Convergence in probability of conditional expectation Suppose I have a sequence of random variables $X_n$ and $\sigma$-fields $\mathcal{F_n} \subseteq \mathcal{F}_n'$. Suppose that $\mathbb{E}[X_n \mid \mathcal{F}_n']$ converges to a constant $c$ in probability. Does it follow that $\mathbb{E}[X_n \mid \mathcal{F}_n]$ also converges to $c$ in probability? It seems the answer should be "yes" since $\mathbb{E}[X_n \mid \mathcal{F}_n]$ is "less random" than $\mathbb{E}[X_n \mid \mathcal{F}_n']$, but I can't prove this.
 A: No, the assertion does in general not hold true.
Consider, for instance, a sequence of independent random variables $(X_n)_{n \in \mathbb{N}}$ such that
$$\mathbb{P}(X_n = -n^2+1) = \frac{1}{n^2} \qquad \mathbb{P}(X_n=1) = 1- \frac{1}{n^2}.$$
It follows easily from the Borel-Cantelli lemma that $X_n \to 1$ almost surely (hence in probability). On the other hand, a straight-forward computation shows $\mathbb{E}(X_n)=0$ for all $n \geq 1$. If we define
$$\mathcal{F}_n := \{\emptyset,\Omega\} \quad \text{and} \quad \mathcal{F}_n' := \sigma(X_1,\ldots,X_n)$$
this means that
$$\mathbb{E}(X_n \mid \mathcal{F}_n) \to 0 \quad \text{but} \quad \mathbb{E}(X_n \mid \mathcal{F}_n') \to 1.$$
A: Yes.
Since $X_n$ is $\mathcal{F}_n$-measurable, $X_n$ is also $\mathcal{F}_n'$-measurable. Therefore, $$\mathbb{E}(X_n \mid \mathcal{F}_n) = X_n = \mathbb{E}(X_n \mid \mathcal{F}_n')$$ 
LHS conv in any mode iff RHS conv in the same mode.
Source: saz comment in reply to my noticing that saz' Xn is not saz' Fn-measurable
