I'm studying Bochner's theorem, which can be stated as follows:
$\textbf{Bochner's Theorem:}$ A continuous function $f:\mathbb{R}^{n} \to \mathbb{C}$ is positive definite if and only if there exists a finite valued and non-negative Borel measure $\mu_{f}$ on $\mathbb{R}^{n}$ such that $\hat{\mu}_{f} = f$.
When $f$ is smooth and rapidly decaying, i.e. $f \in \mathcal{S}(\mathbb{R}^{n})$, I imagined that the measure I was looking for was simply $d\mu_{f} = \mathcal{F}^{-1}(f) \, d\mu$, where $\mu$ is the usual Lebesgue measure on $\mathbb{R}^{n}$. This seems to satisfy the required properties, except that there is no reason that $\mathcal{F}^{-1}(f)$ should be positive, let alone real-valued. Hence the measure $\mathcal{F}^{-1}(f) \, d\mu$ is not, in general, the one described in Bochner's theorem. However, in some cases such as the Gaussian $f(x) = e^{-\|x\|^2}$, this procedure turns out to work fine. This leaves me confused, as I thought measures were uniquely determined by their Fourier transforms.
Can someone clear up the misconception I have? Is it possible to find the measure $\mu_{f}$ explicitly given that I know $f \in \mathcal{S}(\mathbb{R}^{n})$ explicitly?