Denote by $M_{sc}(\mathbb{R})$ the set of all continuous singular probability measures on $\mathbb{R}$. Is there $\mu\in M_{sc}(\mathbb{R})$ such that $\lambda(\{x:\int\limits_\mathbb{R} e^{itx}d\mu(t)=0\})>0$, where $\lambda$ is Lebesgue measure?
It is easy to give examples of Fourier transform of absolutely continouos and discrete measures with such property. For example, $(1-|x|)_+$ and its periodic extension from $[-h,h]$, where $h>1$. But I don't know how to approach this question in case of continuous singular measures.
Partial answer:
Let $s$ be a Fourier transform of continuous singular probability measure $\nu$ on $\mathbb{T}$, i.e. $s(n):=\int\limits_\mathbb{T}e^{int}d\nu(t)$ such that $0<\limsup\limits_{n\to\infty} |s(n)|<1$ and let $f$ be a compactly supported characteristic function, $supp f\subset[-1,1]$. Let $h>1$. Consider the function $$ F(x):=\sum\limits_{n\in\mathbb{Z}}s(n)f(x-2nh),\ x\in\mathbb{R}. $$ It's known, that $F$ is a characteristic function (see W. Feller An Introduction to Probability Theory and Its Application, vol. 2, Ch. XIX, Problem 15--16). So it is sufficient to show that a measure which corresponds to $F$ contains a continuous singular component.