= Solution
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 $R_\alpha(x)=x+\alpha\pmod1$ 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 gives
$$
c_k=\int e^{2\pi ikR_\alpha(x)}\,d\nu(x)=e^{2\pi ik\alpha}c_k.
$$
For $k\ne0$, irrationality makes $e^{2\pi ik\alpha}\ne1$, 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
$$
\boxed{R_\alpha\text{ is uniquely ergodic, with invariant measure }m}.
$$
For the general <uniquely ergodic> system, fix $x\in X$ and form the <empirical measures>
$$
\nu_{N,x}=\frac1N\sum_{n=0}^{N-1}\delta_{T^nx}.
$$
On 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 $g$,
$$
\int(g\circ T-g)\,d\nu_{N,x}=\frac{g(T^Nx)-g(x)}N\longrightarrow0.
$$
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 $f$ proves
$$
\boxed{\lim_{N\to\infty}\frac1N\sum_{n=0}^{N-1}f(T^nx)=\int f\,d\mu\quad\text{for every }x\in X}.
$$
In 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 $\{p\}\sqcup\mathbb Z_{\ge0}$, 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 $2^q=10^p=2^p5^p$, 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 when
$$
t_n\in I=[\log_{10}7,\log_{10}8).
$$
The 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 $\varepsilon>0$, choose continuous functions $a_\varepsilon,b_\varepsilon$ on the circle with $0\le a_\varepsilon\le\mathbf1_I\le b_\varepsilon\le1$ and $\int(b_\varepsilon-a_\varepsilon)\,dm<\varepsilon$, 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 $\varepsilon\downarrow0$ proves <everywhere interval frequency under an irrational rotation>. Consequently
$$
\boxed{\lim_{N\to\infty}\frac{|S\cap[0,N-1]|}{N}=m(I)=\log_{10}8-\log_{10}7=\frac{\log8-\log7}{\log10}}.
$$
This is the leading-seven case of <Benford frequencies for powers of an integer>.
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