A martingale and mesurability problem Let $X$ be an integrable random variable and $\mathcal F_t$ a filtration.
$E[X|\mathcal F_t]$ is an $\mathcal F$-martingale by the tower property.
Let $f$ be a bounded mesurable function. Set $Y_t=f(E[X|\mathcal F_t])$. Let $0<s<t \in \mathbb R$.
The question: is $E[Y_t|\mathcal F_s]$ mesurable with regard to $\sigma(E[X |\mathcal F_s])$ ?
When $f$ is the identity, this is immediate by the martingale property.
If it helps, we can suppose that $\mathcal F_t$ is generated by a stochastic process $Z_t$ and that $E[X|\mathcal F_t] = E[X|Z_t]$.
EDIT: If it helps even more, we can assume that $X=Z_1$, that $(Z_t)$ is Markov (with regard to its filtration $\mathcal F_t$), sample-continuous, and that $0<s<t \leq 1$.
 A: That is false in general. The Markov assumption might help, although it is not obvious to me how. Anyways, here is a non-Markov counter-example.
Take $\mathcal{F}_n = \sigma (Z_1, \ldots, Z_n)$, where the $Z_i$ are i.i.d. with $\mathbb{P} (Z_i = 1) = \mathbb{P} (Z_i = -1) = 1/2$. Take $(X_0, X_1, X_2) = (0, 0, (1+Z_1) Z_2)$. Then:

*

*$Y_2 = f((1+Z_1) Z_2)$.

*$\mathbb{E} (Y_2 | \mathcal{F}_1) = \mathbb{E} (f((1+Z_1) Z_2) | Z_1) = f(0)$ if $Z_1 = -1$, and $\frac{f(-2)+f(2)}{2}$ if $Z_1 = 1$.

*$\mathbb{E} (X_2 | \mathcal{F}_1) = 0$.

*$\sigma (\mathbb{E} (X_2 | \mathcal{F}_1))$ is trivial.

Hence, as long as $\frac{f(-2)+f(2)}{2} \neq f(0)$, the equality you are looking for is false.
The problem is that compensations may happens so that $\sigma (\mathbb{E} (X_2 | \mathcal{F}_1))$ is strictly smaller than $\mathcal{F}_1$. If $f$ is the identity, or in general affine, the same compensations happens for $\mathbb{E} (Y_2 | \mathcal{F}_1)$, so that you still get what you want. However, non-linearity in $f$ may break these compensations, making $\sigma(\mathbb{E} (Y_2 | \mathcal{F}_1))$ finer than $\sigma (\mathbb{E} (X_2 | \mathcal{F}_1))$.
