu }_n\to {\int }_{X}f\circ T𝑑\mu ={\int }_{X}fd(T_{*}\mu ),$$

which means that $T_{*}{\mu }_n\to T_{*}\mu$, and therefore the map $T_{*}$ is continuous. We say that $\mu \in ℳ(X)$ is $T$-invariant if $T_{*}\mu =\mu$. Equivalently, $T:(X,{ℬ}_X,\mu )\to (X,{ℬ}_X,\mu )$ is measure-preserving. We denote by ${ℳ}^T(X)$ the set of $T$-invariant $\mu \in ℳ(X)$. The Kryloff-Bogoliouboff theorem states that ${ℳ}^T(X)$ is nonempty. It is immediate that ${ℳ}^T(X)$ is a convex subset of $C{(X)}^{*}$. Let ${\mu }_n\in {ℳ}^T(X)$ converge to some $\mu \in ℳ(X)$. For $f\in C(X)$ we have, because $T_{*}$ is continuous,

$${\int }_{X}fd(T_{*}\mu )=\underset{n\to \mathrm{\infty }}{lim}{\int }_{X}fd(T_{*}{\mu }_n)=\underset{n\to \mathrm{\infty }}{lim}{\int }_{X}f𝑑{\mu }_n={\int }_{X}f𝑑\mu ,$$

which shows that $\mu$ is $T$-invariant. Therefore ${ℳ}^T(X)$ is a closed set in $ℳ(X)$, and we have thus established that ${ℳ}^T(X)$ is a nonempty compact convex set.

A measure $\mu \in {ℳ}^T(X)$ is called ergodic if for any $A\in {ℬ}_X$ with $T^{-1}A=A$ it holds that $\mu (A)=0$ or $\mu (A)=1$. It is proved that $\mu \in {ℳ}^T(X)$ is ergodic if and only if $\mu$ is an extreme point of ${ℳ}^T(X)$. The Krein-Milman theorem states that if $S$ is a nonempty compact convex set in a locally convex space, then $S$ is equal to the closed convex hull of the set of extreme points of $S$. In particular this shows us that there exist extreme points of $S$. Let ${ℰ}^T(X)$ be the set of extreme points of ${ℳ}^T(X)$, and applying the Krein-Milman theorem with ${ℳ}^T(X)$, which is a nonempty compact convex set in the locally convex space $C{(X)}^{*}$, we have that ${ℳ}^T(X)$ is equal to the closed convex hull ${ℰ}^T$. That is, ${ℳ}^T(X)$ is equal to the closed convex hull of the set of ergodic $\mu \in {ℳ}^T(X)$.

Choquet’s theorem tells us that for each $\mu \in {ℳ}^T(X)$ there is a unique Borel probability measure $\lambda$ on the compact metrizable space ${ℳ}^T(X)$ such that

$$\lambda ({ℰ}^T(X))=1$$

and for all $f\in C(X)$,

$${\int }_{X}f𝑑\mu ={\int }_{{ℰ}^T(X)}\left({\int }_{X}f𝑑\nu \right)𝑑\lambda (\nu ).$$

We have established that ${ℳ}^T(X)$ contains at least one element. $T$ is called uniquely ergodic if ${ℳ}^T(X)$ is a singleton. If ${ℳ}^T(X)=\{{\mu }_0\}$ then ${\mu }_0$ is an extreme point of ${ℳ}^T(X)$, hence is ergodic. If ${ℰ}^T(X)=\{{\mu }_0\}$, then for $\mu \in {ℳ}^T(X)$, by Choquet’s theorem there is a unique Borel probability measure $\lambda$ on ${ℳ}^T(X)$ satisfying $\lambda ={\delta }_{{\mu }_0}$ and

$${\int }_{X}f𝑑\mu ={\int }_{\{{\mu }_0\}}\left({\int }_{X}f𝑑\nu \right)𝑑\lambda (\nu ),$$

i.e.

$${\int }_{X}f𝑑\mu ={\int }_{X}f𝑑{\mu }_0,$$

which means that $\mu ={\mu }_0$. Therefore, $T$ is uniquely ergodic if and only if ${ℰ}^T(X)$ is a singleton. It can be proved that $T$ is uniquely ergodic if and only if for each $f\in C(X)$ there is some $C_f$ such that

$$\frac{1}{N}\sum _{n=0}^{N-1}f(T^{n}x)\to C_f$$

uniformly on $X$. This constant $C_f$ is equal to ${\int }_{X}f𝑑\mu$, where ${ℳ}^T(X)=\{\mu \}$.

For a topological group $X$ and for $g\in X$, define $R_g(x)=gx$, which is continuous $X\to X$. For a compact metrizable group, there is a unique Borel probability measure $m_X$ on $X$ that is $R_g$-invariant for every $g\in X$, called the Haar measure on $X$. Thus for each $g\in X$, the Haar measure $m_X$ belongs to ${ℳ}^{R_g}(X)$, and for $R_g$ to be uniquely ergodic means that $m_X$ is the only element of ${ℳ}^{R_g}(X)$. For a locally compact abelian group $X$, let $\widehat{X}$ be its Pontryagin dual. The following theorem gives a condition that is equivalent to a translation being uniquely ergodic.

Theorem 4.

Let $X$ be a compact metrizable group and let $g\mathrm{\in }X$. $R_g$ is uniquely ergodic if and only if $X$ is abelian and $\chi \mathit{}\mathrm{(}g\mathrm{)}\mathrm{\ne }\mathrm{1}$ for all nontrivial $\chi \mathrm{\in }\widehat{X}$.

Let $𝕋=ℝ/\mathbb{Z}$, let $X={𝕋}^d={ℝ}^d/{\mathbb{Z}}^d$, which is a compact abelian group, and let $g=({\alpha }_1,\mathrm{\dots },{\alpha }_d)\in {ℝ}^d$. For $\chi \in \widehat{X}={\mathbb{Z}}^d$, $\chi =(k_1,\mathrm{\dots },k_d)$,

$$\chi (g)=\exp \left(2\pi i\sum _{j=1}^d{k}_j{\alpha }_j\right).$$

$\chi (g)=1$ if and only if ${\sum }_{j=1}^d{k}_j{\alpha }_j\in \mathbb{Z}$ if and only if there is some $k_{d+1}\in \mathbb{Z}$ such that $k_1{\alpha }_1+\mathrm{\cdots }+k_d{\alpha }_d+k_{d+1}=0$. Therefore for ${\alpha }_1,\mathrm{\dots },{\alpha }_d\in ℝ$, the set $\{{\alpha }_1,\mathrm{\dots },{\alpha }_d,1\}$ is linearly independent over $\mathbb{Z}$ if and only if for $g=({\alpha }_1,\mathrm{\dots },{\alpha }_d)$, the map $R_g(x)=x+g$, ${𝕋}^d\to {𝕋}^d$, is uniquely ergodic.