Doe this imply absolute continuity? Let $\mu$ be a complex Borel measure on $\mathbb R$ such that $\int_a ^{b} e^{itx} d\mu (x) \to 0$ as $|t| \to \infty$ whenever $a<b$. Can we conclude that $\mu$ is absolutely continuous with respect to Lebesgue measure?
 A: No. The counterexamples are the Rajchman measures. I don't know a simple construction, but the Wikipedia article has some references.
If you know a construction of a Rajchman measure on the unit circle, you can transfer it to $\mathbb R$ by extending periodically then multiplying by an integrable function with compact Fourier transform.
We then need that if $\mu$ is Rajchman then so is $1_{[a,b]}\mu.$
In general, given a Rajchman measure $\mu,$ any $\nu\ll\mu$ is also Rajchman, because we can take a sequence $f_n$ with compactly supported Fourier transforms such that $f_n\to d\mu/d\nu$ in $L^1(\mu).$
A: Partial answer: Let $\mu$ be the Cantor measure; that is, the "uniform" measure on the Cantor set. It would be enough to show that $\hat\mu(t)\to0$ at infinity.
Why? Fix $a,b$ and let $d\nu=\chi_{(a,b)}d\mu$. Since $(a,b)$ is the union of the triadic intervals contained in $(a,b)$ it follows that $||\nu-\nu_n||\to0$, where each $\nu_n$ is a finite linear combination of translates of $\mu$. Hence $\hat\nu_n(t)\to0$, and $\hat\nu_n\to\hat\nu$ uniformly, so $\hat\nu\to0$.
Edit: Alas $\hat\mu(t)\not\to0$ at infinity, because $\hat\mu(t)=\prod_k\cos(2\pi t/3^k)$, which makes it clear that $\hat\mu(3^n)\not\to0$. I'm going to leave this here anyway, because one can imagine using the same argument with a different "fractal measure" in place of $\mu$.
