Irrational circle rotations are uniquely ergodic

Fix $f\in C(\mathbb{T},\mathbb{R})$. Compactness makes $f$ bounded, hence $f\in L^1(\lambda )$. By [F2]–[F3], $A_{n}f$ converges on a conull set $Y_f$ to a constant $c_f$. Since $\mathrm{\mid }{A}_{n}f\mathrm{\mid }\le \parallel f{\parallel }_{\mathrm{\infty }}$, dominated convergence and invariance give $c_f=\int c_{f}d\lambda ={\lim }_n\int A_{n}fd\lambda =\int fd\lambda \mathrm{.}$

Given $\epsilon >0$, uniform continuity of $f$ gives $\delta >0$ such that $d(x,y)<\delta$ implies $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\epsilon \mathrm{/}3$. Rotations are isometries, so the same $\delta$ gives $\mathrm{\mid }{A}_{n}f(x)-A_{n}f(y)\mathrm{\mid }<\epsilon \mathrm{/}3$ for every $n$ whenever $d(x,y)<\delta$.

The set $Y_f$ is dense. Indeed, every nonempty circle-open set contains a nondegenerate ordinary interval, possibly on one side of the cut, and such an interval has positive Lebesgue measure by the interval-measure formula. A conull set must meet it.

Compactness supplies a finite $\delta \mathrm{/}3$-net $x_1,\dots ,x_m$. By density choose, using only finite choice, $y_i\in Y_f$ with $d(x_i,y_i)<\delta \mathrm{/}3$. For all sufficiently large $n$, every $\mathrm{\mid }{A}_{n}f(y_i)-c_f\mathrm{\mid }<\epsilon \mathrm{/}3$. Given $x$, choose $i$ with $d(x,x_i)<\delta \mathrm{/}3$; then $d(x,y_i)<2\delta \mathrm{/}3$, and step 1.2 gives $\mathrm{\mid }{A}_{n}f(x)-c_f\mathrm{\mid }<2\epsilon \mathrm{/}3$. Thus $A_{n}f\to c_f$ uniformly.

The argument applies to every real continuous $f$, with $c_f=\int fd\lambda$. By [F5], $R_{\alpha }$ is uniquely ergodic and its unique invariant probability is the already invariant $\lambda$. Countable choice is inherited from [F2] and [F5]; the finite net and finite choices in step 3.1 require no stronger principle. No Fourier series or Weyl criterion was used.