Suppose $B_n\uparrow B_\infty$ and ${Y_n, n \in \bar{N}}$ is a sequence of random variables such that $Y_n \to Y_\infty$.
(a) If $|Y_n|\le Z \in L_1$, then show a.s. that $$E(Y_n|B_n)\to E(Y_\infty|B_\infty).$$ (b) If $Y_n \to Y_\infty$ in $L_1$ $$E(Y_n|B_n)\to E(Y_\infty|B_\infty).$$
Easy forms, but hard for me to solve... any hints?
I will write $E_n[\cdot]$ for $E[\cdot | B_n]$ for convenience. Firstly note $Y_\infty \in L^1$. Define $Z_m = \sup_{r \geq m} |Y_r-Y_\infty|$. Note $Z_m \to 0$. We can then estimate $|E_nY_n - E_\infty Y_\infty| \leq |E_nY_n - E_n Y_\infty| + |E_n Y_\infty - E_\infty Y_\infty| \leq E_n Z_m + |E_n Y_\infty - E_\infty Y_\infty|$ for $m \leq n$. Levy upward then tells us $\limsup |E_nY_n - E_\infty Y_\infty| \leq E_\infty Z_m$. Then since $|Z_m| \leq 2Y_\infty$ is dominated, we have $ |E_nY_n - E_\infty Y_\infty| \to 0$. This shows part (a).
For part (b) assume for contradiction that we do not have $E_n Y_n \to E_\infty Y_\infty$, say $|E_{n_k} Y_{n_k} - E_\infty Y_\infty| > \epsilon$ for all $n$. Then choose a subsequence $Y_{n'_k}$ of $Y_{n_k}$ on which we have a.s. convergence. By part (a) we will have $E_{n'_k} Y_{n'_k} \to E_\infty Y_\infty$, contradicting that $|E_{n_k} Y_{n_k} - E_\infty Y_\infty| > \epsilon$ for all $n$.