We must show that $X$ itself is a version of the conditional expectation $\mathbb{E}(X \mid \mathcal{G})$ by confirming that $X \in \mathcal{G}$ and that $\mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) I_G) = \mathbb{E}(E I_G)$ for all $G \in \mathcal{G}$. The first requirement is true by hypothesis. Moving on to the next one, fix an event $G \in \mathcal{G}$ is an event. Now result
R1196: Indicator function measurable iff underlying set is measurable shows that $I_G$ is measurable in $\mathcal{G}$, whence we may apply the results
to conclude
\begin{equation}
\mathbb{E}(X I_G)
= \mathbb{E}(\mathbb{E}(X I_G \mid \mathcal{G}))
= \mathbb{E}(\mathbb{E}(X \mid \mathcal{G}) I_G)
\end{equation}
This completes the proof. $\square$