I have a stochastic process $X(t,\omega)$ which is a martingale. It is showed that there exists a r.v. $X(\infty)$ such that in $L^1(\Omega)$, $\lim_{t \rightarrow \infty}X(t) = X(\infty)$. In my book it is then stated that:
$X(t) =_{a.s} E[\lim _{s\rightarrow \infty}X(s+t))|\mathcal{F}_t]=E[X(\infty)|\mathcal{F}_t]$. It is said that this is the case because the conditional expectation is continuous in $L^1(\Omega)$.
But I don't really see how. I get that In $L_1$ convergence we have that $E[X(s+t)|\mathcal{F}_t] \rightarrow_{s \rightarrow \infty} E[X(\infty)|\mathcal{F}_t]$. This follows from $E[|E[X(s+t)|\mathcal{F}_t]-E[X(\infty)|\mathcal{F}_t|]\le E[|X(s+t)-X(\infty)|]$, and the last converges to 0 by the $L^1$ convergence.
But how do we get the a.s. convergence?
Here is the argument in my book, I scanned it for you(For clarity $\mathcal{H}^2$ in the argument is the space of stochastic processes where $E[(\sup_t|X_t|)^2]<\infty$):.
By the $L^1$-convergence, we have
$$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = \mathbb{E} \left( X(\infty) \mid \mathcal{F}_t \right). \tag{1}$$
On the other hand, as $(X(t))_{t \geq 0}$ is a martingale, we have
$$\mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t)$$
for all $s \geq 0$. Hence, trivially,
$$L^1-\lim_{s \to \infty} \mathbb{E}(X(t+s) \mid \mathcal{F}_t) = X(t). \tag{2}$$
Since $L^1$-limits are unique (up to a null set), we find by combining $(1)$ and $(2)$ that
$$X(t) = \mathbb{E}(X(\infty) \mid \mathcal{F}_t) \quad \text{almost surely}$$
... and that's exactly what is claimed in your textbook.