Let $d\sigma$ denote the surface measure on $\mathbb{S}^{n-1}$. To compute its Fourier transform $$ \hat{d\sigma}(\xi)=\int e^{-i x\cdot \xi}\, d\sigma(x), $$ a standard technique (cfr. Folland's "Real Analysis" first edition, exercise 19, chapter 8: "Topics in Fourier Analysis") is based on the observation that $$\tag{1} \lvert x\rvert^2\,d\sigma = d\sigma, $$ because $d\sigma$ is supported on the unit sphere. Fourier transforming identity (1) one gets $$\tag{2} -\Delta_\xi \hat{d\sigma}=\hat{d\sigma}, $$ so the sought Fourier transform satisfies the Helmholtz equation. Moreover, $d\sigma$ is spherically symmetric, so equation (2) can be explicitly integrated with initial conditions $$\hat{d\sigma}(0)=\lvert \mathbb{S}^{n-1}\rvert\quad \text{and}\quad \frac{\partial\hat{d\sigma}}{\partial \rho}(0)=0,\quad \text{where}\ \rho=\lvert\xi\rvert.$$
Question. The measure $d\mu=g\,d\sigma$, where $g\ge 0$ is a smooth function on the sphere, is supported on $\mathbb{S}^{n-1}$. Applying the same computation as above we get for its Fourier transform that $$\tag{!!}\hat{d\mu}= \frac{\|g\|_{L^1(\mathbb{S}^{n-1})}}{\lvert \mathbb{S}^{n-1}\rvert} \hat{d\sigma}, $$ which, by uniqueness of the Fourier transform, implies that $g$ must be constant. This is absurd. Where is the fallacy in this reasoning?
