How to interpretate the conditional expectation on a sub-$\sigma$-algebra? How to interpretate the conditional expectation on a sub-$\sigma$-algebra?
$X \in \mathcal{L}^1(\Omega,\mathcal{F},\mathbb{P})$, $\mathcal{A} \subset \mathcal{F}$ sub-$\sigma$-alegbra $\Rightarrow$ $\mathbb{E}[X| \mathcal{A}] (\omega)$ is an almost surely defined random variable that is
i) $\mathcal{A}$-measurable
ii) $\int_A X d\mathbb{P}=\int_A \mathbb{E}[X|\mathcal{A}]$ for all $A \in \mathcal{A}$
So what does it mean that the sub-$\sigma$-algebra can be interpreted as "given information"?
If we have a partition of $(\Omega,\mathcal{F})$, i.e. $A_1,A_2,\dots \in \mathcal{F}$ pairwise disjoint with $\bigcup_{n \in \mathbb{N}} A_n=\Omega$ and we consider the induced $\sigma$-algebra $\mathcal{A}:=\{\bigcup_{i \in I} A_i| I\subset \mathbb{N}\}$, then the "given information" means that we already know which event  $A_i$ will happen.
But what do we know if $\mathcal{A}$ is a more complex sub-$\sigma$-alegbra? Do we know about each event $A \in \mathcal{A}$ whether it happend or not?
 A: Your interpretation of knowing whether $A$ happened or not for all $A \in \mathcal A$ is essentially correct.  However, things get a little tricky in that interpretation because $\mathcal A$ is often augmented by the null sets of $\mathcal F$.  For example, if we are looking at a filtration $(\mathcal F_t)$ generated by a Brownian motion $W$, we often augment $\mathcal F_t$ by the null sets of $\mathcal F_\infty$.  This means that events like $\{W_2 = x\} \in \mathcal F_1$ for all $x \in \mathbb{R}$.  If we wanted to interpret $\mathbb{E}[W_2|\mathcal F_1]$ as knowing whether or not $A$ happened for all $A \in \mathcal A$, this would make it seem like we should know the value of $W_2$, since we know whether or not $W_2 = x$ for all $x \in \mathbb{R}$.
Instead, I would interpret conditioning on $\mathcal A$ as being able to ask whether or not $A$ occurred for all $A \in \mathcal A$.  Since $\mathcal A$ is closed under countable unions, we can ask about countably many events.  In the example about the Brownian motion, this makes it so that while we can ask whether $W_2 = x$ for any $x \in \mathbb{R}$, there's no point because we already know that (with probability one) the answer will be "no."
