# Kronecker’s theorem

Jordan Bell
August 29, 2015

## 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 $\theta_{1},\ldots,\theta_{k},1$ are real numbers that are linearly independent over $\mathbb{Z}$, $\alpha_{1},\ldots,\alpha_{k}$ are real numbers, and $N$ and $\epsilon$ are positive real numbers, then there are integers $n>N$ and $p_{1},\ldots,p_{k}$ such that for $m=1,\ldots,k$,

 $|n\theta_{m}-p_{m}-\alpha_{m}|<\epsilon.$
###### Theorem 2.

If $\theta_{1},\ldots,\theta_{k}$ are real numbers that are linearly independent over $\mathbb{Z}$, $\alpha_{1},\ldots,\alpha_{k}$ are real numbers, and $T$ and $\epsilon$ are positive real numbers, then there is a real number $t>T$ and integers $p_{1},\ldots,p_{k}$ such that for $m=1,\ldots,k$,

 $|t\theta_{m}-p_{m}-\alpha_{m}|<\epsilon.$

We now prove that the above two statements are equivalent.11 1 K. Chandrasekharan, Introduction to Analytic Number Theory, pp. 92–93, Chapter VIII, §5.

###### Lemma 3.

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

###### Proof.

Assume that Theorem 2 is true and let $\theta_{1}^{\prime},\ldots,\theta_{k}^{\prime},1$ be real numbers that are linearly independent over $\mathbb{Z}$, let $\alpha_{1},\ldots,\alpha_{k}$ be real numbers, let $N>0$ and let $0<\epsilon<1$. Let $\theta_{m}=\theta_{m}^{\prime}-q_{m}$ with $0<\theta_{m}\leq 1$. Because $\theta_{1}^{\prime},\ldots,\theta_{k}^{\prime},1$ are linearly independent over $\mathbb{Z}$, so are $\theta_{1},\ldots,\theta_{k},1$. Using Theorem 2 with $k+1$ instead of $k$, $N+1$ instead of $T$, $\frac{1}{2}\epsilon$ instead of $\epsilon$, applied with

 $\theta_{1},\ldots,\theta_{k},1,\qquad\alpha_{1},\ldots,\alpha_{k},0,$

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

 $|t\theta_{m}-p_{m}-\alpha_{m}|<\frac{1}{2}\epsilon,$

and

 $|t-p_{k+1}|<\frac{1}{2}\epsilon.$

Then $p_{k+1}>t-\frac{1}{2}\epsilon>t-\frac{1}{2}>N$, and for $m=1,\ldots,k$, because $0<\theta_{m}\leq 1$,

 $\displaystyle|p_{k+1}\theta_{m}-p_{m}-\alpha_{m}|$ $\displaystyle=|p_{k+1}\theta_{m}-p_{m}+t\theta_{m}-t\theta_{m}-\alpha_{m}|$ $\displaystyle\leq|t\theta_{m}-p_{m}-\alpha_{m}|+|(p_{k+1}-t)\theta_{m}|$ $\displaystyle\leq|t\theta_{m}-p_{m}-\alpha_{m}|+|p_{k+1}-t|$ $\displaystyle<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon.$

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

 $|n\theta_{m}^{\prime}-(nq_{m}+p_{m})-\alpha|=|n\theta_{m}-p_{m}-\alpha_{m}|<\epsilon,$

proving Theorem 1.

Assume that Theorem 1 is true. The claim of Theorem 2 is immediate when $k=1$. For $k>1$, let $\theta_{1}^{\prime},\ldots,\theta_{k}^{\prime}$ be linearly independent over $\mathbb{Z}$, let $\alpha_{1},\ldots,\alpha_{k}$ be real numbers, and let $T$ and $\epsilon$ be positive real numbers. Let $\theta_{m}=|\theta_{m}^{\prime}|>0$, and because $\theta_{1}^{\prime},\ldots,\theta_{k}^{\prime}$ are linearly independent over $\mathbb{Z}$, so are $\theta_{1},\ldots,\theta_{k}$, and then

