order to prove (\[item:sigmaN-2\]), suppose first that $\lim_{N\to \infty} \sigma_N(\omega) = 0$. This implies that $\lim_{N\to \infty}S_N(\xi_\mathbf{k}, \omega) = 0$ for all $\mathbf{k} \in \Lambda^*$. The class ${\mathcal{F}}$ is u.d.d., hence $\omega$ is uniformly distributed in $[0,1)^s$.

To prove the converse, assume that $\omega$ is uniformly distributed in $[0,1)^s$. For any $\epsilon > 0$, there exists a positive integer $K_0 = K_0(\epsilon)$ such that $\rho({\ensuremath{\mathbf k}}) < \epsilon$ for all $\mathbf{k}$ with $||\mathbf{k}|| > K_0$. As in the proof of part (\[item:sigmaN-1\]), this gives $B_{N,K_0}\le \epsilon$, and, due to (\[eqn:sigmaN\]), $$\sigma_N(\omega) \le A_{N,K_0} + \epsilon.$$ The number $A_{N,K_0}$ is a maximum and the class ${\mathcal{F}}$ is u.d.d.. This implies the existence of $N_0=N_0(\epsilon)\in {\ensuremath{\mathbb{N}}}$ with the property $$\forall\ N \ge N_0(\epsilon):
    \ A_{N,K_0} = \max \left\{
        \rho({\ensuremath{\mathbf k}}) \left| S_N( \xi_\mathbf{k}, \omega) \right|:
        0< ||\mathbf{k}|| \le K_0
        \right\}
< \epsilon.$$

We deduce the relation $$\forall\ N \ge N_0(\epsilon): \ \sigma_N(\omega) < 2 \epsilon.$$ This proves the theorem.

Let ${\mathcal{F}}$, $\rho$ and $\sigma_N(\omega)$ be as in Definition \[definition:generalspectraltest\] and let $K$ denote an arbitrary positive integer. Then we have the following inequality of [Erdös-Turán-Koksma]{} for the spectral test: $$\sigma_N(\omega) \le \max \Big\{
    \max_{0<||\mathbf{k}|| \le K} \left\{
        \rho({\ensuremath{\mathbf k}}) \left| S_N( \xi_\mathbf{k}, 