E(Xn) -> E(X) implies E(|Xn - X|) -> 0??

Let X be a random variable. Let Xn be a sequence of random variables.
Then E(|Xn - X|) -> 0 as n->∞ implies E(Xn) -> E(X) as n->∞.
[I can prove this by |E(W)|≤E(|W|) and squeeze theorem.]

Is it also true that E(Xn) -> E(X) implies E(|Xn - X|) -> 0??
How can we prove it?

Any help is much appreciated!

Hello,

Let be such that

And let almost surely.

Then we have and
So the condition is satisfied.

But almost surely. Thus and can't have a limit equal to 0 ;)