Show that $E[X\vert \sigma(\mathcal{H},\mathcal{G})]=E[X\vert \mathcal{G}]$ where $ X \in L^1(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G},\mathcal{H}$ are sub-sigma algebras of $\mathcal{F}$ and the sigma algebra $\mathcal{H}$ is independent of $\sigma(\sigma(X),\mathcal{G})$
My Attempt:
To show that $E[X\vert \mathcal{G}]=E[X\vert \sigma(\mathcal{H},\mathcal{G})]$, its enough to show that $E[X\vert \sigma(\mathcal{H},\mathcal{G})]$ is $\mathcal{G}$-measurable and that for $A \in \mathcal{G}$ the following is true
$E[X \mathcal{1}_{A}]=EE[X\vert \sigma(\mathcal{H},\mathcal{G})] \mathcal{1}_A]$
This is true since $EE[X\vert \sigma(\mathcal{H},\mathcal{G})] \mathcal{1}_A]=EE[X\mathcal{1}_A\vert \sigma(\mathcal{H},\mathcal{G})] ]=E[X \mathcal{1}_A]$
I cant show precisely that $E[X\vert \sigma(\mathcal{H},\mathcal{G})]$ is also $\mathcal{G}$ measurable instead of being measurable just wrt the larger sigma field $\sigma(\mathcal{H},\mathcal{G})$.
Can somebody help?
You can check this on page number 159. He has proved it with great explanation