Does convergence of means imply convergence in mean? For a sequence of nonnegative integrable random variables, 
I know that convergence in mean (aka convergence in $L^{1}$ where $E[|X_{n} - X|]$ approaches zero as n gets large), implies convergence of means (where $E[X_{n}]$ approaches $E[X]$ as n gets large).
My question is: Is the converse necessarily true? and why?
 A: If you impose the additional assumption that $X_n\to X$ almost surely, then the converse is true, because the $\{X_n\}$ will now be uniformly integrable, and it's well known that uniform integrability + convergence in probability implies convergence in $L^1$.
To demonstrate uniform integrability of the $\{X_n\}$, define for each $k$ the function $\phi_k(x):= xI(x>k)$. Argue that if $k$ is a continuity point of $X$, then $\phi_k(X_n)\to\phi_k(X)$ almost surely. Apply Fatou's lemma to get  $$ E(\phi_k(X_n))\to E(\phi_k(X)).\tag1$$ Now for any sequence ${k_1, k_2, \ldots}$ tending to infinity, use the monotone convergence theorem to conclude that $$E(\phi_{k_n}(X))\to0\tag2.$$ 
Finally, let $\epsilon>0$. To prove uniform integrability of $\{X_n\}$ we have to find $K$ so that
$$E(|X_n|I(|X_n|>K)) < \epsilon\tag3
$$
for all sufficiently large $n$. Wlog, we can require $K$ to be a continuity point of $X$. But the LHS of (3) is $E(\phi_K(X_n))$, which tends by (1) to $E(\phi_K(X))$. By (2), we can make $E(\phi_K(X))<\epsilon/2$ by picking $K$ large enough. Given this $K$, use (1) to show that $E(\phi_K(X_n))$ can be made arbitrarily close to $E(\phi_K(X))$ for sufficiently large $n$, and we're done.
A: The converse is not true. Take $X=1$ and $X_n = 0$ with probability 1/2 and $X_n=2$ with probability 1/2. $E[X] = E[X_n] = 1$ but $E[|X-X_n|] = 1\neq 0$
