I've got a question concerning mollifiers. If $\Omega \subset \mathbb{R}^N$ is open and $\mu = (\mu_1,..., \mu_m)$ is a Radon measure in $\Omega$. Let $(\rho_{\epsilon})_{\epsilon > 0}$ be a family of mollifiers. Why does $\mu_{\epsilon} := \mu * \rho_{\epsilon} \mathcal{L}^N$ locally weakly* converge in $\Omega$ to $\mu$ as $\epsilon$ goes to zero? I tried it using Fubini, but couldn't really see how it works out?
Whereas a sequence of Radon measures $(\mu_h)_{h \in \mathbb{N}}$ is called weak* convergent if there exists a Radon measure $\mu$ such that for all $u \in C_0(\Omega)$ \begin{equation} \lim\limits_{n \rightarrow \infty}{\int_{\Omega}} u d\mu_h = \int_{\Omega} u d\mu \end{equation} holds. The convolution of a measure $\mu$ and a continiuos function $f$ is defined as \begin{equation} \mu * f (x) := \int_{\Omega} f(x-y) d\mu(y) \end{equation}
Thanks in advance!
For $m=1$, by Fubini, $$\int_{\mathbb{R}^N} u(x)(\mu\ast\rho_\epsilon)(x) \,dx = \int_\Omega\int_{\mathbb{R}^N} u(x)\rho_\epsilon(x-y) \,dx\,d\mu(y) = \langle\mu,u\ast\tilde{\rho}_\epsilon\rangle_{C_0(\Omega)^\prime,C_0(\Omega)}.$$
Here, $\tilde{\rho}_\epsilon(x):=\rho_\epsilon(-x)$ is just a reflection of $\rho_\epsilon$, which has the same properties. The convolution (of functions) satisfies $u\ast\tilde{\rho}_\epsilon\to u$ in $C(\Omega)$ as $\epsilon\to0$ (cf. Hörmander – The Analysis of Linear Partial Differential Operators I, Theorem 1.3.2). Because $\mu$ defines a continuous functional on $C_0(\Omega)$, $$\langle\mu,u\ast\tilde{\rho}_\epsilon\rangle_{C_0(\Omega)^\prime,C_0(\Omega)} \to \langle\mu,u\rangle_{C_0(\Omega)^\prime,C_0(\Omega)} = \int_\Omega u(x) \,d\mu(x).$$