Portmanteau lemma: Prove that 4 implies 2 If lim inf $E[f(X_n)] \geq E[f(X)]$ for any non-negative, continuous function $f(.)$ then $E[g(X_n)]=E[g(X)]$ for all bounded, continuous functions $g(.)$., where $X_n$ define a sequence of bounded probability measures. The question talks about the version in Asymptotic Statistics in A.W. van der Vaart. I know how to go from 1 to 4 that is prove lim inf $E[f(X_n)] \geq E[f(X)]$ for any non-negative, continuous function $f(.)$ if we have weak convergence using Fatou's Lemma. I am unsure how to show any of the other statements if 4 holds.

Let $f$ be any bounded continuous function and $M$ be its supremum. Then $M-f$ is a non-negative continous function so $\lim \inf E(M-f(X_n)) \geq E(M-f(X))$. Cancelling $M$ we get $\lim \sup Ef(X_n) \leq Ef(X)$. Changing $f$ to $-f$ we get $\lim \inf Ef(X_n) \geq Ef(X)$. Hence $\lim Ef(X_n)=Ef(X)$.