 $\frac{\theta_{1}}{\theta_{k}},\frac{\theta_{2}}{\theta_{k}},\ldots,\frac{% \theta_{k-1}}{\theta_{k}},1$

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

 $\frac{\theta_{1}}{\theta_{k}},\frac{\theta_{2}}{\theta_{k}},\ldots,\frac{% \theta_{k-1}}{\theta_{k}},\qquad\mathrm{sgn}\,\theta_{1}^{\prime}\cdot\alpha_{% 1},\ldots,\mathrm{sgn}\,\theta_{k-1}^{\prime}\cdot\alpha_{k-1},$

we get that there are integers $n>T\theta_{k}$ and $p_{1},\ldots,p_{k-1}$ such that for $m=1,\ldots,k-1$,

 $\left|n\frac{\theta_{m}}{\theta_{k}}-p_{m}-\mathrm{sgn}\,\theta_{m}^{\prime}% \cdot\alpha_{m}\right|<\frac{1}{2}\epsilon.$

Let $t=\frac{n}{\theta_{k}}$. Then $t>T$ and for $m=1,\ldots,k-1$,

 $|t\theta_{m}-p_{m}-\mathrm{sgn}\,\theta_{m}^{\prime}\cdot\alpha_{m}|=\left|n% \frac{\theta_{m}}{\theta_{k}}-p_{m}-\mathrm{sgn}\,\theta_{m}^{\prime}\cdot% \alpha_{m}\right|<\frac{1}{2}\epsilon,$

and

 $|t\theta_{k}-n|=0<\frac{1}{2}\epsilon.$

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

 $\theta_{1},\ldots,\theta_{k},\qquad 0,\ldots,0,\mathrm{sgn}\,\theta_{k}^{% \prime}\cdot\alpha_{k},$

we get that there are integers $\nu>T$ and $q_{1},\ldots,q_{k}$ such that for $m=1,\ldots,k-1$,

 $|\nu\theta_{m}-q_{m}|<\frac{1}{2}\epsilon$

and

 $|\nu\theta_{k}-q_{k}-\mathrm{sgn}\,\theta_{k}^{\prime}\cdot\alpha_{k}|<\frac{1% }{2}\epsilon.$

For $m=1,\ldots,k-1$,

 $\displaystyle|(t+\nu)\theta_{m}-(p_{m}+q_{m})-\mathrm{sgn}\,\theta_{m}^{\prime% }\cdot\alpha_{m}|$ $\displaystyle\leq|t\theta_{m}-p_{m}-\mathrm{sgn}\,\theta_{m}^{\prime}\cdot% \alpha_{m}|+|\nu\theta_{m}-q_{m}|$ $\displaystyle<\frac{1}{2}\epsilon+\frac{1}{2}\epsilon$

and

 $\displaystyle|(t+\nu)\theta_{k}-(p_{k}+q_{k})-\mathrm{sgn}\,\theta_{k}^{\prime% }\cdot\alpha_{k}|$ $\displaystyle\leq|t\theta_{k}-p_{k}|+|\nu\theta_{k}-q_{k}-\mathrm{sgn}\,\theta% _{k}^{\prime}\cdot\alpha_{k}|$ $\displaystyle<\frac{1}{2}\epsilon+\frac{1}{2}.$

Therefore for $m=1,\ldots,k$,

 $\begin{split}&\displaystyle|(t+\nu)\theta_{m}^{\prime}-\mathrm{sgn}\,\theta_{m% }^{\prime}\cdot(p_{m}+q_{m})-\alpha_{m}|\\ \displaystyle=&\displaystyle|\mathrm{sgn}\,\theta_{m}^{\prime}\cdot(t+\nu)% \theta_{m}-\mathrm{sgn}\,\theta_{m}^{\prime}\cdot(p_{m}+q_{m})-\alpha_{m}|\\ \displaystyle=&\displaystyle|(t+\nu)\theta_{m}-(p_{m}+q_{m})-\mathrm{sgn}\,% \theta_{m}^{\prime}\cdot\alpha_{m}|\\ \displaystyle<&\displaystyle\epsilon,\end{split}$

which proves Theorem 2. ∎

## 2 Proof of Kronecker’s theorem

We now prove Theorem 2.22 2 K. Chandrasekharan, Introduction to Analytic Number Theory, pp. 93–96, Chapter VIII, §5.

