Convergence of $max(X_n, X)$ where $X_n \rightarrow X$ a.s. Suppose $X_n \ge 0$, $X_n \rightarrow X$ almost surely for $X_n, X \in L^1$ (both integrable) and that $E(X_n) \rightarrow E(X)$.  I want to prove that $max(X_n, X) \rightarrow X$ in $L^1$.
That is, to show that $E(|max(X_n,X) - X|) \rightarrow 0$
I tried to start with the fact that $E(|max(X_n,X) - X|) = E(|X_n - X|1(X_n>X))$ and that the almost sure convergence implies convergence in measure, but I'm not sure if this is the right approach given that it hasn't led me anywhere.  Any help would be greatly appreciated. Thanks!
Edit: In attempting to solve myself:
Since we have $X_n \ge 0$, 
$$E(|X_n - X|1(X_n>X)) = E(X_n * 1(X_n>X)) - E(X*1(X_n>X)$$
$$\leq E(X_n) - E(X*1(X_n>X))$$
Now I'm still lost but it feels closer.
 A: Assume that $X_n, X$ are non-negative integrable RVs such that
$$ X_n \to X \quad \text{a.s.}, \tag{1} $$
$$ \mathbb{E}[X_n] \to \mathbb{E}[X] \tag{2}$$
hold. We also note that $\left|\max\{X, X_n\} - X\right| \leq X_n$. Then by the Fatou's Lemma,
\begin{align*}
\mathbb{E}[X]
&= \mathbb{E}\left[\liminf_{n\to\infty} (X_n - \left|\max\{X, X_n\} - X\right|)\right] \tag{by (1)} \\
&\leq \liminf_{n\to\infty} \mathbb{E}\left[ X_n - \left|\max\{X, X_n\} - X\right| \right] \tag{by Fatou} \\
&= \mathbb{E}[X] - \limsup_{n\to\infty} \mathbb{E}\left[ \left|\max\{X, X_n\} - X\right| \right] \tag{by (2)} \\
\end{align*}
This shows that $\limsup_{n\to\infty} \mathbb{E}\left[ \left|\max\{X, X_n\} - X\right| \right] = 0$, which in turn implies the desired $L^1$-convergence.
A: Note: $X_n \geq 0$ was added later. 
This is false. Let $X_n=nI_{(0,\frac 1 n)} -nI_{(-\frac 1 n, 0)}$ and $X=0$. Then $X_n \to X$ almost surely, $EX_n=0$ for all $n$, so $EX_n \to EX$ but $\max \{X_n,X\}=nI_{(0,\frac 1 n)}$ which does not tend to $0$ in $L^{1}$. 
A: Sorry for the incorrect answer earlier. Let me provide a new solution here. The main result I apply here is:

Let $0<p<\infty$, if $(X_n)$ is a sequence of random variables in $L^p$, and $X_n \overset{\mathcal{P}}{\to}X$. Then the following are equivalent:
   1. $|X_n|^p$ is uniformly integrable;
   2. $X_n\to X$ in $L^p$;
   3. $\mathbb{E}|X_n|^p\to \mathbb{E}|X|^p<\infty$.

Since $X_n\to X$ a.s., then $\max\{X_n, X\}\to X$ a.s.. From $\mathbb{E}(X_n)\to \mathbb{E}(X)$ and $X_n\ge 0$, by the result cited above, we have $X_n\to X$ in $L^1$. Thus $\max\{X_n, X\} = \frac{1}{2}(X_n+X+|X_n-X|)\in L^1$. To show $\max\{X_n, X\}\to X$ in $L^1$. It suffices to show $\mathbb{E}\left(\max\{X_n, X\}\right) \to \mathbb{E}(X)$, this is clear because
\begin{align}
\mathbb{E}(\max\{X_m,X\}) = \frac{1}{2}\left[\mathbb{E}(X_n)+\mathbb{E}(X)+\mathbb{E}|X_n-X|\right] \to \mathbb{E}(X).
\end{align}
The result I cited here is from Theorem 4.5.4 in Kailai Chung's A Course in Probability Theory.
