Let $\mathbb{T}$ be the 1-torus, let $\mathcal{B}$ be the Borel subsets of $\mathbb{T}$ and let $\frak{M}(\mathbb{T})$ be the set of the Borel complex measures of $\mathbb{T}$. If $\mu,\tau\in\frak{M}(\mathbb{T})$, define: $$\mu*\tau:\mathcal{B}\to\mathbb{C}, A\mapsto\frac{1}{2\pi}\int_{\mathbb{T}\times\mathbb{T}}\chi_{A}(x+y)\operatorname{d}(\mu\otimes\tau)(x,y),$$ so that $\mu*\tau\in\frak{M}(\mathbb{T}).$
Define the Fourier transform as: $$\mathcal{F}: {\frak{M}} (\mathbb{T}) \to l^\infty(\mathbb{Z}), \mu \mapsto \left(n\mapsto\frac{1}{2\pi}\int_{\mathbb{T}}e^{-int}\operatorname{d}\mu(t)\right).$$ How can we prove that: $$\forall \mu,\tau\in {\frak{M}}(\mathbb{T}), \mathcal{F}(\mu*\tau)=\mathcal{F}(\mu)\mathcal{F}(\tau)?$$ I know how to prove the corresponding result for $L^1(\mathbb{T})$ functions, i.e. via Fubini theorem, but actually I've some trouble to prove the corresponding result for measure...
Any help?
Begin by noting that $$ \int_\mathbb{T} fd(\mu*\tau) = \int_{\mathbb{T}\times\mathbb{T}}f(x+y)d(\mu\otimes\tau)(x,y) $$ for all bounded measurable $f$. Hence $$\begin{eqnarray} \mathcal{F}(\mu*\tau)_n &= &\int_{\mathbb{T}\times\mathbb{T}}e^{-in(x+y)}d(\mu\otimes\tau)(x,y)\\ &=&\int_{\mathbb{T}\times\mathbb{T}}e^{-in(x+y)}d\mu(x)d\tau(y)\\ &=&\int_{\mathbb{T}}\left[\int_\mathbb{T}e^{-inx}d\mu(x)\right]e^{-iny}d\tau(y)\\ &=&\mathcal{F}(\mu)_n\mathcal{F}(\tau)_n. \end{eqnarray}$$