$X$ is a random variable in ($\Omega,\mathcal{F},\mathbb{P}$),$\mathcal{G,H}$ are two sub $\sigma$-algebra and $X$ is independent of $\mathcal{G}$ . I don't know whether the following equality is true or not.(if not, I can add the condition that $\mathcal{G,H}$ are independent)
$$\mathbb{E}(X|\mathcal{G,H})=\mathbb{E}(X|\mathcal{H})$$
In intuition,it seems to be true since $\mathcal{G}$ doesn't provide any information.Rigorously，it is sufficient to prove that $\mathbb{E}(X|\mathcal{G,H})$ is $\mathcal{H}$-measurable.I am stuck in this step.Follow the definition of measurable,I want to analyze the structure of $\mathbb{E}(X|\mathcal{G,H})^{-1}(B)$ ($B\in\mathcal{H}$),I don't know what to do next.
The question above comes from Durrett's textbook http://www.math.duke.edu/~rtd/PTE/PTE4_1.pdf
page 226:example 5.6.1 using backwards martingales to prove the SLLN. in the fifth line
$$\mathbb{E}(\xi_j|\mathcal{F}_{-n-1})=\mathbb{E}(\xi_k|\mathcal{F}_{-n-1})$$
the author said it's true by symmetry but I want to prove it rigorously. I know how to prove
$$\mathbb{E}(\xi_j|S_{n+1})=\mathbb{E}(\xi_k|S_{n+1})$$
but I don't know how to prove
$$\mathbb{E}(\xi_j|\mathcal{F}_{-n-1})=\mathbb{E}(\xi_j|S_{n+1})$$
