Is the conditional expectation of a random variable measurable with respect to it? Suppose $X$ is an integrable random variable on $(\Omega,\mathcal G, P)$ and $\mathcal F\subseteq\mathcal G$ is a sub-$\sigma$-algebra. Is it true that
$$\sigma(E[X\mid\mathcal F])\subseteq\sigma(X)?$$
Intuitively I would say "yes" because $E[X\mid\mathcal F]$ collects all the information that $\mathcal F$ supplies to find out the value of $X$ and at most that can be $\mathcal F = \sigma(X)$ because then one has $X$ (i.e. $E[X\mid X]=X$) all the other information in $\mathcal F$ appears to be irrelevant.

No. let $\Omega = [0,1]$, $X= 1_{[{1 \over 2},1]}$, ${\cal F}$ the field generated by $[0,{1 \over 3}), [{1 \over 3}, {2 \over 3}), [{2 \over 3}, 1]$.
Then $E[X|{\cal F}](\omega)  = \begin{cases} 0, & \omega \in [0,{1 \over 3})\\
{1 \over 2}, & \omega \in [{1 \over 3}, {2 \over 3}) \\ 1, & \omega \in [{2 \over 3}, 1]\end{cases}$, so $\sigma(E[X|{\cal F}]) = {\cal F}$, but
$\sigma(X) = \sigma(\{ [0,{ 1\over 2}), [{1 \over 2},1] \})$.
