Comparing $X^{-1}(E(X| \mathscr{G})(A))$ and $A$

Question

Is it true that $X^{-1}(E(X| \mathscr{G})(A)) = A$ where $A \in \mathscr{F} $ and $A\notin \mathscr{G}$ where $(\Omega, \mathscr{F}, P)$ is the probability space?

Answer 1

No, this is not true.

First of all, in general $E[X \mid \mathscr{G}]$ is only defined up to sets of measure zero, so $E[X \mid \mathscr{G}](A)$ is not a well defined set.

But setting this aside, try a probability space with two points: $\Omega = \{a,b\}$ with $\mathscr{F} = 2^\Omega$ and $P(\{a\}) = P(\{b\}) = 1/2$. Set $X(a) = 1$ and $X(b) = -1$ so that $X$ is a fair coin flip. Take $\mathscr{G} = \{\Omega, \emptyset\}$ to be the trivial $\sigma$-field; then $E[X \mid \mathscr{G}]$ is the constant random variable equal to $E[X] = 0$. Now let $A = \{a\}$. Since $E[X \mid \mathscr{G}]$ is constant we have $E[X \mid \mathscr{G}](A) = \{0\}$. But $X^{-1}(\{0\}) = \emptyset \ne A$.