Consider real RVs $X_n$ that converge to $X$ a.s. and are dominated by an integrable $Y$, so $\lvert X_n\rvert\le Y$ a.s. for all $n$ and $E(\lvert Y\rvert)<\infty$. Now let $\mathcal F_n$ be a filtration. I want to prove that $$\lim_{n\rightarrow\infty}E(X_n-X\vert\mathcal F_n)=0$$ both in $L^1$ and almost surely. I have no clue how to handle the fact that the $\sigma$-Algebra in the conditional expectation changes as well over $n$. I'm not even sure I can pull in the limit at all in this case.
Is there a tool/trick to work around the fact that the conditional changes as well?
By the triangle inequality, we have
$$|E(X_n-X \mid \mathcal{F}_n)| \leq E(|X_n-X| \mid \mathcal{F}_n).$$
Using the tower property, we find that
$$E\big[ \big|E(X_n-X \mid \mathcal{F}_n)\big| \big] \leq E(|X_n-X|).$$
Because of the assumptions on $X_n$, the dominated convergence theorem shows that the right-hand side converges to $0$, and so $E(X_n-X \mid \mathcal{F}_n) \to 0$ in $L^1$.
To prove almost sure convergence, we use Lévy's upward theorem. Fix $m \in \mathbb{N}$. Clearly,
$$|E(X_n-X \mid \mathcal{F}_n)| \leq E \left( \sup_{k \geq m} |X_k-X| \mid \mathcal{F}_n \right)$$
for $n \gg 1$. Set $\mathcal{F} := \sigma(\mathcal{F}_n; n \in \mathbb{N})$, then Lévy's upward theorem yields
$$\limsup_{n \to \infty} |E(X_n-X \mid \mathcal{F}_n)| \leq E\left( \sup_{k \geq m} |X_k-X| \mid \mathcal{F} \right) \tag{1}$$
almost surely. Since
$$\sup_{k \geq m} |X_k-X| \xrightarrow[]{m \to \infty} 0$$
and
$$\sup_{k \geq m} |X_k-X| \leq 2 |Y| \in L^1,$$
the dominated convergence theorem (for conditional expectations) entails that the right-hand side of $(1)$ converges almost surely to $0$ as $m \to \infty$. This finishes the proof.
In case that you are not familiar with the dominated convergence theorem for conditional expectations, you can look it up for instance, in the probability book by Chow & Teicher (Theorem 2 in Section 7.1) or in the measure theory book by Schilling (Corollary 27.15, 2nd ed).