comparison of the limits of the expectations of two sequences of r.v. Let $X_{n}$ and $Y_{n}$ be two sequences of non-negative random variables. Assume that 
$$
X_{n}\stackrel{d}{\to}X
$$ 
and
$$
E[X_{n}]\to E[X].
$$
Next, assume that 
$$
P(\underset{n\to\infty}{\liminf} \{Y_{n} \leq X_{n} \}) = 1.
$$
Is the following correct then
$$
\underset{n\to\infty}{\limsup}E[Y_{n}] \leq E[X]
$$
? Or we need more assumptions on the original sequences?
 A: The conclusion does not hold under the mentioned assumption because 


*

*first there is no guarantee that $Y_n$ is integrable. For example, if $Z$ is a non-integrable non-negative random variable, let $Y_n:=X_n+Z\cdot \mathbf 1\left\{Z\gt n\right\}$. For each $\omega$, if $n\geqslant Z\left(\omega\right)$, we have $X_n\left(\omega\right)=Y_n\left(\omega\right)$ but $\mathbb E\left[Y_n\right]$ is infinite for all $n$. 

*Second, assuming $Y_n$ integrable for all $n$ is not enough. Indeed, let $Y_n:=X_n+n^2\cdot \mathbf 1\left\{Z\gt n\right\}$ where $Z$ is such that $n\mathbb P\left\{Z\gt n\right\}\gt 1$ for all $n$. Then for each $\omega$, if $n\geqslant Z\left(\omega\right)$, we have $X_n\left(\omega\right)=Y_n\left(\omega\right)$ but $\mathbb E\left[Y_n\right]=\mathbb E\left[X_n\right]+n^2\mathbb P\left\{Z\gt n\right\}$ which goes to infinity as $n$ goes to infinity.
However, if we assume that $\mathbb E\left[Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right]\to 0$, then the wanted conclusion holds as a consequence of the decomposition 
$$
\mathbb E\left[Y_n\right]=\mathbb E\left[Y_n\mathbf 1\left\{Y_n\leqslant X_n\right\}\right]+\mathbb E\left[Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right]
$$
because the first term of the right hand side do not exceed $\mathbb E\left[X_n\right]$. 
For example, if the sequence $\left(Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right)_{n\geqslant 1}$ is uniformly integrable, then $\mathbb E\left[Y_n\mathbf 1\left\{Y_n\gt X_n\right\}\right]\to 0$ (there is a pointwise convergence to $0$).
