Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$\mathcal{G} \subseteq \mathcal{F}$ is a D470: Subsigma-algebra of $\mathcal{F}$ on $\Omega$
(ii)	$X : \Omega \to \mathbb{R}^D$ is a D4383: Random euclidean real number on $P$
(iii)	\begin{equation} \mathbb{E} |X| < \infty \end{equation}
(iv)	\begin{equation} X \in \mathcal{G} \end{equation}

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

(i)	R2150: Expectation of conditional expectation for a random euclidean real number
(ii)	R2549: Conditional expectation of random complex product when factor is known

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. $\square$