Aliasing when spectral support has positive-measure overlap with a reciprocal translate

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $h>0$ and let $E\subseteq \mathbb{R}$ be Lebesgue measurable. Positive-measure overlap $E\cap (E+m\mathrm{/}h)$ for some $m\in \mathbb{Z}\setminus \{0\}$ gives a nonzero $L^2$ signal supported spectrally in $E$ whose samples on $h\mathbb{Z}$ all vanish. This is the failure mechanism behind reciprocal-lattice periodisation in Sampling at a lattice produces periodisation of the spectrum over the dual lattice.

To see this, partition $\mathbb{R}$ into half-open intervals of length $\mathrm{\mid }m\mathrm{\mid }\mathrm{/}(2h)$. Since the overlap has positive measure, countable subadditivity (Finite and countable subadditivity of measures) gives one interval $J$ for which $D:=J\cap E\cap (E+m\mathrm{/}h)$ has positive measure. It has finite measure by the box formula (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included), and $D$ and $D-m\mathrm{/}h$ are disjoint because $J$ is shorter than $\mathrm{\mid }m\mathrm{\mid }\mathrm{/}h$. Therefore $g:={\mathbf{1}}_D-{\mathbf{1}}_{D-m\mathrm{/}h}$ is a nonzero $L^1\cap L^2$ function vanishing off $E$. Its inverse Plancherel transform $f$ is nonzero (Plancherel theorem) and has the continuous representative $f(x)=\int g(\xi )e^{2\pi ix\xi }d\xi$ (L2 Fourier inversion, Agreement of the integral and L2 transforms, The L1 transform is bounded and uniformly continuous). Translation substitution (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions) and the exponential addition and kernel laws ($\exp (z+w)=\exp z\exp w$, and the complex exponential extends the real exponential, $\ker (\exp )=2\pi i\mathbb{Z}$, and $\exp z=\exp w$ exactly when $z-w\in 2\pi i\mathbb{Z}$) give $f(hk)=(1-e^{-2\pi imk}){\int }_D{e}^{2\pi ihk\xi }d\xi =0$ for every integer $k$. Thus $f$ and the zero signal have identical samples. No extension of the Schwartz-only distributional sampling formula is needed for this witness.

In the classical interval-band case, an interval of length greater than $1\mathrm{/}h$ has positive-measure overlap with its shift by $1\mathrm{/}h$. This corresponds to bandwidth beyond the cutoff $1\mathrm{/}(2h)$ at fixed sampling spacing, rather than a sampling rate above the Nyquist requirement. Being too wide to fit in an interval of length $1\mathrm{/}h$ is insufficient by itself for disconnected $E$: when $h=1$, the set $E=(0,1\mathrm{/}10)\cup (11\mathrm{/}10,6\mathrm{/}5)$ does not fit essentially in such an interval, but its fractional parts lie in the disjoint intervals $(0,1\mathrm{/}10)$ and $(1\mathrm{/}10,1\mathrm{/}5)$. Within each interval the fractional-part map is injective, so no distinct points of $E$ differ by an integer. Its integer translates are therefore pairwise disjoint. The centered-band reconstruction and its convergence modes remain those of Shannon sampling for band-limited $L^2$ functions.