I'm trying to show the following.
Proposition. Let $T$ be the unit circle in the complex plane. Let $\mu$ be a complex Borel measure on $T$. Define the sequence $\hat\mu$ by $$ \hat\mu(n) := \int e^{-int} \ d\mu(t) $$ for each $n\in\mathbb Z$. If $\hat\mu(n) \to 0$ as $n\to +\infty$, then $$ \lim_{n\to\infty}\int e^{-int} f(t) \ d\mu(t) = 0 $$ for every bounded Borel-measurable function $f$ on $T$.
(Reference: RCA by W. Rudin, Exercise 6.7)
What I have done. I have proved that the conclusion of the proposition holds if $f$ is continuous. And I think it is now enough to check the case of characteristic functions on Borel sets. Lusin's theorem provides a connection between continuous functions and measurable functions, but it requires the measure to be regular, which is not guaranteed right now.
So I rely on Riesz representation theorem. The linear functional on $C(T)$, defined by $$ f \mapsto \int f \ d\mu $$ is bounded since $$ \left| \int f \ d\mu \right| \le \|f\|_\infty \cdot \|\mu\| $$ and $\mu$ is of BV, being a complex measure. So I get a unique complex regular Borel measure $\phi$ such that $$ \int f \ d\phi = \int f\ d\mu $$ for every continuous $f$.
Now we may use Lusin's theorem. Let $E$ be a Borel set in $T$ and $\epsilon>0$. Then we get a continuous function $f$ such that $|\phi|(S) < \epsilon$ for $S := \{ t: f(t) - \chi_E(t) \ne 0 \}$ and $\|f\|_\infty \le 1$. Then $$ \left| \int e^{-int} f(t) \ d\phi (t) - \int e^{-int} \chi_E(t) \ d\phi (t) \right| \le \int_S |f-\chi_E|\ d|\phi| \le \int_S (\|f\|_\infty + \|\chi_E\|_\infty)\ d|\phi| \le 2\epsilon. $$ Using this inequality, it is easy to show that $$ \lim_{n\to\infty} \int e^{-int} \chi_E(t) \ d\phi(t) = 0. $$ But how to restore the measure to $\mu$? This is my question.