Fix $n \in \mathbb{N}$ such that $\omega \in E_n$ for some outcome $\omega \in \Omega$. We show that $\frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)}$ satisfies the properties required of a conditional expectation of $X$ given $\mathcal{G}$. Interpreting the real number $\frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)}$ as a constant map, result R1177: Constant map is always measurable guarantees that it is measurable in $\mathcal{G}$. Next, applying results

(i)	R4652: Real-linearity of real expectation
(ii)	R2089: Unsigned basic expectation is compatible with probability measure

we have \begin{equation} \mathbb{E} \left( \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} I_{E_n} \right) = \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} \mathbb{E} \left( I_{E_n} \right) = \frac{\mathbb{E} (X I_{E_n})}{\mathbb{P}(E_n)} \mathbb{P}(E_n) = \mathbb{E} (X I_{E_n}) \end{equation} This finishes the proof. $\square$