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I want to prove the following:

Let $X\in L^1(\Omega,\mathcal{F},\mathbb{P})$ be a random variable and suppose $\mathcal{G,M}\subset\mathcal{F}$ are sub–$\sigma$–algebras such that $\mathcal{G}$ is independent of $X$ and $\mathcal{M}$. Prove that $$\mathbb{E}(X\mid\sigma\,(\mathcal{G},\mathcal{M}))=\mathbb{E}(X\mid\mathcal{M}).$$

Following hints from my textbook, I have proved that for all $G\in\mathcal{G}$ and $M\in\mathcal{M}$ $$\int_{G\cap M}\mathbb{E}(X\mid\mathcal{M})\,\text{d}\mathbb{P}=\int_{G\cap M}X\,\text{d}\mathbb{P}.$$ However, there's number of issues I do not understand and I hope you could clarify them to me.

I know that $\mathbb{E}(X\mid\sigma\,(\mathcal{G},\mathcal{M}))$ is the only r.v. $\eta$ such that
(i) $\eta$ is $\sigma\,(\mathcal{G},\mathcal{M})$–measurable,
(ii) $\forall_{A\in\sigma\,(\mathcal{G},\mathcal{M})} \int_A\eta=\int_AX.$
By showing that RHS of the equality in the problem satisfies (i) and (ii), I'm done. However, my questions are:

$1)$ Why is $\mathbb{E}(X\mid\mathcal{M})$ a $\sigma\,(\mathcal{G},\mathcal{M})$–measurable r.v.?
$2)$ So far, I proved (ii) for $A$'s of the form $A_1\cap A_2$ where $A_1\in\mathcal{G}$, $A_2\in\mathcal{M}$. Does the claim follows from that? If so, why?

Thanks for help!

Edit: Ad question 2): I've just found that it is sufficient to verify (ii) for every set $A$ in some $\pi$–system that contains $\Omega$ and generates $\sigma\,(\mathcal{G},\mathcal{M})$. Is it true that $\{G\cap M\mid G\in\mathcal{G}, M\in\mathcal{M}\}$ is a generator of $\sigma\,(\mathcal{G},\mathcal{M})$? I will strongly appreciate any stories that may possibly improve my understanding of $\sigma$–algebras.

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up vote 1 down vote accepted
  1. Because $E[X\mid \cal M]$ is by definition $\cal M$-measurable and $\cal M\subset \sigma(\cal G,\cal M)$.
  2. The idea which consists to use $\pi$-system is good. Let $\mathcal{P}:=\{G\cap M,G\in\mathcal{G},M\in\cal M\}$. Then $\cal P\supset \cal M,\cal G$ as $\Omega\in\cal M\cap\cal G$. If $G\in \cal G$ and $M\in\cal M$, then $G\in\sigma(\cal G,\cal M)$ and $M\in\sigma(\cal G,\cal M)$, and their intersection still will be in $\sigma(\cal G,\cal M)$, proving that $\sigma(\cal P)=\sigma(\cal G,\cal M)$.

Now we conclude showing that the collection of subsets $S$ of $\Omega$ satisfying $$\int_SE[X\mid \mathcal M]\mathrm{d}\Bbb P=\int_SX\mathrm{d}\Bbb P$$ is a $\lambda$-system.

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