> [!definition] > > Let $G$ be a [[Topological Group|topological group]], $\mu$ and $\nu$ be [[Borel Measure|Borel measures]] on $G$, then their **convolution** is > $ > \begin{align*} > (\mu * \nu)(E) &= \int_G \int_G \chi_{E}(xy) \ d\mu(x) d\nu(y) \\ > &= \int_G\mu(E \cdot x^{-1})d\nu(x) > \end{align*} > $ > In other words, it is the [[Push-Forward|push-forward]] of the [[Product Measure|product measure]] $\mu \times \nu$ with respect to the composition map $G^2 \to G$. > > If the group is commutative, then $\mu * \nu = \nu * \mu$, and $(\mu * \nu) * \rho = \mu * (\nu * \rho)$. > [!theorem] > > Let $\nu$ be a probability measure on $\real^d$. If there exists a probability measure $\mu$ such that $\mu * \nu = \mu$, then $\nu = \delta_0$ is the point mass at $0$. > > *Proof*. Using [[Characteristic Function|characteristic functions]], > $ > \wh \mu(\xi) \cdot \wh \nu(\xi) = \wh \mu(\xi) > $ > By continuity of $\wh\mu$, there exists $r > 0$ such that $\wh \nu|_{B(0, r)} = 1$. Since $\wh \nu$ has no imaginary part around $0$, > $ > \int_{\real^d}\cos(x, \xi)d\nu(x) = 1 > $ > so $\angles{x, \xi} = 0$ modulo $2\pi$ $\nu$-almost surely. Let $e$ be a unit vector, and $\xi_1, \xi_2 \in B(0, r)$ be parallel to $e$ such that $\xi_1 = c\xi_2$ where $c \not\in \rational$. Therefore $\angles{x, \xi_1} = c\angles{x, \xi_2} = 0$ modulo $2\pi$, which only happens if both are equal to $0$, and $\angles{x, \xi} = 0$ $\nu$-almost surely, and $\nu = \delta_{0}$.