Weak convergence and convergence of moments Consider a random variable $X$ defined on the probability space $(\Omega, \mathcal{F}, P)$ such that $X:\Omega\rightarrow \mathbb{R}$. 
Suppose that $X\sim N(\mu, \sigma^2)$. 
Consider a random function $T(X):\mathbb{R}\rightarrow \mathbb{R}$. 
Let $\{Y_n\}$ be a sequence of random variables all defined on the probability space $(\Omega_n, \mathcal{F}_n, P_n)$ such that $Y_ n:\Omega_n\rightarrow \mathbb{R}$ $\forall n$. 
Assume $Y_n=O_{P_n}(1)$ and $Y_n\rightarrow_d T(X)$ as $n\rightarrow \infty$ where the meaning of $O_{P_n}(\cdot)$ is described here. 
Can I conclude that $\lim_{n\rightarrow \infty}E_{P_n}(Y_n)=E_P(T(X))$? In negative case, which additional assumptions would be sufficient?
This question is from proof of Theorem 15.1 in van der Vaart "Asymptotic Statistics" p.216 when the author writes "Because $\phi_n$ are unformly bounded $E_h\phi_n\rightarrow E_hT$".
 A: *

*It does not really matter that $Y_n$ are defined on different probability spaces: all your assumptions and statements are related to their distribution on $\mathbb{R}$. So you can freely assume that everything is defined on the same probability space.

*The normal distribution of $X$ is irrelevant too, since $T(X)$ can have arbitrary distribution.

*The assumption $Y_n = O_{P}(1)$ follows from the weak convergence thanks to the Prokhorov theorem. 

*You want to deduce the convergence of expectations from weak convergence. This is not always possible. There are some extra sufficient conditions: 


*

*uniform integrability: $\sup_n E[Y_n \mathbf{1}_{|Y_n|>C}] \to 0$, $C\to\infty$. This is the strongest one but usually not very convenient;

*de la Vallée-Poussin condition: for some function $V:[0,+\infty)\to[0,+\infty)$ such that $V(x)/x\to +\infty$, $x\to+\infty$, it holds $\sup_n E[V(|Y_n|)]<\infty$, e.g. $\sup_n E[|Y_n|^{1+\varepsilon}]<\infty$ for some $\varepsilon>0$. 

*The simplest condition is boundedness: $\sup_{n} \sup_{\omega}|Y_n(\omega)|<\infty$. This is the condition van der Vaart refers to. 


