Does convergence in probability imply convergence in expected value?

Question

Suppose that $Y_N \overset{p}{\to} Y$ as $N \rightarrow \infty$, i.e., $Y_N$ converges to $Y$ in probability. How do I prove that $$ E(Y_N) \rightarrow E(Y) $$ as $N \rightarrow \infty$? This is easy when $Y_N \overset{a.s.}{\to} Y$ as one can apply dominated convergence theorem (since integration over a set of measure zero yields zero). My definition for the expected value is $$ E(X)=\int_{\Omega} X dP, $$ where $\Omega$ is the space and $P$ probability measure.

Answer 1

This is not true. Even if $Y_n\to Y$ almost-surely we do not in general have $\mathbb EY_n\to\mathbb E Y$.

As a typical example of this, consider the random variables $Y_n$ defined by $$\mathbb P[Y_n=n]=1/n, \mathbb P[Y_n=0]=1-1/n.$$ Then clearly $Y_n\to 0$ but $$\mathbb EY_n = 1, \hspace{0.5cm} \text{ for all } n \in\mathbb N.$$

In fact, knowing $\textbf{when}$ we can deduce convergence of expected values is entirely the point of the convergence theorems (dominated convergence, monotone convergence...).