for $f:W\to \mathbb{R}$, $$\begin{split} \mathbb{E}\left[\sum_{{\mathbf{x}}\in \beta\Lambda_{p_j}^{\prime}(C)} f({\mathbf{x}}) \right] &= \mathbb{E}\left[\sum_{{{\mathbf{x}}\in \beta\Lambda_{p_j}^{\prime}(C)}\above 0pt {\phi_{p_j}({\mathbf{x}}/\beta)=0}} f( {\mathbf{x}}) \right] + \mathbb{E}\left[\sum_{{{\mathbf{x}}\in \beta\Lambda_{p_j}^\prime(C)}\above 0pt {\phi_{p_j}({\mathbf{x}}/\beta)\neq0}} f({\mathbf{x}}) \right]. \\
    \end{split}
    \label{eq:splitAverage}$$ From the assumption on $f$, $$\left|\sum_{{{\mathbf{x}}\in \Lambda_{p_j}^{\prime}(C)}\above 0pt {\phi_{p_j}({\mathbf{x}})=0}} f(\beta {\mathbf{x}})\right|=\left|\sum_{{{\mathbf{x}}\in \Lambda_{p_j}^{\prime}}} f(\beta {\mathbf{x}})\right| \leq \sum_{{{\mathbf{x}}\in \Lambda_{p_j}\backslash\left\{\mathbf{0}\right\}}} \frac{b}{(1+\left\| \beta {\mathbf{x}}\right\|)^{m+\delta}}.$$ Since the lattice $\Lambda_p(C)$ has rank $m \geq 1$, the series on the right-hand-side of the above inequality is absolutely convergent for any $p_j$. Moreover, since, by assumption $$\left\| \beta \mathbf{x} \right\| \geq \beta \lambda_1(\Lambda_{p_j})\geq c V^{1/m} (V(\Lambda))^{-1/m} p_j^{\alpha} \to \infty,$$ each individual term of the last sum tends to zero, as $p_j \to \infty$ and, by dominated convergence, the sum tends to zero. Let $\gamma = (p_j^k-1)/(p_j^n-1)$. For the second term of Equation , we have: $$\begin{split} \mathbb{E}\left[\sum_{{{\mathbf{x}}\in \beta\Lambda_{p_j}^{\prime}(C)}\above 0pt {\phi_{p_j}({\mathbf{x}})\neq0}} f(\beta {\mathbf{x}}) \right]  \stackrel{(a)}{=} \gamma \sum_{{\mathbf{x}}\in \Lambda^{\prime}}f(\beta{\mathbf{x}})  
    \stackrel{(b)}{=}  \sum_{r = 1}^{\infty} \frac{\mu(r)}{r^m} \sum_{{\mathbf{x}}\in \Lambda{\backslash\left\{0\right\}}}r^{m} \gamma f(r\beta {\mathbf{x}})
    \end{split}$$ where $\mu$ denotes the Möbius function (see, e.g. [@Cassels Sec. VI. 3.2]). In the above, (a) follows from and (b) is the Möbius function inversion formula (see, e.g., [@Cassels Sec. VI. 3.2]). The theorem follows by using the property $$\sum_{r=1}^{\infty}\frac{\mu(r)}{r^m} = \frac{1}{\zeta(m)}$$ and observing that for the inner sum satisfies $$\lim_{p_j \to \infty} \sum_{{\mathbf{x}}\in 