###### Proof of Theorem 2.

Let $\theta_{1},\ldots,\theta_{k}$ be real numbers that are linearly independent over $\mathbb{Z}$, let $\alpha_{1},\ldots,\alpha_{k}$ be real numbers, and let $T$ and $\epsilon$ be positive real numbers.

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

 $\lim_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{\tau}e^{ict}dt=\begin{cases}0&c% \neq 0\\ 1&c=0.\end{cases}$

For $c_{1},\ldots,c_{r}\in\mathbb{R}$ with $c_{m}\neq c_{n}$ for $m\neq n$, and for $b_{\nu}\in\mathbb{C}$, let

 $\chi(t)=\sum_{\nu=1}^{r}b_{\nu}e^{ic_{\nu}t}.$

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

 $\lim_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{\tau}\chi(t)e^{-ic_{\mu}t}dt=\sum_% {\nu=1}^{r}b_{\nu}\lim_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{\tau}e^{i(c_{\nu% }-c_{\mu})t}dt=b_{\mu}.$

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

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

which satisfies $0\leq\phi(t)\leq k+1$.

Define $\phi:\mathbb{R}^{k}\to\mathbb{R}$ by

 $\psi(x_{1},\ldots,x_{k})=1+x_{1}+\cdots+x_{k}$

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

 $\displaystyle\psi^{p}$ $\displaystyle=(1+x_{1}+\cdots+x_{k})^{p}$ $\displaystyle=\sum_{\nu_{0}+\nu_{1}+\cdots+\nu_{k}=p}\binom{p}{\nu_{0},\nu_{1}% ,\ldots,\nu_{k}}x_{1}^{\nu_{1}}\cdots x_{k}^{\nu_{k}}$ $\displaystyle=\sum_{\nu}a_{\nu_{1},\ldots,\nu_{k}}x_{1}^{\nu_{1}}\cdots x_{k}^% {\nu_{k}},$

for which

 $\sum_{\nu}a_{\nu_{1},\ldots,\nu_{k}}=(k+1)^{p}$

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

 $F(t)=\psi(e^{2\pi i(t\theta_{1}-\alpha_{1})},\ldots,e^{2\pi i(t\theta_{k}-% \alpha_{k})}).$

Then

 $F(t)^{p}=\sum a_{\nu_{1},\ldots,\nu_{k}}\exp\left(\sum_{m=1}^{k}\nu_{m}\cdot 2% \pi i(t\theta_{m}-\alpha_{m})\right).$

Because $\theta_{1},\ldots,\theta_{k}$ are linearly independent over $\mathbb{Z}$, for $\nu\neq\mu$ it is the case that $2\pi\sum_{m=1}^{k}\nu_{m}\theta_{m}\neq 2\pi\sum_{m=1}^{k}\mu_{m}\theta_{m}$. Write $c_{\nu}=2\pi\nu\cdot\theta$ and

 $b_{\nu}=a_{\nu_{1},\ldots,\nu_{k}}\exp\left(-2\pi i\sum_{m=1}^{k}\nu_{m}\alpha% _{m}\right),$

with which

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

Then for each multi-index $\mu$,

 $\lim_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{\tau}F(t)^{p}e^{-ic_{\mu}t}dt=b_{% \mu}.$ (1)

 $\limsup_{t\to\infty}\phi(t)

