Prove that if $X_n \to X$ in probability, then $X_n \to X$ in $L^1$. 
Suppose $|X_n| \leqslant Z$ for all $n$, where $E(Z) < \infty$. Prove that if $X_n \to X$ in probability, then $X_n \to X$ in $L^1$.

So let's assume that $X_n \to X$ in probability. Since $X_n \leqslant Z$, the latter will also hold for $X$. Therefore $Z_n = |X_n - X|$ satisfies $Z_n \leqslant 2Z$ a.s.
In addition:
$$
E|Z_n|
\begin{aligned}[t]
& = E(Z_n \cdot 1_{[Z_n \leqslant \epsilon]}) + E(Z_n \cdot 1_{[Z_n > \epsilon]}) \\
& \leqslant \epsilon + 2E(Z \cdot 1_{[Z_n > \epsilon]})
\end{aligned}
$$
As $n \to \infty$, $P(|Z_n| > \epsilon) \to 0$ and the last term tends, therefore, to zero.

Well I'd like to understand why is this possible. I suppose that for a certain reason, I could write the latter in order to obtain $E[1_{[Z_n>\epsilon]}]$, which is the same as computing the probability. Well, but why?
 A: Let $Z'$ be a random variable taking only finitely many values. Then
$$
\mathbb E\left(Z\mathbf{1}_{Z_n>\varepsilon}\right)=\mathbb E\left((Z-Z')\mathbf{1}_{Z_n>\varepsilon}\right)+\mathbb E\left(Z'\mathbf{1}_{Z_n>\varepsilon}\right)\leqslant \mathbb E\left(\lvert Z-Z'\rvert\right)+
\sup_{\omega\in\Omega}\lvert Z'(\omega)\rvert\mathbb P\left(Z_n>\varepsilon\right). 
$$
Consequently,
$$
\limsup_{n\to\infty}\mathbb E\left(Z\mathbf{1}_{Z_n>\varepsilon}\right)\leqslant \mathbb E\left(\lvert Z-Z'\rvert\right)
$$
and we know that by definition of the Lebesgue integral, this can be made as small as we wish.
A: Since $E(Z) < \infty$, there exists $M > 0$ such that $E(ZI_{[Z > M]}) < \epsilon$.  It then follows that \begin{align}
& E(ZI_{[Z_n > \epsilon]}) \\
=& E(ZI_{[Z_n > \epsilon, Z > M]}) + E(ZI_{[Z_n > \epsilon, Z \leq M]}) \\
\leq & E(ZI_{[Z > M]}) + MP(Z_n > \epsilon) \\
<& \epsilon + MP(Z_n > \epsilon).
\end{align}
Now let $n \to \infty$ to make the second term in the right hand side of the above inequality vanish.
