Do convergence in distribution along with uniform integrability imply convergence in mean? There are at least 2 places in Wikipedia saying that $X_n$ converges to $X$ in mean in and only if $X_n$ converges to $X$ in probability and $X_n$ is uniformly integrable. See the following link for example:
http://en.wikipedia.org/wiki/Convergence_of_random_variables#Properties_4

However, I found a reference in Billingsley's book saying that we only need convergence in distribution along with uniform integrability to imply the convergence in mean. Is Wikipedia wrong, or am I missing something here?

Thanks.
I don't have Billingsley's book here, but I guess what he says is that uniform integrability and convergence in distribution imply convergence of the means (and not in mean), i.e. $E[X_n]$ converges to $E[X]$. 

One simple (though not truely elementary) way to see it is to use Skorokhod's theorem stating that convergence in distribution is equivalent to almost-sure convergence of copies of the random variables in some abstract probability state. Then we just apply the usual Vitali theorem about uniform integrability for these copies. It gives convergence in mean of the copies of $X_n$ to the copy of $X$, and therefore convergence of the mean of (the copy of) $X_n$ to that of (the copy of) $X$. I put parentheses since the mean only depends on the law and hence is the same for $X_n$ or a copy of it. This is the result.
Convergence of mean (or any other feature of the distribution for that matter) does not imply weak convergence of the random variables.

Take the example of an infinite sequence of iid random variables of any non degenerate distribution and see what is going wrong.