It is known that if $X, \ Y $ are independent random variables then the distribution of $X+Y $, which we may get by convoluting the distributions of $X $ and $Y $, has characteristic function $\phi_{X+Y } = \phi_X \phi_Y$.
If we define the characteristic function of a finite measure $\mu $ on $\mathbb {R}^d $ as $\phi_{\mu } (t) = \int \exp(\mathrm{i}\langle t,x\rangle\mu(\mathrm dx) $, where $\langle t,x\rangle$ denotes the inner product, do we have similarly that $\phi_{\mu } (t)\phi_{\lambda }(t) = \int \exp(\mathrm{i}\langle t,x\rangle)(\mu\star\lambda)(\mathrm dx) $, where $\mu\star\lambda$ denotes the convolution of $\mu $ and $\lambda $?
Thanks in advance!
Yes and you get it from the case of probability measures by considering $\mu_1=\frac {\mu} {\mu(\Omega)}, \nu_1=\frac {\nu} {\nu(\Omega)}$ or directly by repeating the argument for the case or probbaility measures.