Then there is some $\lambda and some $t_{0}$ such that when $t\geq t_{0}$,

 $|F(t)|=\phi(t)\leq\lambda.$

Thus for $p$ a positive integer,

 $\displaystyle\limsup_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{\tau}|F(t)|^{p}dt$ $\displaystyle\leq\limsup_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{t_{0}}|F(t)|^{% p}dt+\limsup_{\tau\to\infty}\frac{1}{\tau}\int_{t_{0}}^{\tau}|F(t)|^{p}dt$ $\displaystyle=\limsup_{\tau\to\infty}\frac{1}{\tau}\int_{t_{0}}^{\tau}|F(t)|^{% p}dt$ $\displaystyle\leq\limsup_{\tau\to\infty}\frac{1}{\tau}\lambda^{p}(\tau-t_{0})$ $\displaystyle=\lambda^{p}.$

But then by (1),

 $|b_{\mu}|\leq\limsup_{\tau\to\infty}\frac{1}{\tau}\int_{0}^{\tau}|F(t)|^{p}dt% \leq\lambda^{p},$

and then

 $\displaystyle(k+1)^{p}$ $\displaystyle=\sum_{\nu}a_{\nu_{1},\ldots,\nu_{k}}$ $\displaystyle=\sum_{\nu}|b_{\nu}|$ $\displaystyle\leq\sum_{\nu}\lambda^{p}$ $\displaystyle\leq\lambda^{p}\cdot\binom{p+k}{k}.$

Let $r=\frac{\lambda}{k+1}$, for which $0, and so for each positive integer $p$ it holds that

 $1\leq r^{p}\cdot\binom{p+k}{k}.$ (2)

Now,

 $\binom{p+k}{k}=\binom{p+k}{p}=\frac{p^{k}}{\Gamma(k+1)}\left(1+\frac{k(k+1)}{2% p}+O(p^{-2})\right),\qquad p\to\infty.$

In particular,

 $r^{p}\cdot\binom{p+k}{k}=O(r^{p}\cdot p^{k}),\qquad p\to\infty,$

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

 $\limsup_{t\to\infty}\phi(t)\geq k+1,$

and because $\phi(t)\leq k+1$,

 $\limsup_{t\to\infty}\phi(t)=k+1.$ (3)

Now let $0<\eta<1$. By (3) there is some $t\geq T$ for which $\phi(t)\geq k+1-\eta$. For $1\leq m\leq k$, write

 $z_{m}=e^{2\pi i(t\theta_{m}-\alpha_{m})}=x_{m}+iy_{m}.$

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

 $k+1-\eta\leq\phi(t)\leq(k-1)+|1+e^{2\pi i(t\theta_{m}-\alpha_{m})}|,$

which yields

 $2\geq|1+e^{2\pi i(t\theta_{m}-\alpha_{m})}|\geq 2-\eta.$

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

 $2+2x_{m}\geq(2-\eta)^{2}=4-4\eta+\eta^{2}>4-4\eta,$

so

 $1-2\eta

Furthermore,

 $y_{m}^{2}=1-x_{m}^{2}=(1-x_{m})(1+x_{m})\leq 2(1-x_{m})<2\cdot 2\eta=4\eta.$

Therefore

 $|z_{m}-1|^{2}=(x_{m}-1)^{2}+y_{m}^{2}<4\eta^{2}+4\eta<8\eta,$

hence

 $2|\sin\pi(t\theta_{m}-\alpha_{m})|=|e^{2\pi i(t\theta_{m}-\alpha_{m})}-1|<8^{1% /2}\eta^{1/2}<4\eta^{1/2}.$

For $x\in\mathbb{R}$, denote by $\left\|x\right\|$ the distance from $x$ to the nearest integer. We check that

 $|\sin(\pi x)|=\sin(\pi\left\|x\right\|)\geq\frac{2}{\pi}\cdot\pi\left\|x\right% \|=2\left\|x\right\|.$

Thus, for each $m=1,\ldots,k$,

 $\left\|t\theta_{m}-\alpha_{m}\right\|<\eta^{1/2}.$

We have taken $t\geq T$ .Take $\eta^{1/2}=\epsilon$, i.e. $\eta=\epsilon^{2}$, and take $p_{m}$ to be the nearest integer to $t\theta_{m}-\alpha_{m}$, for which $|t\theta_{m}-p_{m}-\alpha_{m}|<\epsilon$, proving the claim. ∎

