Equivalence of conditional expectations wrt two sigma-algebras Let $X$ be a random variable and let $\mathcal{G}$, $\mathcal{H}$ be two sub-$\sigma$-algebras. Consider the equation $$\mathbb{E}(\mathbb{E}(X|\mathcal{G})|\mathcal{H}) = \mathbb{E}(X|\mathcal{G}\cap\mathcal{H}) \text{ almost surely.}$$ I am tasked to show that the above holds in the following three cases: $\mathcal{G}\subseteq\mathcal{H}$, $\mathcal{H}\subseteq\mathcal{G}$, $\mathcal{G}$ and $\mathcal{H}$ are independent.
If we let $Y = \mathbb{E}(\mathbb{E}(X|\mathcal{G})|\mathcal{H})$ and take $A\in\mathcal{G}\cap\mathcal{H}$, it appears (unless I am making a mistake) that $\mathbb{E}(Y\mathbb{I}(A))=\mathbb{E}(X\mathbb{I}(A))$ (forgive the ugly indicator notation) holds generally, not just in the three cases above. Thus it remains to show that $Y$ is $(\mathcal{G}\cap\mathcal{H})$-measurable in the three cases, but I only see how to do this for the case $\mathcal{H}\subseteq\mathcal{G}$, as then $\mathcal{H}=\mathcal{G}\cap\mathcal{H}$ and $Y$ is of course $\mathcal{H}$-measurable. Any help for the other two cases would be greatly appreciated! Also perhaps an example of why this does not hold generally would be useful if someone could provide one.
 A: If $\mathcal G \subseteq \mathcal H$, then $\mathbb E(X | \mathcal G)$ is $\mathcal H$-measurable, so $Y = \mathbb E(X | \mathcal G)$, and as $\mathcal G = \mathcal H \cap \mathcal G$, $Y$ is $\mathcal H \cap \mathcal G$-measurable.
If $\mathcal G$ and $\mathcal H$ are independent, then $\mathbb E(X | \mathcal G)$ is independent of $\mathcal H$, and so $Y$ is constant - thus, of course, $\mathcal H \cap \mathcal G$-measurable.
This should give an idea on how to construct counterexample in general case: take $\mathcal G$ and $\mathcal H$ with trivial intersection, but non-trivial dependencies.
I think the simplest example will be like this: let our probability space be  $\{1, 2, 3\}$ with uniform measure, $X$ be identical, let $\mathcal G$ be generated by event $X = 1$ and $\mathcal H$ by $X = 2$.
Let $u = \mathbb E(X | \mathcal G)$. Then $u(1) = 1$ and $u(2) = u(3) = 2.5$.
From this we have $Y(1) = Y(3) = 1.75$ and $Y(2) = 2.5$.
On the right side, however, we have constant variable with value $2$.
