Let $\mu$ be a probability measure on $\mathbb R$ and the Fourier transform of $\mu$ is defined by $$ \widehat{\mu}(\xi) = \int_{\mathbb{R}} e^{-2\pi i \xi x} ~ d \mu(x). $$ Let $\mathcal{Z}(\widehat{\mu})=\{ \xi \in \mathbb R: \widehat{\mu}(\xi)=0 \}$.
Question: how about the set $\mathcal{Z}(\widehat{\mu})$? Can we show that the set $\mathcal{Z}(\widehat{\mu})$ is at most countable?
Try: when $\mu$ has compact support, we can extend $\widehat{\mu}$ to the whole complex plane by $$ \widehat{\mu}(z) = \int_{\mathbb{R}} e^{-2\pi i z x} ~ d \mu(x). $$ Next, using compact support, we can show $\widehat{\mu}(z)$ is an entire function. By the theorem in complex analysis, we can obtain that the set $\mathcal{Z}(\widehat{\mu})$ has no limit point.
I have no idea of the case when the support of $\mu$ is not compact.
The function $\phi (t)=1-|t|$ for $|t| \leq 1$ and $0$ for $|t| >1$ is the Fourier transform of a probablity measure. This is proved in Feller's book: An Introduction to Probability Theory and its Applications. See the section on 'Special densities. Mixtures' in the chapter on 'Characterisitic Functions'.