Conditional expectation with respect to two sigma algebras 
So the problem is to define two sigma algebras, a stochastic variable,
  specify a probabilty measure on the sample space $\Omega$ and show
  that the following relation doesn't equal:
$$E[E[X|\mathcal{G}]|\mathcal{F}]\neq E[E[X|\mathcal{F}]|\mathcal{G}]$$
So I started by defining the sigma algebras:
$$\Omega=\{a,b,c\}$$ $$\mathcal{F}=\{\emptyset,\Omega,a,(b,c)\}$$
  $$\mathcal{G}=\{\emptyset,\Omega,b,(a,c)\}$$ With the stochastic
  variable $X$: $$X(a)=0,X(b)=1,X(c)=2$$ Probability measures.
P on $(\Omega,\mathcal{F})$:
$$P(\emptyset)=0,P(\Omega)=1,P(a)=p,P(b\cup c)=1-p$$
P on $(\Omega,\mathcal{G})$:
$$P(\emptyset)=0,P(\Omega)=1,P(b)=p,P(a\cup c)=1-p$$

So far I assume that I haven't messed up.
The expectation part though...I'm stuck.
I tried the following:
$$Y=E[X|\mathcal{G}]$$
$$Z=E[X|\mathcal{F}]$$
$$Y=E[X|\mathcal{G}]=E[X|\mathcal{F}]$$
$$Y=E[X|\mathcal{G}]=E[X|b]1_b+E[X|a\cup c]1_{a\cup c}$$
$$Z=E[X|\mathcal{F}]=E[X|a]1_b+E[X|b\cup c]1_{b\cup c}$$
But after that I can't see how I should progress.
 A: Ok I will try to solve it now.

So the problem is to define two sigma algebras, a stochastic variable,
  specify a probabilty measure on the sample space $\Omega$ and show
  that the following relation doesn't equal:
$$E[E[X|\mathcal{G}]|\mathcal{F}]\neq E[E[X|\mathcal{F}]|\mathcal{G}]$$
So I started by defining the sigma algebras:
$$\Omega=\{a,b,c\}$$ $$\mathcal{F}=\{\emptyset,\Omega,a,(b,c)\}$$
  $$\mathcal{G}=\{\emptyset,\Omega,b,(a,c)\}$$ With the stochastic
  variable $X$: $$X(a)=0,X(b)=1,X(c)=2$$ Probability measures.
P on $(\Omega,\mathcal{F})$:
$$P(\emptyset)=0,P(\Omega)=1,P(a)=p,P(b\cup c)=P(B)+P(C)=p+p$$
Thus $B$ and $C$ are disjoint
P on $(\Omega,\mathcal{G})$:
$$P(\emptyset)=0,P(\Omega)=1,P(b)=p,P(a\cup c)=P(A)+P(C)=p+p$$
Thus $A$ and $C$ are disjoint 

$$Y=E[X|\mathcal{G}]$$
$$Z=E[X|\mathcal{F}]$$
$$E[Z|\mathcal{G}]=E[Y|\mathcal{F}]$$
$$Y=E[X|\mathcal{G}]=E[X|b]1_b+E[X|a\cup c]1_{a\cup c}$$
$$Z=E[X|\mathcal{F}]=E[X|a]1_a+E[X|b\cup c]1_{b\cup c}$$
Solving Y:
$$Y=E[X|\mathcal{G}]=E[X|b]1_b+E[X|a\cup c]1_{a\cup c}$$
$$Y=1*1_b+1*1_{a\cup c} = 1_b+1_{a\cup c}$$
Solving Z:
$$Z=E[X|\mathcal{F}]=E[X|a]1_a+E[X|b\cup c]1_{b\cup c}$$
$$Z=0*1_a+\frac{1*P(b)+2*P(C)}{P(b)+P(c)}$$
$$Z=\frac{3p}{2p}=\frac{3}{2}1_{b \cup c}$$
So far so good I guess?
Thus I end up with:
$$E[1_b+1_{a\cup c}|\mathcal{F}]$$
$$E[\frac{3}{2}1_{b \cup c}|\mathcal{G}]$$
The expected value of a constant, no matter what sigma algebra it's conditioned with ends up being a constant.
$$E[\frac{3}{2}1_{b \cup c}|\mathcal{G}]=\frac{3}{2}E[1_{b \cup c}|\mathcal(G)]$$
And as the events $b$ and $a\cup c$ are disjoint one should be able to seperate them.
$$E[1_b+1_{a\cup c}|\mathcal{F}]=E[1_b|\mathcal{F}]+E[1_{a\cup c}|\mathcal{F}]$$
The expectation of a indicator function is just the probability of the expected value.
The union between $a\cup c$ isn't included in the $\mathcal{F}$ and $b \cup c$ isn't included in $\mathcal{G}$, I guess one says that it isn't $\mathcal{F}$-measurable and $\mathcal{G}$-measurable respectively?. Thus I assume that one ends up with:
$$E[1_b+1_{a\cup c}|\mathcal{F}]=\frac{P(b)}{P(b)+P(C)}+0$$
$$E[1_b+1_{a\cup c}|\mathcal{F}]=\frac{1}{2}$$
$$\frac{3}{2}E[1_{b \cup c}|\mathcal(G)]=0$$
Thus:
$$E[E[X|\mathcal{G}]|\mathcal{F}]=\frac{1}{2}$$
$$E[E[X|\mathcal{F}]|\mathcal{G}]=0$$
So we get:
$$\frac{1}{2}\neq 0$$
Hopefully I haven't messed up.
EDIT:
If I remember correctly the union of an event $A$ and $B$ is defined as the event $A$ or $B$ happening or both happening. I also looked up what the indicator function of a union is and got:
$$1_{A\cup B}=1_A+1_B-1_A1_B$$
Substituting the indicatior of the unions I get:
$$ \frac{3}{2}E[1_{b \cup c}|\mathcal{G}]=\frac{3}{2}(E[1_b+1_c-1_b1_c|b]1_b+E[1_b+1_c-1_b1_c|a\cup c]1_{a \cup c})$$
Which then gives:
$$\frac{3}{2}E[1_{b \cup c}|\mathcal{G}]=\frac{3}{2}(1_b+\frac{1}{2}1_{a \cup c})=\frac{3}{2}1_b+\frac{3}{4}1_{a \cup c}$$
The last step is then to evaluate:
$$E[1_b+1_{a \cup c}|\mathcal{F}]=1_a+1_{b \cup c}$$
In the end:
$$1_a+1_{b \cup c} \neq \frac{3}{2}1_b+\frac{3}{4}1_{a \cup c}$$
I assume this holds because if you look at the indicator functions they can't really equate.
