Will the generated sigma algebra have this property? Lets say you have a measurable space $(\Omega, \mathcal{A})$. And a measurable function $X: (\Omega, \mathcal{A})\rightarrow(\mathbb{R},\mathcal{B}(\mathbb{R}))$. We then know that for the sigma algebra generated by this function : $\sigma(X)=\{ X^{-1}(B)| B \in \mathcal{B}(\mathbb{R})\}$ has these properties:
The sets $X^{-1}(\{r\})$ are disjoint and if a person we do not know chooses an $\omega$ but only tells us the value of of $X(\omega)$ but not $\omega$ itself, we can still find out for any $A\in \sigma(X)$ contains $\omega$ or not, like this: Lets say the stranger tells us that $X(\omega)=a$. Look at $X^{-1}(\{a\})\cap A$, if $X^{-1}(\{a\})\cap A=\emptyset$, then A did not happen. If $X^{-1}(\{a\})\cap A\ne \emptyset$, then A must have happened, because then we must have had $X^{-1}(\{a\})\subset A$.(Because $X^{-1}(\{a\})\cap A=X^{-1}(\{a\})\cap X^{-1}(B)=X^{-1}(\{a\}\cap B)$).
My question is: Does this property extend when we have a collection of functions?: $X_c,c \in C$, where C is an index set. And we then look at the sigma algebra $\mathcal{F}$ which we define to be the sigma algebra generated by $\cup_{c \in C}\sigma(X_c). $ If a stranger now picks an $\omega$, but does not tell you the $\omega$, but he tells you all the values of $X_c(\omega), c \in C$, and you have an arbitrary set $A \in \mathcal{F}$, can you then say if you can find out if $\omega \in A$, and how would you find it out?
PS: This question is ofcourse related to the fact that in regard to random variables, and stochastic processes etc., the sigma-algebra(or filtration) is said to be containing the information, but almost none of the books I have seen says how the information is recovered. 
 A: For the case of just one r.v. I like to think this way: given $A\in \sigma(X)$, if $A=X^{-1}(B)$ for some $B\in \mathcal B(\mathbb R)$ and let's say that the result of the experiment is $\omega_0 \in \Omega$ and that $X(\omega_0)=x_0$. Now, by definition, we have that
$$\omega_0 \in X^{-1}(B) \quad \iff \quad X(\omega_0) \in B,$$
that is
$$\omega_0 \in A \quad \iff \quad x_0 \in B.$$
So: given $A \in \sigma(X)$ , I look for a borelian set whose pre-image by $X$ is precisely $A$. Now, $A$ occurs if and only if the value taken by $X$ is in that borelian set.
The important remark is that this is possible for any $A \in \sigma(X)$, since actually
$$\sigma(X)=\sigma(\{X^{-1}(B): B\in \mathcal B (\mathbb R)\})=\{X^{-1}(B): B\in \mathcal B (\mathbb R)\},$$
which can be easily proven by checking that this last set is indeed a $\sigma$-algebra itself (here are useful properties of the functional preimage such as $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$, and so for $\cap$ and $\setminus$.)
Now, for the general case you propose, there is a similar situation:
$$\mathcal F=\sigma(\{X_c\}_{c\in \mathcal C})=\sigma(\{X_c^{-1}(B): c \in \mathcal C, \; B\in \mathcal B (\mathbb R)\}),$$
but this time it might be the case that $\sigma(\{X_c\}_{c\in \mathcal C}) \supsetneq \{X_c^{-1}(B): c \in \mathcal C, \; B\in \mathcal B (\mathbb R)\}$, since for $c_1 \neq c_2$ (both from $\mathcal C$) and $B_1,B_2\in \mathcal B (\mathbb R)$ (both equal or not), while $X_{c_1}^{-1}(B_1)$ and $X_{c_2}^{-1}(B_2)$ are both in that last mentioned set (and of course in the $\sigma$-algebra it generates), $X_{c_1}^{-1}(B_1) \cup X_{c_2}^{-1}(B_2)$ has to be in the $\sigma$-algebra, but need not be in the generating set (and if it does, is because it is the preimage of some borelian through some $X_c$.)
Now, given $A \in \mathcal F$, just as an example let's supose we have
$$A=\left[X_{c_0}^{-1}(B_0)\cap \left(\bigcup_{n=1}^{\infty} X_{c_n}^{-1}(B_n)\right)\right]\cup \left(\bigcap_{n=0}^{\infty} X_{c_{2n}}^{-1}(B_0)\right)$$
(which has to belong to $\mathcal F$ since each $X_{c_i}^{-1}(B_j)$ does).
To see that it is enough (at least for this case) to know $x_c=X_c(\omega_0)$ for each $c\in \mathcal C$, we can check that
$$\omega_0 \in A$$
$$\Updownarrow$$
$$\left[\omega_0 \in X_{c_0}^{-1}(B_0) \wedge \exists n\left(n \in \mathbb N \wedge \omega_0 \in X_{c_n}^{-1}(B_n)\right)\right] \vee \forall n\left(n\in \mathbb N_0 \implies \omega_0 \in X_{c_{2n}}^{-1}(B_0)\right) $$
$$\Updownarrow$$
$$\left[x_{c_0} \in B_0 \wedge \exists n\left(n \in \mathbb N \wedge x_{c_n} \in B_n\right)\right] \vee \forall n \left(n\in \mathbb N_0 \implies x_{c_{2n}} \in B_0 \right).$$
It's quite complicated to present a generic member of $\mathcal F$ (I believe it will get you at least to consider techniques of transfinite induction, if it's possible at all), so you shouldn't expect to get a constructive proof.
I think your approach might work very well with a proper formalization.
If we go back to our last example, we see that given $A$ and $\omega_0$:


*

*we did not know $\omega_0$, but we assumed we knew $x_c=X_c(\omega_0) \in \mathbb R$ for each $c \in \mathcal C$;

*we assumed that we could decide in each case whether $x_{c_i}\in B_j$, for $i,j \in \mathbb N_0$. To deal with the general situation, lets assume that we can decide $\forall c\in \mathcal C$ and $\forall B \in \mathcal B(\mathbb R)$ whether it is true or not that
$$x_c \in B.$$


Furthermore, let's say that $A\in \mathcal F$ is decidable if having all that information for a particular $\omega\in\Omega$ we can conclude without ambiguity that $\omega \in A$ or that $\omega \notin A$. More precisely, $A$ is decidable iff
$$X_c(\omega_1)=X_c(\omega_2)\quad \forall c\in \mathcal C$$
implies that
$$\omega_1 \in A \quad \iff \quad \omega_2 \in A.$$
Lets call $\mathcal D \subset \mathcal F$ to the collection of all events of the $\mathcal F$ which are decidable. Then, it is not difficult to show that $\mathcal D$ contains all the sets $X_c^{-1}(B)$ and that it is a $\sigma$-algebra, so that $\mathcal D \supset \mathcal F$. This proves that $\mathcal D = \mathcal F$.
