Is is true that if $E|X_n - X| \to 0$ then $E[X_n] \to E[X] $? My question is motivated by the following problem:
Show that if $|X_n - X| \le Y_n$ and $E[Y_n] \to 0$ then $E[X_n] \to E[X]$.
I started off by saying that since
$$|X_n - X|\ge 0 $$
then $$E[|X_n - X|]\ge 0 $$ At the same time 
$$E[|X_n - X|]\le E[Y_n] $$
and so 
$$0 \le E[|X_n - X|] \le E[Y_n]$$
By the squeeze theorem then $E[|X_n - X|] \to 0$. I don't know how to proceed from here. 
I know that if $E[|X_n - X|] = 0$ then I  can set up a contradiction, like so:
Suppose that $|X_n - X| = c$, $c \ne 0$ and $E[|X_n - X|] = 0$. Then $$E[|X_n - X|] = E[c] = c \ne 0. $$ 
But I don't know how to show this would hold in the limit. I am planning to use this to show that $X_n \to X$ and therefore $E[X_n] \to E[X]$ 

$\vert E(X) - E(X_n) \vert =\vert E(X-X_n) \vert  \leq E(|X-X_n|) \to 0$
