Property of complex borel measures with absolutely convergent Fourier series (Wiener algebra) Let $\mu$ be a complex Borel measure on the circle $\mathbb{R}/\mathbb{Z}$ with
$$
\sum_{n \in \mathbb{Z}} \lvert\hat{\mu}(n)\rvert < \infty.
$$
How does it follow that $d\mu(x) = f(x) dx$ for $f$ continuous on the circle?
I suppose it is due to the fact that the hypothesis is equivalent to $\mu \ast D_n$ being continuous and converging uniformly, where $D_n$ the the Dirichlet kernel. But how do I know that $\mu \ast D_n$ relates to $\mu$ in any way with no convergence results?
This is the start of an exercise (1.2) from the beginning of Classical and Multilinear Harmonic Analysis Vol 1. by Muscalu and Schlag.
More generally, I am somewhat struggling with the use of complex Borel measures in statements where I am accustomed to seeing functions that are $L^p$ with respect to some positive measure. I understand that they are related via Radon-Nikodyn, but I struggle with how to think about certain statements: in particular, the convolution of a continuous function and a complex Borel measure, among others. Any advice/reading recommendations on getting more familiar with the heavy use of these measures and their interaction with functions?
Thanks.

Define $$f(t)=\sum_{n=-\infty}^\infty \hat\mu(n)e^{int}.$$The series converges uniformly so $f$ is continuous. The uniform convergence also shows that $$\hat f(n)=\hat\mu(n).$$So uniqueness (for complex measures) shows that $\mu=f$, or more carefully $d\mu=f\,dt$.
If you don't buy the uniqueness for complex measures bit:
Lemma: If $\nu$ is a complex measure and $\hat\nu(n)=0$ for all $n$ then $\nu=0$.
Proof: $\int p\,d\nu=0$ for every trigonometric polynomial $p$. Hence by uniform approximation $\int \phi\,d\nu=0$ for every continuous function $\phi$. Hence $\nu=0$ (gotta assume we know something).
