Convolution of measures We say that a family of measures $\mu_{t}\to \mu$ weakly if for any $g\in C_{0}$, $\int g d\mu_{t} \to \int g d\mu$. Show that if $\mu_{t}\to \mu$ weakly, then $\nu*\mu_{t}\to \nu*\mu$ weakly, where $\nu*\mu$ denotes the convolution of the measures $\mu$ and $\nu$.
If the measures are finite, then for $g\in C_0$
$$\int g(x)\,(\nu*\mu_t)(dx)=\int\int g(x+y)\,\mu_t(dy)\,\nu(dx).$$
For each $x$, the weak convergence $\mu_t\to \mu$ gives
$$\int g(x+y)\,\mu_t(dy)\to \int g(x+y)\,\mu(dy).$$
Then by the dominated convergence theorem,
$$\int\int g(x+y)\,\mu_t(dy)\,\nu(dx)\to\int\int g(x+y)\,\mu(dy)\,\nu(dx).$$
By definition, this is the weak convergence $\nu*\mu_t\to \nu*\mu$.
For a counterexample, consider the shifted $\delta$-masses on $\mathbb{R}$:
$$ \int f(x) \ \delta_t(x) = f(t).$$
It's clear that $\delta_{t} \to 0$ vaguely as $t \to \infty$, while
$\delta_t * dx = dx$ as measures by the translation invariance of Lebesgue measure.