$X=\mathbb{E}[X\mid\mathcal{F}] $ implying $\mathcal{F}$-measurability of $X$

Question

For a random variable $X$ we know, that if $X$ is measurable w.r.t some filtration $\mathcal{F}$ it satisfies $X=\mathbb{E}[X\mid\mathcal{F}]$.

I wonder whether one can reverse this property and state, if $X=\mathbb{E}[X\mid\mathcal{F}]$ then $X$ is $\mathcal{F}$-measurable. This would accumulate to $X$ is $\mathcal{F}$ measurable if and only if $X=\mathbb{E}[X\mid\mathcal{F}]$.

I guess it makes sense by definition of the conditional expectation:

Definition of the conditional expectation

The conditional expectation of $X$ given $\mathcal{F}$, denoted by $\mathbb{E}[X\mid\mathcal{F}]$, is defined as any random variable Y which satisfies

1. $Y$ is $\mathcal{F}$-measurable
2. $\int_A X d\mathbb{P} = \int_A Y d\mathbb{P}$ for all $A \in \mathcal{F}$

Does the logic work? Thanks in advance!

Answer 1

Since $\mathbb E[X\mid \mathcal F]$ is $\mathcal F-$measurable, the answer is obviouly yes.