Conditional expectation and an independent $\sigma$-algebra Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$ be sub-$\sigma$-algebras of $\mathcal{F}$ and let $X$ be a real-valued random variable that is independent of $\mathcal{H}$. Does
$$
\mathbb{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\mathbb{E}[X\mid\sigma(\mathcal{G})]
$$
hold?
Intuitively, I think this should hold, by I can not think of a proof for this. I would be grateful for help.

This is not true. Let $U,V$ be i.i.d. $N(0,1)$, $X=U+V, Y=U$ and $Z=U-V$. It is well known (and easy to prove) that $X$ and $Z$ are independent. But $E[X|Y,Z]=U+V$ and $E[X|Y]=E[U+V|U]=U$. So $\mathcal G=\sigma (Y), \mathcal H=\sigma (Z)$ provides a counter-example.
