Assume that $\mu^n$ are probability measures on $R$ that convergence weakly(-*) to $\mu$, i.e for all $f \in C_b (R)$ (bounded and continuous), we have that $\int f(x) \mu^n(dx) \rightarrow \int f(x) \mu(dx)$.
Assume that $g \in C(R)$ is an unbounded function (for example $g(x)=x$) but assume that the integral of $\mu^n$ wrt to $|g|$ is uniformly bounded, i.e. $\sup_n \int |g(x)| \mu^n(dx) \leq C$
Let us call $g_k(x)$ the function such that $g_k(x)=g(x)$ for $x \in [-k,k]$ and $g_k(x) = g(k)$ for $x>k$ and similar for $x<-k$.
Hence $g_k$ is bounded and continuous, hence $\int g_k(x) \mu^n(dx) \rightarrow \int g_k(x) \mu(dx)$ for each $k$.
Further we have by dominant converging theorem that for each $n$ we have $\int g_k(x) \mu^n(dx) \rightarrow \int g(x) \mu^n(dx)$
Question
Does this also imply that $\int g(x) \mu^n(dx) \rightarrow \int g(x) \mu(dx)$ ?
Alternative question
If thats wrong for general $g$ is it true for the example $g(x)=x$?
take $g(x)=x, \ \mu_n=(1-\frac{1+(-1)^n}{n})\delta_0+\frac{1+(-1)^n}{n}\delta_n$.