limit of functionals on weak convergent random variables Suppose real value random variables satisfy
$X_{n} \Rightarrow X$ (convergence in distribution)
as $n\to \infty$ in the same probability space 
$(\Omega, \mathcal F, \mathbb P)$.
It is well known that $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$ for all continuous
bounded real functions $f:\mathbb R \to \mathbb R$.
[Q.] If $f$ is continuous and linear growth, i.e. $|f(x)| < K(1 + |x|)$ for 
some constant $K$, can you find counter-example for $\lim_{n\to \infty} 
\mathbb E f(X_{n}) = \mathbb E f(X)$?
What additional conditions are needed to still
have $\lim_{n\to \infty} \mathbb E  f(X_{n}) = \mathbb E  f(X)$?
 A: Since $f(X_{n}) \Rightarrow f(X)$ as long as $f$ is continuous, 
by redefining $X_{n}:= f(X_{n})$, it is 
equivalent to ask following question:
[Q1.] If $X_{n}\Rightarrow X$, then what additional condition is needed to
have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$?
[Ex.] Let $X_{n}: [0,1] \mapsto \mathbb R$ be given by
$X_{n} (\omega) = n I_{(0,1/n)}(\omega)$ and $\mathbb P$ be 
Lesbegue meaure. Then, $X_{n} \Rightarrow X:=0$,
since it is indeed almost sure convergence. However, it violates
$\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$.
So one immediate sufficient condition needed to
have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$ is that,
$X_{n} \to X$ almost surely, and satisfies other conditions of
Dominated (or Monotone) Convergence Theorem.
A: $\lim_{n \rightarrow \infty}\mathbb{E}[f(X_n)] = \mathbb{E}[f(X)]$ for all continuous $f$ of linear growth if and only if $X_n \Rightarrow X$ and $\lim_{n \rightarrow \infty}\mathbb{E}[|X_n|] = \mathbb{E}[|X|]$. This is exactly convergence in (first order) Wasserstein distance.
