Conditional independence of sigma-algebras If ${\mathcal{H}_1}$ and ${\mathcal{H}_2}$ are conditionally independent given $\mathcal{G} \subseteq {\mathcal{H}_2}$, are they conditionally independent given $\mathcal{F}$ such that  $\mathcal{G} \subseteq \mathcal{F} \subseteq {\mathcal{H}_2}$? Does the same hold for arbitrary $\mathcal{G},\mathcal{F}$ such that $\mathcal{G} \subseteq \mathcal{F}$?
Intuitively, it seems to me that having more information in the condition should't change conditional independence of ${\mathcal{H}_1}$ and ${\mathcal{H}_2}$, but when trying to prove it formally, I can see no guarantee that something like that would hold.
If the statement doesn't hold, a hint for a counterexample would be welcome.
EDIT: ${\mathcal{H}_1},{\mathcal{H}_2},\mathcal{G},\mathcal{F}$ are all assumed to be $\sigma $-algebras in the text above.
EDIT 2: I think the above statement would imply: If $X$ is a random variable on $\left( {\Omega ,\mathcal{A},\mathbb{P}} \right)$ and ${{\rm H}_1},{{\rm H}_2},\Gamma  \subseteq \Omega $ such that ${\left. X \right|_{{{\rm H}_1}}}$ and ${\left. X \right|_{{{\rm H}_2}}}$ are conditionally independent given ${\left. X \right|_\Gamma }$ such that $\Gamma  \subseteq {{\rm H}_2}$, then ${\left. X \right|_{{{\rm H}_1}}}$ and ${\left. X \right|_{{{\rm H}_2}}}$ are conditionally independent given ${\left. X \right|_\Phi }$ in case that $\Gamma  \subseteq \Phi  \subseteq {{\rm H}_2}$.
If the former statement doesn't imply the latter or if it doesn't hold, I'm really only interested in the latter statement. ${\left. X \right|_{\rm H}}$ denotes the restriction of $X$ to ${\rm H}$.
 A: Using this statement, it follows that it suffices to prove that $\mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{G}} \right] = \mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{F}} \right] = \mathbb{E}\left[ {{1_{{H_1}}}|{\mathcal{H}_2}} \right]$.
From the assumption it follows that $\mathbb{E}\left[ {{1_{{H_1}}}|{\mathcal{H}_2}} \right] = \mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{G}} \right],\forall {H_1} \in {\mathcal{H}_1}$. We also have $\mathbb{E}\left[ {\mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{F}} \right]{1_F}} \right] = \mathbb{E}\left[ {{1_{{H_1}}}{1_F}} \right]$ from the definition of conditional expectation. It also holds that $\mathbb{E}\left[ {{1_{{H_1}}}{1_F}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{1_{{H_1}}}|{\mathcal{H}_2}} \right]{1_F}} \right]$ since ${\mathbb{E}\left[ {{1_{{H_1}}}|{\mathcal{H}_2}} \right]}$ is $\mathcal{F}$-measurable because $\mathcal{F} \subseteq {\mathcal{H}_2}$. From the assumption it then follows that $\mathbb{E}\left[ {\mathbb{E}\left[ {{1_{{H_1}}}|{\mathcal{H}_2}} \right]{1_F}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{G}} \right]{1_F}} \right]$, so $\mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{G}} \right] = \mathbb{E}\left[ {{1_{{H_1}}}|\mathcal{F}} \right] = \mathbb{E}\left[ {{1_{{H_1}}}|{\mathcal{H}_2}} \right]$.
This proves the statement.
