Can we conclude that $X$ is $\mathcal{G}$-mensurable if $X=\mathbb{E}[X|\mathcal{G}]$ a.e.?

Question

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G\subseteq F}$ be a sub-$\sigma$-algebra and $X:\Omega\to \mathbb{R}$ be an integrable random variable. Can we conclude that $X$ is $\mathcal{G}$-mensurable if $X=\mathbb{E}[X|\mathcal{G}]$ a.e.?

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I don't know if it's true or false, however, according to some arguments in statistics (as you can see in this link, for instance), that proposition should be true.

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EDIT: The following is the definition of conditional expectation I'm using.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G\subseteq F}$ be a sub-$\sigma$-algebra and $X:\Omega\to \mathbb{R}$ be an integrable random variable. We say that $f:\Omega\to \mathbb{R}$ is a conditional expectation of $X$ given $\mathcal{G}$ if the following propositions are true:

1. $f$ is $\mathcal{G}$-measurable and $\mathbb{P}$-integrable;
2. $\int _Gfd\mathbb{P}=\int _GXd\mathbb{P}$ for all $G\in\mathcal{G}$.

Answer 1

It follows immediately from condition 1 in your definition that $\mathbb{E}(X | \mathcal{G})$ is $\mathcal{G}$-measurable. In probability theory, one usually assumes that filtrations are complete, so all the null sets in $\mathcal{F}$ are also contained in $\mathcal{G}$. This allows you to modify the conditional expectation on a null set (i.e. pick a version of the conditional expectation) to obtain equality with $X$.

Answer 2

To expand a little bit, if you assume that all zero sets of $\mathcal F$ are contained in $\mathcal G$ (a pretty normal assumption, as mentioned in the earlier answer), and that the measure is complete, you can make a very direct answer.

Pick any Borel set $B \subset \mathbb R$; further let $S$ be the set where $X$ and $Y = E[X \mid \mathcal G]$ agree. Then $$X^{-1}(B) = \{\omega \in \Omega \mid X(\omega) \in B\} = (X^{-1}(B) \cap S) \cup (X^{-1}(B) \cap S^C).$$ Now $X^{-1}(B) \cap S^C$ is measure $0$ and by assumption in $\mathcal G$. On the other hand, $$X^{-1}(B) \cap S^C = Y^{-1}(B) \cap S^C$$ clearly differs from the $\mathcal G$-measurable set $Y^{-1}(B)$ on a null set, and by completeness is also in $\mathcal G$. So $X^{-1}(B)$ is in $\mathcal G$.