## 3 Uniform distribution modulo $1$

For $x\in\mathbb{R}$ let $[x]$ be the greatest integer $\leq x$, and let $\{x\}=x-[x]$, called the fractional part of $x$. For $P=(x_{1},\ldots,x_{d})\in\mathbb{R}^{d}$ let $\{P\}=(\{x_{1}\},\ldots,\{x_{d}\})$, which belongs to the set $Q=[0,1)^{d}$. Let $P_{j}=(x_{j,1},\ldots,x_{j,d})$, $j\geq 1$, be a sequence in $\mathbb{R}^{d}$, and for $A\subset Q$ let

 $\phi_{n}(A)=\{k:1\leq k\leq n,\{P_{j}\}\in A\}.$

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

 $\lim_{n\to\infty}\frac{\phi_{n}(V)}{n}=\lambda(V),$

where $\lambda$ is Lebesgue measure on $\mathbb{R}^{d}$: for $V=[a_{1},b_{1}]\times\cdots[a_{d},b_{d}]$, $\lambda(V)=\prod_{j=1}^{d}(b_{j}-a_{j})$.

We have proved that if $\theta_{1},\ldots,\theta_{k},1$ are linearly independent over $\mathbb{Z}$, then the sequence $\{n\theta\}=(\{n\theta_{1}\},\ldots,\{n\theta_{k}\})$ is dense in $Q$.a It can in fact be proved that $(n\theta)$ is uniformly distributed modulo $1$.33 3 Giancarlo Travaglini, Number Theory, Fourier Analysis and Geometric Discrepancy, p. 108, Theorem 6.3.

## 4 Unique ergodicity

Let $X$ be a compact metric space, let $C(X)$ be the Banach space of continuous functions $X\to\mathbb{R}$, and let $\mathscr{M}(X)$ be the space of Borel probability measures on $X$, with the subspace topology inherited from $C(X)^{*}$ with the weak-* topology.44 4 This is the same as the narrow topology on $\mathscr{M}(X)$. One proves that $\mu$ and $\nu$ in $\mathscr{M}(X)$ are equal if and only if $\int_{X}fd\mu=\int_{X}fd\nu$ for all $f\in C(X)$. $\mathscr{M}(X)$ 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 $\mathscr{M}(X)$ is itself compact. $C(X)^{*}$, with the weak-* topology, is not metrizable, but it is the case that $\mathscr{M}(X)$ with the subspace topology inherited from $C(X)^{*}$ is metrizable.

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

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

for Borel sets $A$ in $X$. For $\mu_{n}\to\mu$ in $\mathscr{M}(X)$ 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 $\mu\in\mathscr{M}(X)$ is $T$-invariant if $T_{*}\mu=\mu$. Equivalently, $T:(X,\mathscr{B}_{X},\mu)\to(X,\mathscr{B}_{X},\mu)$ is measure-preserving. We denote by $\mathscr{M}^{T}(X)$ the set of $T$-invariant $\mu\in\mathscr{M}(X)$. The Kryloff-Bogoliouboff theorem states that $\mathscr{M}^{T}(X)$ is nonempty. It is immediate that $\mathscr{M}^{T}(X)$ is a convex subset of $C(X)^{*}$. Let $\mu_{n}\in\mathscr{M}^{T}(X)$ converge to some $\mu\in\mathscr{M}(X)$. For $f\in C(X)$ we have, because $T_{*}$ is continuous,

 $\int_{X}fd(T_{*}\mu)=\lim_{n\to\infty}\int_{X}fd(T_{*}\mu_{n})=\lim_{n\to% \infty}\int_{X}fd\mu_{n}=\int_{X}fd\mu,$

