Suppose $u_c \in L^1(\mathbb{R}^D)$ has the Fourier transform $\widehat{u_c}$, and that we're sampling it on a multidimensional lattice $\Lambda$, with reciprocal lattice $\Lambda^*$. The sampled function is $u: \Lambda \to \mathbb{R}$ given by $u(x) = u_c(x)$ for all $x \in \Lambda$. By the inverse Fourier transform, $$u_c(x) = \int_{\mathbb{R}^D} \widehat{u_c}(\xi) \exp(2\pi i \xi \cdot x) d\xi$$ That means we can evaluate $u(x)$ the same way for $x \in \Lambda$ \begin{align} u(x) &= \int_{\mathbb{R}^D} \widehat{u_c}(\xi) \exp(2\pi i \xi \cdot x) d\xi \\ &= \int_P \sum_{r \in \Lambda^*} \widehat{u_c}(\xi + r) \exp(2\pi i (\xi + r) \cdot x) d\xi \\ &= \int_P \left[ \sum_{r \in \Lambda^*} \widehat{u_c}(\xi + r) \right] \exp(2\pi i (\xi) \cdot x) d\xi \end{align} where $P$ is a unit cell of $\Lambda^*$ (meaning it tiles $\mathbb{R}^D$ by translations by elements of $\Lambda^*$), and the last equality follows from the definition of the repiprocal ($r \cdot x$ is an integer for all $x \in \Lambda, r \in \Lambda^*$). All this I understand. Then, Dubois claims here that by taking the Fourier transform of the sampled signal $u(x)$, we get $$ \widehat{u}(\xi) = \frac{1}{d(\Lambda)} \sum_{r \in \Lambda^*} \widehat{u_c}(\xi + r)$$ which I am just not seeing at all. When I take the Fourier transform of $u(x)$, I get
\begin{align} \hat{u}(\xi) &= \sum_{x \in \Lambda} u(x) \exp(-2\pi i \xi \cdot x) \\ &= \sum_{x \in \Lambda} \int_P \sum_{r \in \Lambda^*}\widehat{u_c}(\xi + r) d\xi \end{align} because of the cancellation from the exponential terms. But I don't see at all how to eliminate the summation over $x \in \Lambda$, or resolve the integration of $\widehat{u_c}$ over $P$.
It works the same way as in dimension $1$ : Fourier series of periodization.
For $L= A \Bbb{Z}^n$ a lattice ($A\in GL_n(\Bbb{R})$) and $g$ is say Schwartz. Then the Fourier transform of $$h(x) = g(x)\sum_{l\in L}\delta(x-l)$$ is $$H(\xi) = \int_{\Bbb{R}^n}e^{-2i\pi <\xi,x>} g(x)\sum_{l\in L}\delta(x-l) dx = \sum_{l\in L} e^{-2i\pi <\xi,l>} g(l)$$ Then look at $$U(\xi) = \sum_{k\in A^{-\top}\Bbb{Z}^n} G(\xi-k)$$ It is equal to its Fourier series $$U(\xi ) = \frac1{|\det(A^{-\top})|}\sum_{l\in L} e^{2i\pi <\xi,l>} c_l$$ $$c_l = \int_{A^{-\top} [0,1]^n} U(\xi) e^{-2i\pi <\xi,l>} d\xi = \int_{\Bbb{R}^n} G(\xi) e^{-2i\pi <\xi,l>} d\xi= g(-l)$$
ie. $$U(\xi) =\frac1{|\det(A^{-\top})|}H(\xi)$$