Let $X,Y,Z$ be integrable random variables on the same probability space.
Is it true that $\sigma(E[X|Y], E[X|Z]) \subseteq \sigma(E[X|Y,Z])$ ?
My intuition says yes: the approximation of $X$ on $\sigma(Y,Z)$ is richer in information than the one on $\sigma(Y)$ and the one on $\sigma(Z)$. How to prove this?
If it is not true, what if we add $\sigma(Y) \subseteq \sigma(Z)$ ? Does this make the above true?
No, not even with the added condition. The issue is that conditioning on a smaller $\sigma$-field may be able to "split apart" two events that $X$ does not otherwise distinguish.
Consider a probability space with 4 outcomes, $\Omega = \{a,b,c,d\}$, each having probability $1/4$. We take $\mathcal{F} = 2^\Omega$ so all sets are measurable.
Let $Y = 1_{\{a,b\}}$ and let $Z$ be injective, so that $\sigma(Z) = \mathcal{F}$ and in particular $\sigma(Y) \subset \sigma(Z)$. Define $X$ as follows: $$\begin{array}{c|c} \omega & X(\omega) \\ \hline a & 0 \\ b & 2 \\ c & 2 \\ d & 8 \end{array} $$ We have $E[X \mid Z] = E[X \mid Y,Z] = X$. So $\sigma(E[X \mid Y,Z]) = \sigma(X)$ is generated by the events $\{a\}, \{b,c\}, \{d\}$ and is not all of $\mathcal{F}$. In particular $\{b\} \notin \sigma(E[X \mid Y,Z])$.
However, you may check that $E[X \mid Y]$ is given by $$\begin{array}{c|c} \omega & E[X \mid Y](\omega) \\ \hline a & 1 \\ b & 1 \\ c & 5 \\ d & 5 \end{array} $$
Thus $\{a,c\} \in \sigma(E[X \mid Y])$. So $\sigma(E[X \mid Y], E[X \mid Z])$ contains both $\{a,b\}$ and $\{b,c\}$ so it is all of $\mathcal{F}$. In particular, it contains $\{b\} = \{ E[X \mid Z] = 2 \} \cap \{ E[X \mid Y] = 1\}$.
