Suppose $\{\mu_n\}$ is a sequence of Borel probability measures on $\mathbb{C}$ such that
$$
\lim_{n \to \infty} \int fd\mu_n
$$
exists for all $f \in C_b(\mathbb{C}, \mathbb{R})$. Is it true that there exists a measure $\mu$ such that
$$
\int f d\mu = \lim_{n \to \infty} \int fd\mu_n?
$$
In other words, is there a measure $\mu$ such that $\mu_n \to \mu$ weakly?
In my problem, we can assume the support of the $\mu_n$ is contained in the unit ball, if it helps.
Best Answer
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.