Convergence in probability and convergence of $\sup_{n\geq m}E(X_n\mid\mathcal{F_m}).$ Suppose $X_n$ is adapted to the filtration $(\mathcal{F}_n)_{n=1,2,\ldots}$, $X_n$ is positive and bounded above by some fixed $M$. If $X_n$ converges to a constant $x$ in probability, does this imply that $\sup_{n\geq m}E(X_n\mid\mathcal{F_m})\rightarrow x$ in probability as $m\rightarrow\infty$?
I think a.e. convergence gives yes. But I think this might fail for convergence in probability.  Any counterexamples?
 A: Take $(A_j)_j$ a sequence of independent events such that $\mu(A_j)\to 0$ and $\sum_j\mu(A_j)=\infty$. By the Borel-Cantelli lemma, $\mu(\limsup_n A_n)=1$, hence we can find a sequence $(m_k)_k$ of integers such that $\mu\left(\bigcup_{j=m_k+1}^{m_{k+1}}A_j\right)>1/2$ for each $k$. 
We define $\mathcal F_{m_k}:=\sigma(A_j,1\leqslant j\leqslant m_{k+1})=:\mathcal F_l$ for $m_{k}\leqslant l\lt m_{k+1}$. In this way, we have 


*

*$X_n:=\chi(A_n)\to 0$ in probability, $0\leqslant X_n\leqslant 1$ almost surely;

*let $l\in\mathbb N$; then $m_{k}\leqslant l\lt m_{k+1}$ for some integer $k$. In this case $\mathcal F_l=\sigma(A_j,k\leqslant m_{k+1})$ hence $\chi_{A_l}\in\mathcal F_l$.

*$\sup_{n\geqslant m_k}\mathbb E[X_n\mid \mathcal F_{m_k}]\geqslant 
\chi\left(\bigcup_{j=m_k+1}^{m_{k+1}}A_j\right)$, hence $\sup_{n\geqslant m_k}\mathbb E[X_n\mid \mathcal F_{m_k}]$ does not converge to $0$ in probability. 


However, if $X_n\to x$ almost surely, then 
$$\sup_{n\geqslant m}\mathbb E[|X_n-x|\mid\mathcal F_m]\leqslant\mathbb E[\sup_{n\geqslant m}|X_n-x|\mid\mathcal F_m],$$
hence 
$$\mathbb E[\sup_{n\geqslant m}\mathbb E[|X_n-x|\mid\mathcal F_m]]\leqslant \mathbb E[\sup_{n\geqslant m}|X_n-x|]$$
and this converge to $0$ using monotone convergence. 
