We show that $\mathbb{E} X$ is the conditional expectation of $X$ given $\mathcal{G}$ by confirming that it satisfies the required properties. To start off, result
R1177: Constant map is always measurable guarantees that $\mathbb{E} X \in \mathcal{G}$.
Next, fix $G \in \mathcal{G}$ and let $I_G$ denote the indicator function for $G$ within $\Omega$. Since $X$ and $\mathcal{G}$ are independent, result
R2524: Independent sigma-algebra produces independent indicator functions shows that so are $X$ and $I_G$. Thus, we may apply result results
to conclude
\begin{equation}
\mathbb{E}(X I_G)
= \mathbb{E} X \mathbb{E} I_G
= \mathbb{E}(\mathbb{E}(X) I_G)
\end{equation}
$\square$
