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For irrational $\alpha$, the irrational skew shift is the skew product $T_{\alpha }(x,y)=(x+\alpha ,y+x)$ on the torus $(\mathbb{R}/\mathbb{Z}{)}^2$. It preserves normalized Lebesgue measure and has iterates $T_{\alpha }^n(x,y)=(x+n\alpha ,y+nx+\alpha n(n-1)/2)$ modulo one. Its Koopman operator sends the Fourier basis character $e_{r,s}=e^{2\pi i(rx+sy)}$ to $e^{2\pi ir\alpha }{e}_{r+s,s}$. The resulting infinite chains of Fourier coefficients show that it is an ergodic transformation. Moreover, uniform equidistribution of an irrational skew shift proves unique ergodicity.

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.

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.