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A continuous map $F$ of a compact metric space is uniquely ergodic if it has exactly one invariant measure that is a Borel probability measure. The uniqueness requirement ranges over all invariant Borel probability measures.

For an irrational rotation of the circle, let $R_{\alpha }x=x+\alpha$ on $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, and let $\nu$ be any invariant Borel probability measure. For each integer $r$, set ${\chi }_r(x)=e^{2\pi irx}$. Invariance givesFor $r\ne 0$, irrationality forces $e^{2\pi ir\alpha }\ne 1$, so the corresponding Fourier coefficient is zero. For $r=0$ it is one. These are exactly the Fourier coefficients of normalized Lebesgue measure $m$. By the Stone-Weierstrass theorem, trigonometric polynomials are uniformly dense in the continuous functions on the circle group; hence $\nu$ and $m$ integrate every continuous function equally and are the same Borel probability measure. Since $m$ is invariant, the irrational rotation of the circle is uniquely ergodic, with unique measure $m$.

For the irrational skew shift on the two-dimensional torus, writeThe map is invertible, with $T^{-1}(x,y)=(x-\alpha ,y-x+\alpha )$ modulo one, and it preserves $m_2$ as allowed in the question. We first prove the ergodic transformation property by the invariant-function characterization of ergodicity.

Let $F\in L^2(m_2)$ satisfy $F\circ T=F$. Its expansion in the Fourier basis is $F={\sum }_{r,s}{c}_{r,s}{e}_{r,s}$ in $L^2$. Direct calculation of the Koopman operator givesUniqueness of the Fourier coefficients therefore impliesFor $s\ne 0$, the magnitudes of the Fourier coefficients along all distinct indices $(r+js,s)$, $j\in \mathbb{Z}$, are equal. By the Bessel inequality they are square summable, so every such coefficient must be zero. When $s=0$, the relation becomes $c_{r,0}=e^{2\pi ir\alpha }{c}_{r,0}$, which forces $c_{r,0}=0$ for $r\ne 0$. Only $c_{0,0}$ remains. Thus every invariant $L^2$ function is constant, and

To prove unique ergodicity, we will establish uniform averages for every continuous function directly. No theorem on unique ergodicity of skew products is needed. Induction on $n$, using the old first coordinate in the second coordinate of $T$, yieldsIn particular, for a Fourier basis element,If $s=0$ and $r\ne 0$, this is a geometric series in $n$, anduniformly in $(x,y)$.

For $s\ne 0$, we give the finite Van der Corput inequality for finite scalar sequences and its proof. For $|u_n|\le 1$, extend $u_n$ by zero outside $0\le n<N$. Fix an integer $1\le H\le N$ and set $C_h={\sum }_{n=0}^{N-h-1}{u}_{n+h}\overline{u_n}$. Every original summand appears $H$ times in the identityThe Cauchy-Schwarz inequality, followed by expansion of the squared window sums, givesConsequently the Van der Corput inequality for finite scalar sequences isThe finite prefactor is important; the order of limits will be $N\to \infty$ with $H$ fixed, followed by $H\to \infty$.

For the irrational skew shift character sequence above, differencing cancels the quadratic term:For every fixed $h\ge 1$, $s\alpha h$ is irrational, so another geometric series estimate givesuniformly in $(x,y)$. With fixed $H$, the Van der Corput inequality for finite scalar sequences therefore impliesLetting $H\to \infty$ proves that the averages of every nonconstant Fourier basis element converge uniformly to zero. The constant character has average one. This establishes uniform equidistribution of an irrational skew shift on all trigonometric polynomials.

The Stone-Weierstrass theorem makes these trigonometric polynomials uniformly dense in $C({\mathbb{T}}^2)$. If $p$ approximates a continuous $g$ with $\Vert g-p{\Vert }_{\infty }<\epsilon$, thenTaking $N\to \infty$ and then $\epsilon \downarrow 0$ proves uniform convergence to $\int gd{m}_2$ for every continuous $g$.

Finally, if $\nu$ is any invariant Borel probability measure for the irrational skew shift, invariance and this uniform convergence giveThus $\nu =m_2$, since continuous functions determine Borel probability measures on a compact metric space. We concludeThe uniform-average proof also shows that every starting point has the same limiting continuous-function averages, a stronger conclusion than the almost-everywhere assertion provided by the pointwise ergodic theorem.

A continuous transformation $T$ of a compact metric space $X$ is uniquely ergodic when there is exactly one $T$-invariant Borel probability measure on $X$. We use the usual compact-space convention for unique ergodicity; the compactness and continuity hypotheses matter in the assertion about all starting points.

