d}}L_1\cdots{{\mathrm d}}L_v \nonumber \\
&=&\left(\frac{1}{\Delta L}{\int_{{\bar{L}}}^{{\bar{L}}+\Delta L}}\exp({{\mathrm i}}\zeta\tan kL){{\mathrm d}}L
\right)^v.\end{aligned}$$ The subscript ${\rm L}$ indicates that the expectation is with respect to an average over bond lengths. Since the map $t\mapsto t^v$ is continuous, lemma \[lem:1\] together with [(\[eqn:10\])]{} allows us to deduce that $$\lim_{k\to\infty}{\mathbb E}_{\rm L}(\exp({{\mathrm i}}\zeta Z(k,{{\mathbf L}})))=
{{\mathrm e}}^{-v|\zeta|}.
\label{eqn:11}$$ The limiting density corresponding to the characteristic function on the right hand side is given by $$P_Z(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(-{{\mathrm i}}\zeta x-v|\zeta|)
{{\mathrm d}}\zeta =\frac{1}{\pi}\frac{v}{v^2+x^2}.$$ The theorem follows now from the classical continuity theorem for characteristic functions ([@fel:ipt] chapter XV).

$\square$

The proof of theorem \[thm:2\] uses Weyl’s Equidistribution theorem. This celebrated result [@wey:ube] has numerous applications in analysis and number theory. We state here the form most convenient for application to our current work. Let ${\mathbb T}^v$ be the $v$-dimensional torus, with sides of length $\pi$.

\[thm:weyl\] Let $f\in C({\mathbb T}^v)$, and let the components of ${{\mathbf L}}$ be linearly independent over ${\mathbb Q}$. Then $$\lim_{K\to\infty}\frac{1}{K}\int_0^K f(L_1k,\ldots,L_vk){{\mathrm d}}k= 
\frac{1}{\pi^v}\int_{{\mathbb T}^v} f({{\mathbf x}}){{\mathrm d}}{{\mathbf x}}$$ where $\displaystyle {{\mathrm d}}{{\mathbf x}}={{\mathrm d}}x_1\cdots{{\mathrm d}}x_v$ denotes Lebesgue measure.

We shall use Weyl’s theorem as our main tool to relate $k$-averages to bond length averages. It is for this reason that it is 