Consider $X,Y$ be two independent random variable
I want to know under what sigma-algebra $\mathcal{F}$, we can say the conditional expectation $E[X|\mathcal{F}]$ is independent of $E[Y|\mathcal{F}]$
Here is my guess: Call the sigma-algebra $\mathcal{G}:=\sigma(X)$ and $\mathcal{H}:=\sigma(Y)$, and $\mathcal{G}$ is independent of $\mathcal{H}$ since $X$ is independent of $Y$. now if I take $\mathcal{F}:=\sigma(\mathcal{G},\mathcal{H})$, then the remaining is to check the following independence property holds: $$E\left[ {\;E[X|{{\cal F}}] \cdot E[Y|{{\cal F}}]\;} \right] = \underbrace {E[E[X|{{\cal F}}]] \cdot E[E[Y|{{\cal F}}]]}_{ = EXEY}$$ So observe the LHS and by using the assumption that $X,Y$ are independent, we have $$ E\left[ {E[X|{{\cal F}}] \cdot E[Y|{{\cal F}}]} \right] = E\left[ {E[X|\sigma \left( {\mathcal{G},\mathcal{H}} \right)] \cdot E[Y|\sigma \left( {\mathcal{G},\mathcal{H}} \right)]} \right] = E\left[ {XY} \right] = EXEY$$
If we define $\mathcal{F} := \sigma(\mathcal{G},\mathcal{H})$, then $\mathcal{F}$ contains by definition both $\mathcal{G}$ and $\mathcal{H}$ and therefore
$$\mathbb{E}(X \mid \mathcal{F}) = X \qquad \text{and} \qquad \mathbb{E}(Y \mid \mathcal{F}) = Y.$$
So, obviously the conditional expectations are again independent - there is no need for further calculations. (Note that $\mathbb{E}(Z_1 \cdot Z_2) = \mathbb{E}(Z_1) \cdot \mathbb{E}(Z_2)$ does in general not imply independence of $Z_1$ and $Z_2$ for two integrable random variables $Z_1$, $Z_2$.)