${{1\!\!1}}_{\Sigma_{\epsilon,{{\mathbf L}}}}({{\mathbf x}})$ is piecewise constant.

By the differentiability properties of $\tilde{f}$ we can write for every ${{\mathbf x}}\in\Sigma$ $$f({{\mathbf x}})=\tilde{f}({{\mathbf x}})=
\frac{1}{2\epsilon}\int_{-\epsilon}^{\epsilon}
\tilde{f}(k{{\mathbf L}}+{{\mathbf x}}){{\mathrm d}}k+{{\mathrm O}}(\epsilon)$$ as $\epsilon\to 0$. The implied constant does not depend on ${{\mathbf x}}$. Setting ${{\mathbf x}}=k_n{{\mathbf L}}$ gives $$\frac{1}{N}\sum_{n=1}^N f(k_n{{\mathbf L}})=\frac{1}{2\epsilon N}\sum_{n=1}^N
\int_{-\epsilon}^{\epsilon}\tilde{f}((k+k_n){{\mathbf L}}){{\mathrm d}}k+{{\mathrm O}}(\epsilon).$$ Let the mean density of zeros of $Z(k,{{\mathbf L}})$ be $\bar{d}$: $$\bar{d}:=\lim_{K\to\infty}\frac{\#\{n:k_n\leq K\}}{K}.$$ Then $$\begin{aligned}
  \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(k_n{{\mathbf L}})&=&\frac{1}
{2\epsilon\bar{d}}\lim_{K\to\infty}\frac{1}{K}\int_0^{K} \tilde{f}(k{{\mathbf L}})
{{1\!\!1}}_{\Sigma_{\epsilon,{{\mathbf L}}}}(k{{\mathbf L}}){{\mathrm d}}k+{{\mathrm O}}(\epsilon) \\
&=&\frac{1}{\pi^v \bar{d}}\int_{{\mathbb T}^v} \tilde{f}({{\mathbf x}})
{{1\!\!1}}_{\Sigma_{\epsilon,{{\mathbf L}}}}({{\mathbf x}}){{\mathrm d}}{{\mathbf x}} +{{\mathrm O}}(\epsilon),\end{aligned}$$ applying theorem \[thm:weyl\] to the piecewise continuous function $\tilde{f}({{\mathbf x}})
{{1\!\!1}}_{\Sigma_{\epsilon,{{\mathbf L}}}}({{\mathbf x}})$. Changing to the system of coordinates $(t,{{\boldsymbol \xi}})$ on ${\mathbb T}^v$ via the change of variables $$x_i=L_i t+s_i({{\boldsymbol \xi}}),$$ gives $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(k_n{{\mathbf L}})=\frac{1}
{\pi^v\bar{d}}\int_{\Sigma}\frac{1}{2\epsilon}\int_{-\epsilon}^{\epsilon} 
\tilde{f}(t,{{\boldsymbol \xi}})J({{\boldsymbol \xi}}){{\mathrm d}}t{{\mathrm d}}{{\boldsymbol \xi}}+{{\mathrm O}}(\epsilon).$$ Since this is true for all $\epsilon>0$, we deduce that $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(k_n{{\mathbf L}})=\frac{1}
{\pi^v\bar{d}}\int_{\Sigma}{f}({{\boldsymbol \xi}})J({{\boldsymbol \xi}}){{\mathrm d}}{{\boldsymbol \xi}}.$$ By setting $f=1$, we see that $$\bar{d}=\frac{1}{\pi^v}\int_{\Sigma}J({{\boldsymbol \xi}}){{\mathrm d}}{{\boldsymbol \xi}},$$ to complete the proof.

$\square$

We note incidentally that $\bar{d}$ can be evaluated using spectral methods [@kot:pot] to give $$\bar{d}=\frac{1}{\pi}\sum_{j=1}^v L_j=\frac{v{\bar{L}}}{\pi}+{{\mathrm O}}(v\Delta L)
\qquad\mbox{as $v\Delta L\to 0$.}$$

We observe that the right hand side of equation 