Convergence in $L^1$ and uniform integrability 
Suppose that $X_n\leq Y_n\leq Z_n$ where $X_n\to X$, $Y_n\to Y$, $Z_n\to Z$ in probability. If $E(X_n)\to E(X)$ and $E(Z_n)\to E(Z)$, show that $E(Y_n)\to E(Y)$.

I want to solve the task above. 
I think the following theorem (which we have proven) is useful here:
Let $X_n\in L^1$ be a sequence of random variables and $X\in L^1$ then the following holds:
$\mathbb E(|X_n-X|)\rightarrow0 \iff  X_n\rightarrow X$ in probability and $(X_n)$ is uniformly integrable.
By assumption we have $Y_n\rightarrow Y$ in probability, so we just need to shot that $Y_n$ is uniformly integrable. 
It is also clear, that $Y_n$ is uniformly integrable, if there is a $g\in L^1$, such that $|Y_n|<g$ for all $n$. 
All I need to show now, ist that one of the integrable functions $|X|$ or $|Z|$ has the desired property. (But I don't know how to show this..)
Please read my ideas and help me according to my attempt. If my attempt is totally wrong, I will be very glad if someone can give me a solution, or better : some hints.
Thanks in advance!
 A: We need to show that $\left\{Y_n\right\}$ is uniformly integrable, given that $\left\{X_n\right\}$ and $\left\{Z_n\right\}$ are each uniformly integrable. To do this, there are two defining properties that need to be checked. In what follows, I assume that the random variables are all defined on the same probability space.
Boundedness: Firstly, we show that $\,\,\sup\limits_{n}\left\{\mathbb{E}\left[\left|Y_n\right|\right]\right\}{}<{}\infty$. 
Observe that, since $\,\,X_n(\omega)\leq Y_n(\omega)\leq Z_n(\omega)\,\,$ pointwise, then 
$$
\left|Y_n\right|\leq\max\left\{\left|X_n\right|,\left|Z_n\right|\right\}{}={}\left|X_n\right|{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}{}+{}\left|Z_n\right|{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}.
$$
Consequently, for each $n$,
$$
\begin{eqnarray*}
\mathbb{E}\left[\left|Y_n\right|\right]&{}={}&\mathbb{E}\left[\left|X_n\right|{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}{}+{}\left|Z_n\right|{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}\right]\newline
&&\newline
&{}={}&\mathbb{E}\left[\left|X_n\right|{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}\right]{}+{}\mathbb{E}\left[\left|Z_n\right|{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}\right]\newline
&&\newline
&{}\leq{}&\mathbb{E}\left[\left|X_n\right|\right]{}+{}\mathbb{E}\left[\left|Z_n\right|\right]\newline
&&\newline
&{}\leq{}&\sup\limits_{n}\left\{\mathbb{E}\left[\left|X_n\right|\right]\right\}{}+{}\sup\limits_{n}\left\{\mathbb{E}\left[\left|Z_n\right|\right]\right\}{}<{}\infty\,.
\end{eqnarray*}
$$
But, this is true for all $n$. Therefore, $\,\,\sup\limits_{n}\left\{\mathbb{E}\left[\left|Y_n\right|\right]\right\}{}<{}\infty$.
Arbitrarily Small Integrals: Choose $\epsilon>0$. There exist $\delta_x>0$ and $\delta_z>0$ such that, for all $n$ and events $A$ with $P\left(A\right)<\delta_y:=\min\left\{\delta_x,\delta_z\right\}$, we have
$$
\mathbb{E}\left[{\bf{1}}_{A}\left|X_n\right|\right]{}<{}\dfrac{\epsilon}{2}\,\,\,\,\mbox{ and }\,\,\,\,\mathbb{E}\left[{\bf{1}}_{A}\left|Z_n\right|\right]{}<{}\dfrac{\epsilon}{2}\,.
$$
So, for any $n$, under these circumstances we have,
$$
\begin{eqnarray*}
\mathbb{E}\left[{\bf{1}}_{A}\left|Y_n\right|\right]&{}\leq{}&\mathbb{E}\left[\left|X_n\right|{\bf{1}}_{A}{\bf{1}}_{\left\{\left|X_n\right|\geq\left|Z_n\right|\right\}}\right]{}+{}\mathbb{E}\left[\left|Z_n\right|{\bf{1}}_{A}{\bf{1}}_{\left\{\left|X_n\right|<\left|Z_n\right|\right\}}\right]\newline
&&\newline
&{}\leq{}&\mathbb{E}\left[{\bf{1}}_{A}\left|X_n\right|\right]{}+{}\mathbb{E}\left[{\bf{1}}_{A}\left|Z_n\right|\right]\newline
&&\newline
&{}<{}&\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon\,.
\end{eqnarray*}
$$
Edit:@Kolmo points out, and I agree, that uniform integrability is not needed here. In fact, since $X_n, Y_n$ and $Z_n$ each converge in probability, they also converge in distribution. Consequently, the only thing we need to show is that
$$
\mathbb{E}\left[Y\right]<\infty.
$$
But, this must be the case, since $X$ and $Z$ satisfy 
$$
\lim\limits_{n\to\infty}\mathbb{E}\left[X_n\right]{}={}\mathbb{E}\left[X\right]{}\leq{}\mathbb{E}\left[\left|X\right|\right]<\infty
$$ 
and 
$$
\lim\limits_{n\to\infty}\mathbb{E}\left[Z_n\right]{}={}\mathbb{E}\left[Z\right]{}\leq{}\mathbb{E}\left[\left|Z\right|\right]<\infty,
$$ 
so, using the bound on $\left|Y_n\right|$ given above,
$$
\mathbb{E}\left[Y\right]{}={}\lim\limits_{n\to\infty}\mathbb{E}\left[Y_n\right]{}\leq{}\lim\limits_{n\to\infty}\mathbb{E}\left[\left|Y_n\right|\right]{}\leq{}\mathbb{E}\left[\left|X\right|\right]{}+{}\mathbb{E}\left[\left|Z\right|\right]{}<{}\infty\,.
$$ 
