In order to show the uniqueness of the Fourier coefficients of a signed measure, I need to show that :
For any two finite signed measures $\nu_1, \nu_2$ on $\left([-\pi, \pi], \mathcal{B}_{[-\pi, \pi]}\right)$ such that : $\int f\mathrm d\nu_1=\int f\mathrm d\nu_2$ holds for every $f$ continuous (and bounded), we have: $\nu_1=\nu_2$.
This property holds for finite positive measures (Approximation of bounded measurable functions with continuous functions) because the equality of integrals for a continuous function implies that of integrals of a measurable bounded function.
The answers in this thread are based on the fact that for $\mu, \mu'$ probability measures, $\mathcal{C}=\{B\in\mathcal{B}_{[-\pi, \pi]} : \mu(B)=\mu'(B)\}$ is a $\pi$-system.
Does this even hold here ?
EDIT: It does when $\nu_1([-\pi, \pi])=\nu_2([-\pi, \pi])$, (required to show that $\{B∈\mathcal{B}_{[−π,π]}:ν_1(B)=ν_2(B)\}$ is a Dynkin system), which is true in my case, since $\hat{ν_1}=\hat{ν_2}$.
I suppose that both signed measures are finite, since otherwise we cannot expect that the integral over arbitrary continuous functions exists. You can approximate any interval $[a,b]$ by (for example) the sequence $$g_n(x) := \begin{cases} 1 & \text{ if } x \in [a,b] \\ nx + (1-na) &\text{ if } x \in [a-1/n,a] \\ -nx +(1+nb) & \text{ if } x \in [b,b+1/n] \\ 0 & \text{otherwise} \end{cases}.$$ By the dominated convergence theorem (for signed measures) we get $$\nu_1([a,b]) =\nu_2([a,b]).$$ Thus, both measures are equal on a $\cap$-stable generator of the Borel-$\sigma$-algebra. Now, you need to extend the uniqueness theorem for $\sigma$-finite measures. In fact, the same proof applies. Note that $$\mathcal{C}=\{B\in\mathcal{B}_{[-\pi, \pi]} : \nu_1(B)=\nu_2(B)\}$$ is a $\pi$-system containing a $\cap$-stable generator with a sequence $E_n \in \mathcal{C}$ such that $E_n \uparrow \Omega$.
For completeness, we should give reasons for the application of the dominated convergence theorem: By Jordan's decomposition theorem, you can write for a signed measure $\mu = \mu_1 - \mu_2$ with measures $\mu_1$ and $\mu_2$. (Moreover, there exists a 'minimal decomposition' - called the Jordan decompisition.)