Weyl equidistribution for irrational rotations

If $a=c$, use the zero function; if $(a,c)=(0,1)$, use the constant-one function. Otherwise, for every sufficiently small $\delta >0$, circular distance gives continuous functions $0\le {\mathrm{\ell }}_{\delta }\le {\mathbf{1}}_I\le u_{\delta }\le 1$ as follows: ${\mathrm{\ell }}_{\delta }$ is zero on the closed complementary arc and rises linearly to one within distance $\delta$ inside $I$, while $u_{\delta }$ is one on the closed arc $[a,c]$ and falls linearly to zero within distance $\delta$ outside it. Thus they differ from ${\mathbf{1}}_I$ only in the two boundary arcs of total length at most $4\delta$.

Monotonicity, linearity, and [F3] give $c-a-2\delta \le \int {\mathrm{\ell }}_{\delta }d\lambda \le c-a\le \int u_{\delta }d\lambda \le c-a+2\delta ,$ after decreasing $\delta$ if necessary; estimates truncated at $0$ and $1$ give the same conclusion near a degenerate complementary arc.

Since $R_{\alpha }^k(0)=\{k\alpha \}$, the visit frequency to $I$ is $A_n{\mathbf{1}}_I(0)$. The pointwise sandwiches and [F2] yield $\int {\mathrm{\ell }}_{\delta }d\lambda \le {\mathrm{lim\hspace{0.17em}inf}}_n{A}_n{\mathbf{1}}_I(0)\le {\mathrm{lim\hspace{0.17em}sup}}_n{A}_n{\mathbf{1}}_I(0)\le \int u_{\delta }d\lambda \mathrm{.}$ Letting $\delta \downarrow 0$ and using step 2.1 proves that the limit is $c-a$.

The interval $I$ was arbitrary, so the definition of equidistribution applies. Countable choice is inherited from [F1]; the two approximants for a fixed interval and $\delta$ are explicit. This proof does not use Fourier's Weyl criterion.