A nonzero L1 function and its transform cannot both have compact support

Given: Countable choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), a function $f\in L^1({\mathbb{R}}^n;\mathbb{C})$, compact sets $K_1,K_2\subseteq {\mathbb{R}}^n$ with $f=0$ almost everywhere on ${\mathbb{R}}^n\setminus K_1$ and $\widehat{f}=0$ on ${\mathbb{R}}^n\setminus K_2$.

Countable choice is assumed; it is the hypothesis carried by the $L^1$ uniqueness theorem below (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).

Compact support gives an entire continuation: $F(z):={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi ix\cdot z}dx$ converges absolutely at every $z\in {\mathbb{C}}^n$, has entire coordinate slices, and satisfies $F(x)=\widehat{f}(x)$ for every real $x$ (Compact support gives an entire Fourier-Laplace transform by slices).

A separately holomorphic map on ${\mathbb{C}}^n$ that vanishes on a nondegenerate real box $I_1\times \cdots \times I_n$ is identically zero (Separately holomorphic functions vanishing on a real box are zero).

If $g,h\in L^1({\mathbb{R}}^n;\mathbb{C})$ have equal $L^1$ transforms, then $g=h$ almost everywhere (Uniqueness of the L1 Fourier transform).

A subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded; in particular compact subsets are closed (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line). Consequently a nonempty open set $U\subseteq {\mathbb{R}}^n$ contains a nondegenerate box: if $\xi \in U$ and the ball of radius $\epsilon >0$ around $\xi$ lies in $U$, then ${\prod }_{j=1}^n[{\xi }_j-\epsilon \mathrm{/}(2\sqrt{n}),{\xi }_j+\epsilon \mathrm{/}(2\sqrt{n})]\subseteq U$, since every point of this box is at Euclidean distance at most $\epsilon \mathrm{/}2$ from $\xi$.