Q:

counterexample to conditional expectation

Let F,G be some $\sigma-algebra$ is it true that in general
$E\left(E\left(X\mid G\right)\mid F\right)\neq E\left(X\mid F\cap G\right)$?
I think it's not, however I can't provide a counter example

A:

$\Omega = \{a,b,c\}$, three atoms each of probability $1/3$.
$X(a)=X(b)=0$, $X(c)=1$.
$\mathcal F = \big\{\Omega, \varnothing,\{a,b\}, \{c\}\big\}$
$\mathcal G = \big\{\Omega,\varnothing,\{a\}, \{b,c\}\big\}$.
Compute $E[E[X|\mathcal G]|\mathcal F]$: it has value $1/4$ at $a$ and $b$, value $1/2$ at $c$.
Complute $E[X|\mathcal F \cap \mathcal G]$: it is the constant $1/3$.
added remark 
Define sequence $X_0=X$, $X_{2n+1} = E[X_{2n}|\mathcal F]$, $X_{2n+2} = E[X_{2n+1}|\mathcal G]$.  Then can you show that sequence $X_n$ converges (in some appropriate sense) to $E[X|\mathcal F \cap \mathcal G]$ ??

