Show: $E(|X|)\leqslant\liminf_{n\to\infty} E(|X_n|)$ 

Let $X,X_n$ be random variables with $X_n\to X$ is distribution.
    Show 
    $$
E(|X|)\leqslant\liminf_{n\to\infty} E(|X_n|).
$$


I call the probability space on which $X$ and $X_n$ are defined $(X,A,P)$. Then $X_n\to X$ in distribution means
$$
\lim_n \int f\, dP_{X_n}=\int f\, dP_X
$$
for all bounded and continious functions $f\colon\mathbb{R}\to\mathbb{R}$, right?
Do not know how continue...
 A: If the absolute value function were bounded, this would be easy.  It's not, but we can approximate it by bounded continuous functions.
Let $\phi_k(x) = |x| \wedge k$.  Notice that $\phi_k$ is bounded and continuous, that $\phi_k(x) \le |x|$, and as $k \to \infty$, $\phi_k(x) \uparrow |x|$ pointwise.  Fix $k$ for the moment.  Now for each $n$ we have $$E \phi_k(X_n) \le E |X_n|.$$
Passing to the limit as $n \to \infty$ we have
$$\liminf_{n \to \infty} E \phi_k(X_n) \le \liminf_{n \to \infty} E |X_n|.$$
But since $\phi_k$ is bounded and continuous, the convergence in distribution gives us $E\phi_k(X_n) \to E \phi_k(X)$, so $\liminf_{n \to \infty} E \phi_k(X_n) = E \phi_k(X)$.  Thus we have
$$E \phi_k(X) \le \liminf_{n \to \infty} E |X_n|.$$
Finally let $k \to \infty$ and use the monotone convergence theorem (or Fatou's lemma) on the left side:
$$E |X| \le \liminf_{n \to \infty} E|X_n|.$$
A: Assume that the assertion is wrong. Than there exists a subsequence $X_{n_j}$ such that $lim_{j\rightarrow \infty}E[|X_{n_j}|]<E[|X|]|]$. 
Since $X_n \rightarrow X$ in distribution , there exists a subsubsequence such that $X_{n_{j_k}}\rightarrow X$ almost surely. Now Fatou implies that
$lim_{j\rightarrow \infty}E[|X_{n_j}|]=liminf_{k \rightarrow \infty }E[|X_{n_{j_k}}|]\ge E[liminf_{k \rightarrow \infty }|X_{n_{j_k}}|]=E[|X|]$,
a contradiction.
Edit: This is rubbish.
