Let $X_t$ be a submartingale. Knowing that $\sup_t|X_t| \in L^1$ and that $X_t \to X_{\infty}$ a.s. and in $L^1$, I want to prove that $$X_t\le E[X_\infty|\mathcal F_t]$$ using Fatou's lemma for limsup $$E[X_\infty]=E\big[\limsup_t X_t \big] \ge \limsup_t E[X_t]\ge E[X_t]$$ where the last inequality hold because $E[X_t]$ is an increasing sequence, however I'm not sure of this last step. Indeed it seems to me that (being an increasing sequence) also $$\limsup_t E[X_t]=E[X_\infty]$$ so the inequality cannot be true.
$1)$ So, is it true that $\limsup_t E[X_t]=E[X_\infty]$?
$2)$ how could i prove the statement then?
Yes, it's true. By assumption, $X_t \to X_{\infty}$ in $L^1$ and this implies $\lim_{t \to \infty} \mathbb{E}(X_t) = \mathbb{E}(X_{\infty})$. Since $(X_t)_{t \geq 0}$ is a submartingale, this is equivalent to $\sup_{t \geq 0} \mathbb{E}(X_t) = \mathbb{E}(X_{\infty})$.
You need to apply Fatou's lemma for conditional expectations: \begin{align*} \mathbb{E}(X_{\infty} \mid \mathcal{F}_t) &= \mathbb{E}(\limsup_{T \to \infty} X_T \mid \mathcal{F}_t) \\ &\stackrel{(\star)}{\geq} \limsup_{T \to \infty} \mathbb{E}(X_T \mid \mathcal{F}_t) \\ &\stackrel{(\star \star)}{\geq} X_t\end{align*}
where $(\star)$ follows from Fatou's lemma for conditional expectation and $(\star \star)$ follows from the submartingale property.