How to concceptually think about sub-sigma algebras? I am trying to gain intuition on sub-sigma algebras, and how to think of them in a probabilistic context as information. Let $(\Omega,\mathcal{F})$  be a measurable space and $\mathcal{F'}\subset \mathcal{F}$ be a sub-sigma algebra. Then $\mathcal{F}'$ is conceptually some sort of partial information. For example, if $X$ were a random variable, $\sigma(X)$ would be the sub-sigma algebra generated by this random variable or the one which contains the "information" contained in the random variable $X$.
So if I have some piece of information $\mathcal{F}'$, say the outcome of an experiment, how should I imagine this in my head? What I am thinking is that, if I know $\mathcal{F}'$, then I have some partial information on $\mathcal{F}$. Therefore, when the goddess of chance Tyche selects a $w\in \Omega$, I know for any particular $A\in \mathcal{F}'$ whether or not $\omega$ is in the set or not. Is this the correct way of thinking about it?
I'm also thinking about this in a gambling scenario. Suppose I am in a poker game and I have information on my opponent's emotions. Knowing this piece of information might give me some information on their hand, but I wouldn't be able to definitively say what the hand would be (I only know the information contained in this sub-sigma-algebra, and the sub-sigma-algebra is closed under the operations). So I could say when Tyche selects my $\omega \in \Omega$, then since I am observing my opponents, I know maybe $\omega \in \{\text{Player 1 is nervous}\}$ $\omega \notin \{\text{Player 1 is not nervous}\}$  or for another example $\omega \in \{\text{Player 2 is calm}\}\cap \{\text{Player  1 is nervous}\}$. So I can kind of act like an oracle for this $\sigma$-algebra, telling you whether something happened in this outcome or not, but I can't say anything in a finer setting unless I know that information as well.
Is there a different way to think about this? Any sources I can read to help my understanding? I am trying to not develop the wrong intuition here but I want to learn how to think probabilistically.
 A: I tend to think of a $\sigma$-algebra $\mathcal F$ as a way of expressing how much you are able to distinguish the outcomes. For me the philosophy is that the outcome $\omega\in\Omega$ is in general not directly observable but instead you can only tell, as you mentioned, if it belongs to $A$ for any $A\in\mathcal F$. In that sense $\mathcal F$ measures the amount of information you are able to have on an outcome $\omega\in\Omega$, because the more sets $\mathcal F$ contains, the more you know about $\omega$. The maximal amount of information is attained when $\mathcal F=\mathcal P(\Omega)$. In that case, for any $\omega'\in\Omega$, you have that $\{\omega'\}\in\mathcal F$ so you can tell whether $\omega=\{\omega'\}$, or equivalently whether $\omega=\omega'$. In other words, you know $\omega$. The opposite situation would be the total lack of information: $\mathcal F=\{\emptyset,\Omega\}$. In that case, you can only tell whether $\omega\in\emptyset$ or $\omega\in\Omega$, which tells you nothing about $\omega$.
Let us take a less edge case situation. Suppose you roll a die whose value is $\omega\in\Omega=\{1,2,3,4,5,6\}$. If you are able to just look at its value, then you know $\omega$, so the information you have is represented by $\mathcal P(\Omega)$. But suppose now that someone painted the odd values in red and the even ones in blue. When you roll the die, you can observe the colour but have no way to distinguish between $2$, $4$, $6$, nor between $1$, $3$, $5$. You can only tell whether $\omega\in\{2,4,6\}$ or $\omega\in\{1,3,5\}$, and of course also if it belongs to $\emptyset$ or $\Omega$. Therefore in that case it would be relevant to represent your amount of information by the $\sigma$-algebra $\{\emptyset,\{1,3,5\},\{2,4,6\},\Omega\}$.
Given $\Omega$ endowed with a $\sigma$-algebra $\mathcal F$, you define a probability measure $\mathbb P$ on $\mathcal F$ because you can only affect a probability to the events you can actually observe. In the latter example, you can observe $\{2,4,6\}$ and therefore allocate a probability to that set, that is a quantification of a likelihood that you observe that set. It would make no sense to quantify the likelihood that you observe $\{2\}$, because that it not an event you are able to observe, therefore you don't define $\mathbb P(\{2\})$. If you still did, that would mean that you are extrapolating the probability of an event you are not able to observe, which means that you implicitly place yourself in a bigger $\sigma$-algebra.
I'll finish with a very important example of $\sigma$-algebra $\mathcal F$, that generated by a random variable $X$. In that case, $\mathcal F$ is composed of the sets of the form $\{X\in B\}$ (for any measurable subset $B$ of the space in which $X$ is valued). This means that for any outcome $\omega\in\Omega$ and such $B$ you are able to tell whether $X(\omega)\in B$. It can then be shown that if you endow $\Omega$ with that $\sigma$-algebra $\mathcal F$, then any real-valued random variable $Y$ can be expressed as a (measurable) map $f$ of $X$, that is $Y=f(X)$. In other words, representing your amount of information by $\mathcal F$ is saying that the results of the random experiments you can make are functions of $X$.
A: Short answer: thinking a sub-sigma algebra $\mathcal{F}^{\prime}$ in terms of partial information with respect to a bigger sigma-algebra $\mathcal{F}$ is correct.
The reason is that a smaller sigma-algebra $\mathcal{F}^{\prime} \subset \mathcal{F}$ simply contains less sets. In simple examples, sets can be thought probabilistically  as possible scenarios (see coin flipping example in sub-sigma-algebras section here). A sub-sigma-algebras contains less scenarios.
