Conditional probability and independent sigma algebras

Question

Exercise: Given a random variable $X$ , sigma algebras $\mathcal{G},\mathcal{H}$ such that $\mathcal{H}$ is independent from both $\mathcal{G}$ and $X$ prove that $\operatorname{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\operatorname{E}[X\mid\mathcal{G}]$.

I've proven that $\sigma(\mathcal{G},\mathcal{H})=\sigma(S)$ where $S=\{ H\cap G \vert H\in\mathcal{H},G\in\mathcal{G}\}$ .
Obviously $\operatorname{E}[X\mid\mathcal{G}]$ is $\sigma(\mathcal{G},\mathcal{H})$-mesurable. I've also proven that $\forall A\in S ,\ \operatorname{E}[X\mid\mathcal{G}]$ satisfies the conditional expectation property, i.e. $\int_AX=\int_A\operatorname{E}[X\mid\mathcal{G}]$

I would like to know if i can conclude using a criteria for coincidence of finite measures:
given $A\in\sigma(\mathcal{G},\mathcal{H})$ the maps that send $A \rightarrow \int_AX$ and $A \rightarrow \int_A\operatorname{E}[X\mid\mathcal{G}]$ are well defined finite measures on $\sigma(\mathcal{G},\mathcal{H})$ and both coincide on $S$ which is a set of generators for $\sigma(\mathcal{G},\mathcal{H})$ and it is also stable under finite intersection, thus such measures coincide on $\sigma(\mathcal{G},\mathcal{H})$ which is equivalent with what i want to show.

Thank you

Answer 1

Yes, the criterion you are applying is sometimes known as the Uniqueness Lemma:

Let $\mathcal I$ be a $\pi$-system. If two finite measures agree on $\mathcal I$, then they agree on $\sigma({\mathcal I})$.

A $\pi$-system is a collection of sets that is closed under finite intersection, which is true for your family $S$.