Existence of weak limit of measures

Define $l: C_c(\mathbb{C},\mathbb{R}) \to \mathbb{R}$ by $l(f) = \lim_{n \to \infty} \int fd\mu_n$. Then $l$ is a positive, continuous linear functional, so by Riesz representation theorem, there is some measure $\mu$ with $l(f) = \int fd\mu$.

Yes! Consider the restriction to the space of continuous functions with compact support, denoted by $C_{c}(\mathbb{C}, \mathbb{R})$. Now define $$L(f):= \lim_{n\rightarrow \infty} \int f \, d\mu_{n}\qquad , \, f\in C_{c}(\mathbb{C}, \mathbb{R}).$$

It is straightforward to check that $L$ defines a bounded linear functional on $C_{c}(\mathbb{C},\mathbb{R})$. Indeed, since the $\left\{\mu_{n}\right\}$'s are Borel probability measures on $\mathbb{C}$, we have

$$\lvert L(f) \rvert \leq \lvert \lvert f\rvert \rvert_{\infty}$$ Hence by the Riesz–Markov–Kakutani representation theorem, there exists a unique Borel measure $\mu$,with total variation-norm less than 1, such that $$L(f) = \int f \, d\mu \qquad \forall f\in C_{c}(\mathbb{C},\mathbb{R}).$$ In fact, the integral is well-defined for all $C_{b}(\mathbb{C}, \mathbb{R})$ as well.