Let $X_n , X \in \mathcal L^1$ s.t. $X_n\rightarrow X $ in $\mathcal L^1$. Let $\mathcal g$ be a sub $\sigma$ algebra.
I want to show $$\mathbb E[X_n|\mathcal g] \rightarrow E[X|\mathcal g]\ in \ \mathcal L^1$$
What I have tried:
Use Scheffe lemma to get $\mathbb E[|X_n|]\rightarrow \mathbb E[|X|]\ in \ \mathcal L^1$. Try to show $\mathbb E|X_n|= \mathbb E[|\mathbb E[X_n|\mathcal g]| ]$ and$ \mathbb E|X|= \mathbb E[|\mathbb E[X|\mathcal g]|]$ so that again by Scheffe lemma get the wanted result.
I am stuck at the stage showing $\mathbb E|X_n|= \mathbb E[|\mathbb E[X_n|\mathcal g]|] $. It is enough to show $|X_n|= | \mathbb E[X_n|\mathcal g]| $. Put $Y_n=E[X_n|g]$ and $Y=E[X|g] $. I know by linearity that $Y_n = E[X_n^+|g]-E[X_n^-|g]$ and by positivity that $E[X_n^+|g]\ge 0$ and $E[X_n^-|g]\ge0$. But I don't know how to show that $E[X_n^+|g]$ and $E[X_n^-|g]$ have disjoint support a.s. If so,then everything is set.
Any help is appreciated.
I think it's not that hard.
Since $\mathbb{E}|X-X_n|\to 0$, one has $$ \mathbb{E}\big|\mathbb{E}[X_n|g]-\mathbb{E}[X|g]\big|=\mathbb{E}\big|\mathbb{E}[X_n-X|g]\big|\leq \mathbb{E}\mathbb{E}[|X_n-X||g]=\mathbb{E}[|X_n-X|]\to 0.$$