Compactly supported smooth functions are unique from any open set of Radon angles

The hypothesis and the compact-support Fourier-slice identity [F2] give $\widehat{f}(\sigma \theta )=0$ for every $\theta \in \mathrm{\Theta }$ and $\sigma \in \mathbb{R}$. By [F1], this includes the nonempty open positive-radius sector generated by $J$; choose a product of open real intervals $I_1\times I_2$ inside that sector. For each real ${\xi }_2\in I_2$, the entire function $z_1\mapsto \widehat{f}(z_1,{\xi }_2)$ vanishes on $I_1$, so [F5] makes it zero for every complex $z_1$. Then for every real ${\xi }_1$ the entire second-coordinate slice vanishes on $I_2$, and [F5] makes it zero everywhere. Thus $\widehat{f}$ vanishes on all of ${\mathbb{R}}^2$.

Let $K=\mathrm{supp}f$. If $K=\mathrm{\varnothing }$, then $f=0$. Otherwise, fix a rectangle $B$ whose interior contains $K$ and interpret all integrals below as compact-support Riemann integrals over $B$. Repeated use of the coordinate derivative formula in [F3] shows ${\int }_B{x}^{\alpha }f(x)dx=0$ for every multi-index $\alpha$, since all real-frequency derivatives of $\widehat{f}$ vanish at the origin. The real polynomials restricted to $K$ form a unital subalgebra separating points, because the coordinate functions distinguish distinct points. For an arbitrary $\epsilon >0$, apply [F6] separately to $\mathrm{Re}\bar{f}$ and $\mathrm{Im}\bar{f}$ with error $\epsilon \mathrm{/}2$ and combine the two real polynomials into one complex polynomial $p$ with $\mathrm{\parallel }p-\bar{f}{\mathrm{\parallel }}_{C(K)}<\epsilon$. Every moment is zero, so ${\int }_{B}fp=0$. Since $f=0$ outside $K$, the real Riemann integral bound [F4] on both components gives $0\le {\int }_B\mathrm{\mid }f{\mathrm{\mid }}^2=\mid {\int }_{B}f(\bar{f}-p)\mid \le \sqrt{2}\mathrm{vol}(B)\mathrm{\parallel }f{\mathrm{\parallel }}_{C(B)}\epsilon$. As this holds for every $\epsilon >0$, ${\int }_B\mathrm{\mid }f{\mathrm{\mid }}^2=0$; no sequence of approximants is selected. If $f$ were nonzero at a point, continuity would bound $\mathrm{\mid }f{\mathrm{\mid }}^2$ below by a positive number on a nondegenerate box inside $B$, forcing its nonnegative Riemann integral to be positive. Thus $f=0$ everywhere. This proof uses neither Fourier inversion nor Countable Choice.