characteristic function of a convolution of measures Take the probability measures $\mu,\nu$ on $\mathbb{R}$ and denote $\varphi_{\mu}$ (the same for $\nu$) its characteristic function. Why holds
$$\varphi_{\mu *\nu}(t)=\varphi_{\mu}(t)\cdot\varphi_{\nu}(t)$$
where $\mu*\nu$ denotes the convolution of $\mu$ and $\nu$?

Neither do we need to assume that $\mu$ and $\nu$ admit densities with respect to a common reference measure, nor do we need to restrict this result to $\mathbb R$.
Remember that if $(E,\mathcal E)$ is a measurable group and $$\tau:E^2\to E\;,\;\;\;(x,y)\mapsto xy$$ denotes the group operation, then $\mu\ast\nu$ is the pushforward $\tau(\mu\otimes\nu)$ of the product measure $\mu\otimes\nu$ under $\tau$.
Assuming that $E$ is a $\mathbb R$-Banach space (consider as a group with the group operation being the addition) and $\mathcal E=\mathcal B(E)$, we immediately obtain \begin{equation}\begin{split}\varphi_{\mu\ast\nu}(x')&=\int\tau(\mu\otimes\nu)({\rm d}x)e^{{\rm }i\langle x',\:x\rangle}\\&=\int\mu({\rm d}x)\int\nu({\rm d}y)e^{{\rm }i\langle x',\:\tau(x,\:y)\rangle}\\&=\int\mu({\rm d}x)e^{{\rm }i\langle x',\:x\rangle}\int\nu({\rm d}y)e^{{\rm }i\langle x',\:y\rangle}=\varphi_\mu(x')\varphi_\nu(x')\end{split}\tag1\end{equation} for all $x'\in E'$.
