We work on a probability space $(\Omega,\mathcal{F},P)$, and $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{F}$. Suppose that $X_n\to X$ in $L^1$, i.e. that $E[|X_n-X|] \to 0$. When does this imply that $E[|X_n-X| \bigr| \mathcal{G}] \to 0$ or (weaker) $E[X_n|\mathcal{G}] \to E[X| \mathcal{G}]$ a.s.?
It is an easy application of Jensen to show that if $X_n \to X$ in $L^p$ then $E[X_n|\mathcal{G}] \to E[X|\mathcal{G}]$ in $L^p$. Of course there is also the conditional version of dominated convergence which gives the result if $X_n \to X$ a.s. and they are dominated in $L^1$, this case is not so useful for me. The result also follows in the trivial case that $\mathcal{G}$ is independent of $\sigma(X_n:n \in \mathbb{N})$ and $\sigma(X)$.
Just some intuition of where this question is coming from: I am studying for a probability exam and have been working a lot with uniformly integrable martingales. In particular, this question popped in my head while proving Levy's upward convergence theorem (take whatever filtration $\mathcal{F}_n$ you like):
Theorem (Levy's Upward Convergence Theorem) If $\xi \in L^1$ and $M_n:= E[\xi | \mathcal{F}_n]$, then $M$ is a uniformly integrable martingale, and $M_n \to E[\xi|\mathcal{F}_\infty]$ a.s. and in $L^1$.
I know it might be a little far-fetched to hope that $E[|X_n-X| \bigr| \mathcal{G}] \to 0$ without assuming $X_n \to X$ a.s., but maybe some UI assumption or something else might do the trick.
Jochen Wengenroth's example at MathOverflow provides a counter-example.