Interpretation of $\sigma$-algebra and filtrations (follow-up question) This is a follow-up question to Interpretation of sigma algebra, particularly to Jun Deng's excellent answer. He used the example of two coin tosses to explain some fundamentals of how filtrations and conditional expected values work.
The following three $\sigma$-fields (for times $0,1,2$) have been presented: 
$\mathcal{F}_0=\{\emptyset,\Omega\}$.
$\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}\supset \mathcal{F}_0 $ 
$\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}\supset \mathcal{F}_1$
Now, his intuitive explanations makes perfect sense. What I am after though, is a more rigorous mathematical derivation of these. I understand that these are the pre-images of the random variable, but I can't wrap my mind around it.
Lastly, I have a similar problem with the following:
$$E[X|\mathcal{F_2}](\omega)=X(\omega)\qquad\text{for every}\ \omega $$
Intuitively, this makes sense, mathematically, I am at loss.
 A: 'I understand that these are the pre-images of the random variable'
They COULD be. I mean given any sigma-algebra, you could just make a random variable that had that sigma-algebra for its set of preimages right?
There's no rigour there, I think. When you flip a coin and see the result, there are more things we know almost certainly.
Previously the only events whose probabilities were 0 or 1 were the ones in $\mathscr{F_0}$. After the first flip, we have added 2 events. Depending on the toss, we would know which of those events has probability 1 and which has probability 0.

As for the last one, what you got there on the LHS is a conditional expectation.
Conditional expectation with respect to an event A is defined constructively(*):
Given $X$ on $(\Omega, \mathscr{F}, \mathbb{P})$
$E(X|A) = \frac{1}{P(A)}\int_A X dP$, provided of course $P(A) > 0$ (Would you use conditional probability on an event with probability zero?).
Conditional expectation with respect to a sigma-algebra $\mathscr{G}$ is not defined constructively(*) (which I guess means no explicit formula or something):
It is some random variable $Z$ s.t.
$\sigma(Z) \subseteq \mathscr{G}$
and
$\int_G Z dP = \int_G X dP \ \forall G \in \mathscr{G}$.
We usually denote $Z \doteq E[X|\mathscr{G}]$.
So, the rigorous explanation is that:


*

*$Z = X$ is $\mathscr{F}_2$-measurable (check that $\sigma(X) \ \subseteq \ \mathscr{F}_2$).

*Check that:
$\int_G Z dP = \int_G X dP \ \forall G \in \mathscr{F}_2$ where $Z \doteq E(X|\mathscr{F}_2)$ in this case = $X$.
So just plug in Z = X to get:
$\int_G X dP = \int_G X dP$.
This clearly holds $\forall G \in \mathscr{F}_2$ or any $\mathscr{G}$.
Conditional expectation may not seem intuitive since its definition is not constructive (*). Try proving its properties. What I did above was prove 'Stability'.
P.S. I think non-constructive (*) definitions need existence and uniqueness or something.
Existence of conditional expectation is due to something called the Radon-Nikodym derivative or theorem, which looks like, iirc afaik, the rigorous version of differentiating one probability distribution to another.
Uniqueness involves continuity of measure.

(*)


*

*Wiki:


'A couple of points worth noting about the definition:
This is not a constructive definition; we are merely given the required property that a conditional expectation must satisfy.'
This seems to give a constructive definition.
This doesn't seem to do so.
Also: Definition of "non-constructive proof"


*Apparently, as Did pointed out, construction is possible if the sigma-algebra in question is finite. I think that is what is being pointed out here.