Let $R_{\alpha }(x)=x+\alpha (\mathrm{mod}1)$ on the circle group, with $\alpha$ irrational. Normalized Lebesgue measure $m$ is invariant under this irrational rotation of the circle. If $\nu$ is any invariant Borel probability measure, define its Fourier coefficients by $c_k=\int e^{2\pi ikx}d\nu (x)$. Invariance givesFor $k\ne 0$, irrationality makes $e^{2\pi ik\alpha }\ne 1$, so $c_k=0$; also $c_0=1$. These are the Fourier coefficients of $m$. Hence $\nu$ and $m$ have the same integrals against every trigonometric polynomial. Such polynomials are uniformly dense in the continuous functions by the Stone-Weierstrass theorem, so the measures agree on every continuous test function and therefore agree as Borel measures. Thus

For the general uniquely ergodic system, fix $x\in X$ and form the empirical measuresOn a compact metric space, the Borel probability measures are compact for weak convergence of probability measures. Any subsequential limit $\nu$ is invariant: for every continuous $g$,Here $g$ is bounded and $g\circ T$ is continuous, so the identity passes to the limit. Uniqueness of the invariant Borel probability measure gives $\nu =\mu$. Every subsequential limit is therefore $\mu$, and compactness implies convergence of the entire sequence. Testing against $f$ provesIn fact this proves uniform ergodic convergence for uniquely ergodic systems: if convergence were not uniform in $x$, choose $N_j\to \infty$ and $x_j$ where the discrepancy stays above a fixed positive number. The same compactness and telescoping argument applied to ${\nu }_{N_j,x_j}$ forces a subsequence to converge to $\mu$, a contradiction. This also makes clear why an almost-everywhere Birkhoff ergodic theorem alone would not establish the requested everywhere assertion. Without the compact-space hypothesis the assertion need not hold: on the discrete space $\{p\}\bigsqcup {\mathbb{Z}}_{\ge 0}$, set $T(p)=p$ and $T(n)=n+1$. The only invariant probability is ${\delta }_p$, but the continuous bounded function which is zero at $p$ and one on the integer orbit has orbit average one there.

For the decimal application put $\alpha ={\log }_{10}2$. This is irrational: if $\alpha =p/q$ with positive integers $p,q$, then $2^q=10^p=2^p{5}^p$, contradicting unique prime factorization. Writing $n\alpha =m_n+t_n$ with $m_n=\lfloor n\alpha \rfloor$ and $t_n\in [0,1)$ gives $2^n=10^{m_n}10^{t_n}$. Its leading decimal digit is seven exactly whenThe half-open upper endpoint correctly excludes powers whose leading digit is eight, including $2^3=8$.

The indicator of $I$ is not continuous, so an extra step is needed. For any $\epsilon >0$, choose continuous functions $a_{\epsilon },b_{\epsilon }$ on the circle with $0\le a_{\epsilon }\le 1_I\le b_{\epsilon }\le 1$ and $\int (b_{\epsilon }-a_{\epsilon })dm<\epsilon$, by tapering in small neighbourhoods of the two endpoints. Applying the everywhere averaging result to these functions traps the lower and upper limits of the interval frequency between their integrals. Letting $\epsilon \downarrow 0$ proves everywhere interval frequency under an irrational rotation. ConsequentlyThis is the leading-seven case of Benford frequencies for powers of an integer.

For a Borel probability measure on a Polish space and a continuous map to another Polish space, the image measure has the displayed support. Every neighborhood of an image of a support point has an open inverse image of positive measure. Conversely, the closed image closure has full image measure because the original support has full measure. This transfers a support description for a driving rough path to the solution of a rough differential equation.

A probability measure is tight if compact sets carry arbitrarily nearly all its mass. Every Borel probability measure on a Polish space is tight. The stronger family condition is uniform tightness.

For an irrational skew shift, the averages of every continuous function $g$ converge uniformly in the starting point to $\int gd{m}_2$, with $m_2$ normalized Lebesgue measure. Nonconstant Fourier basis characters have either linear or quadratic phases. Linear phases are bounded geometric series; for quadratic phases the Van der Corput inequality for finite scalar sequences reduces to linear correlations of irrational frequency, uniformly in the starting point. Approximation by trigonometric polynomials proves the assertion and identifies every invariant Borel probability measure as $m_2$.

A continuous self-map of a compact metric space is uniquely ergodic when it has exactly one invariant Borel probability measure. An irrational rotation of the circle is uniquely ergodic: invariance multiplies its nonzero-index Fourier coefficients by nontrivial phases, forcing those coefficients to vanish. The Stone-Weierstrass theorem then identifies the invariant measure as normalized Lebesgue measure.