Conditional expectation in a non-filtered probability space The question I have is in the image below:

If $\mathcal{F}_n \subseteq \mathcal{F}_{n+1}$ this would be a trivial problem, but this is obviously not guaranteed.  I thought of maybe defining $\mathcal{G}_i \equiv \sigma \left(\bigcup_{j=1}^i \mathcal{F}_j\right)$ but this isn't really helpful.  So instead perhaps I thought put $\mathcal{H}_i \equiv \bigcap_{j=1}^i\mathcal{F}_i$ and this yields that $$E(X_i|\mathcal{H}_i) = E(X_1|\mathcal{H}_i) $$ but I'm convinced this isn't helpful either.
 A: You essentially want to show:
$E[[X | Y] | Z] = X$ implies $E[X | Y] = X$ a.s.  If you have this, you can further induct to get your solution.
To show this, note that what you are trying to show is equivalent to showing:
$E[ (E[X | Y] - X)^2 ] = 0$.
Now, let us rewrite that as
$E[ (E[X | Y] - X)(E[X | Y] - X) ] = 0$.
which means
$E[ E[X | Y](E[X | Y] - X) ] - E[X (E[X | Y] - X)] = 0$.
So, the first one is $0$ by the definition of conditioning on $Y$, since E[X | Y] is Y measurable, and the difference X - E[X | Y] has to be orthogonal to all Y measurable functions.
For the second term, we rewrite it as:
$E[X (E[X | Y] - X)] = E[X (E[X | Y] - E[E[X | Y]|Z])]$
but again, noting that $X$ is Z measurable implies that this quantity has to be 0.
For the general case, let's try a different proof strategy:
We have, since squaring is a contraction for conditional expectations,
$$X^2 \leq E[(X | F_n | ... | F_i)^2 | ... | F_1] \leq ... \leq E[X^2 | F_n | ... | F_1]$$
Taking expectations of both sides, we see that the inequalities have to be equalities almost surely, i.e. we have
$$ X^2 = E[(X | F_n | ... | F_2)^2 | F_1]$$
Our goal is to show that
$E[(X - E[(X | F_n | ... | F_2)^2]$ = 0, so we want to have
$$E[ E[(X^2 + E[(X | F_n | ... | F_2)^2 - 2 X E[(X | F_n | ... | F_2) | F_1]]$$
Now, let us note that in $F_1$, the first two terms merge, so we have that that term equals to
$$ 2 E[ E[X | F_n | ... | F_1] (E[X | F_n | ... | F_1) - E[X| ... |F_2)]
However, this is 0 by the $F_1$ measurability of the conditional expectation.
This shows that $X_2 = X_1$ in the above question, so now we can induct on a smaller set.
