We show that the basic real number $\frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)}$ equals the conditional expectation $\mathbb{E}(X \mid \mathcal{G})$ on the event $E_n$. If $\frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)}$ is understood as a constant map, then result
R1177: Constant map is always measurable shows that it is measurable in $\mathcal{G}$, which is the first condition required of a conditional expectation.
Next, applying results
we have
\begin{equation}
\int_{E_n} \frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)} \, d \mathbb{P} = \frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)} \int_{E_n} \, d \mathbb{P} = \frac{\mathbb{E}(X I_{E_n})}{\mathbb{P}(E_n)} \mathbb{P}(E_n) = \mathbb{E}(X I_{E_n}) = \int_{E_n} X \, d \mathbb{P}
\end{equation}
This finishes the proof. $\square$
