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If $a\ge 2$ is an integer with irrational ${\log }_{10}a$, then the leading digit $d$ of $a^n$ has limiting frequency ${\log }_{10}(1+1/d)$. Indeed the logarithmic fractional parts are the orbit of zero under an irrational rotation of the circle, so everywhere interval frequency under an irrational rotation applies. Thus for $a=2$, digit seven has frequency $(\log 8-\log 7)/\log 10$.

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.

Write the interval as $I=(c,d]$, where $0<c<d<1$, to avoid confusing its left endpoint with the rotation parameter $\alpha$. For every sufficiently large integer $k$, define an inner and outer interval on the circle byApply part (b) simultaneously to this countable collection of intervals. The intersection $G$ of the corresponding full-measure sets still has full Lebesgue measure, and is therefore dense in the circle.

Fix an arbitrary $x$. For each sufficiently large $k$, choose $y_k\in G$ with circle distance less than $1/k$ from $x$. An irrational circle rotation preserves this distance, so for every $j\ge 0$,Averaging and using part (b) for $y_k$ givesLetting $k\to \infty$ proves the everywhere interval frequency under an irrational rotation: