I know the following formulation of the Fatou's lemma:
From Jacod-Protter (2004)
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $(X_n)_n$ be a sequence of random variables defined on it.
If the random variables $X_n$ satisfy $X_n\geq Y$ a.s. $(Y\in\mathcal{L}^1)$, all $n$, we have $$E\{\liminf_{n\to\infty}X_n\}\leq\liminf_{n\to\infty}E\{X_n\}\tag{1}$$In particular, if $X_n\geq 0$ a.s. all $n$, then $$E\{\liminf\limits_{n\to\infty}X_n\}\leq\liminf\limits_{n\to\infty}E\{X_n\}\tag{2}$$
Given the above result, how is that possible to show that:
$$\displaystyle\mathbb{E}( {\limsup_{n \mathop \to \infty} X_n}) \ge \limsup_{n \mathop \to \infty} \mathbb{E}({X_n})\tag{3}$$
? If so, how?
The idea is to use the fact that $\liminf (-X_n)=-\limsup X_n.$ So then \begin{align} &\phantom{\implies} E(\liminf (-X_n))=E(-\limsup(X_n))=-E(\limsup(X_n))\\ &\leq \liminf E(-X_n)=\liminf(-E(X_n))=-\limsup(E(X_n)). \end{align} Then just flip the inequalities. Typically the assumption to justify this is that $|X_n|\leq Y$. Then $-Y\leq X_n \leq Y$. In particular, $-X_n\geq -Y$, so Fatou's is justified applied to $-X_n$.