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

A measure $\mu\in\mathscr{M}^{T}(X)$ is called ergodic if for any $A\in\mathscr{B}_{X}$ with $T^{-1}A=A$ it holds that $\mu(A)=0$ or $\mu(A)=1$. It is proved that $\mu\in\mathscr{M}^{T}(X)$ is ergodic if and only if $\mu$ is an extreme point of $\mathscr{M}^{T}(X)$.55 5 Manfred Einsiedler and Thomas Ward, Ergodic Theory with a view towards Number Theory, p. 99, Theorem 4.4. 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 $S$.66 6 Walter Rudin, Functional Analysis, second ed., p. 75, Theorem 3.23. In particular this shows us that there exist extreme points of $S$. Let $\mathscr{E}^{T}(X)$ be the set of extreme points of $\mathscr{M}^{T}(X)$, and applying the Krein-Milman theorem with $\mathscr{M}^{T}(X)$, which is a nonempty compact convex set in the locally convex space $C(X)^{*}$, we have that $\mathscr{M}^{T}(X)$ is equal to the closed convex hull $\mathscr{E}^{T}$. That is, $\mathscr{M}^{T}(X)$ is equal to the closed convex hull of the set of ergodic $\mu\in\mathscr{M}^{T}(X)$.

Choquet’s theorem77 7 Manfred Einsiedler and Thomas Ward, Ergodic Theory with a view towards Number Theory, p. 103, Theorem 4.8. tells us that for each $\mu\in\mathscr{M}^{T}(X)$ there is a unique Borel probability measure $\lambda$ on the compact metrizable space $\mathscr{M}^{T}(X)$ such that

 $\lambda(\mathscr{E}^{T}(X))=1$

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

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

We have established that $\mathscr{M}^{T}(X)$ contains at least one element. $T$ is called uniquely ergodic if $\mathscr{M}^{T}(X)$ is a singleton. If $\mathscr{M}^{T}(X)=\{\mu_{0}\}$ then $\mu_{0}$ is an extreme point of $\mathscr{M}^{T}(X)$, hence is ergodic. If $\mathscr{E}^{T}(X)=\{\mu_{0}\}$, then for $\mu\in\mathscr{M}^{T}(X)$, by Choquet’s theorem there is a unique Borel probability measure $\lambda$ on $\mathscr{M}^{T}(X)$ 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 $\mathscr{E}^{T}(X)$ 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 $X$.88 8 Manfred Einsiedler and Thomas Ward, Ergodic Theory with a view towards Number Theory, p. 105, Theorem 4.10. This constant $C_{f}$ is equal to $\int_{X}fd\mu$, where $\mathscr{M}^{T}(X)=\{\mu\}$.

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 $\mathscr{M}^{R_{g}}(X)$, and for $R_{g}$ to be uniquely ergodic means that $m_{X}$ is the only element of $\mathscr{M}^{R_{g}}(X)$. 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.99 9 Manfred Einsiedler and Thomas Ward, Ergodic Theory with a view towards Number Theory, p. 108, Theorem 4.14.

###### Theorem 4.

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

Let $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, let $X=\mathbb{T}^{d}=\mathbb{R}^{d}/\mathbb{Z}^{d}$, which is a compact abelian group, and let $g=(\alpha_{1},\ldots,\alpha_{d})\in\mathbb{R}^{d}$. For $\chi\in\widehat{X}=\mathbb{Z}^{d}$, $\chi=(k_{1},\ldots,k_{d})$,

 $\chi(g)=\exp\left(2\pi i\sum_{j=1}^{d}k_{j}\alpha_{j}\right).$

$\chi(g)=1$ 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 $k_{1}\alpha_{1}+\cdots+k_{d}\alpha_{d}+k_{d+1}=0$. Therefore for $\alpha_{1},\ldots,\alpha_{d}\in\mathbb{R}$, the set $\{\alpha_{1},\ldots,\alpha_{d},1\}$ is linearly independent over $\mathbb{Z}$ if and only if for $g=(\alpha_{1},\ldots,\alpha_{d})$, the map $R_{g}(x)=x+g$, $\mathbb{T}^{d}\to\mathbb{T}^{d}$, is uniquely ergodic.