Exercise 4, § 3

Let $\mu$ be Lebesgue measure on $\mathbf{R}$, and $\nu$ Lebesgue measure on $[0, +\infty[$. Let $x_1, x_2$ be in $\mathbf{R}$. Show that the convolution product $((\varepsilon_{x_1} - \varepsilon_{x_2}) * \nu) * \mu$ is defined, but that $\mu$ and $\nu$ are not convolvable. Show that the convolution products $\nu * ((\varepsilon_{x_1} - \varepsilon_{x_2}) * \mu)$ and $(\nu * (\varepsilon_{x_1} - \varepsilon_{x_2})) * \mu$ are defined, but are distinct for $x_1 \neq x_2$.