Basic Interpretation of Compostion of Observables and their Measurement Given two (or more) observables $A, B$ which commute one can construct a third observable $C= A \circ B$. If $\psi$ is a common eigenvector of $A, B$ with eigenvalues $\lambda_1, \lambda_2$ then it is clear that the measurement of $C$ of the state $\psi$ gives the measurement result $\lambda =\lambda_1 \lambda_2$, i.e. the result of the measurement of the observable $C$ is the result from the measurement of $A$ times the result from the measurement $B$. But what if $\psi$ is an eigenvector of $C$, but not of $A$ and $B$? Is there any connection between the measurement results of $A$, $B$ and $C$?
Example: Let there be three observers which measure a spin state with the corresponding observables
$A = \sigma_x \otimes \mathbb{I} \otimes \mathbb{I}$,
$B=\mathbb{I} \otimes \sigma_y \otimes \mathbb{I}$ and
$C=\mathbb{I} \otimes \mathbb{I} \otimes \sigma_y$.
They commute and we can construct $D= A \circ B \circ C = \sigma_x \otimes \sigma_y \otimes \sigma_y$.
Now the GHZ-state $\psi = \frac{1}{\sqrt{2}} ( | +z, +z, +z \rangle - | -z, -z, -z\rangle)$ is an eigenvector of $D$ with eigenvalue $\lambda =-1$ but it is not an eigenstate of $A, B$ or $C$.
Each of the observers will get a result $\pm 1$. Is there any connection between this individual results and the eigenvalue of $\psi$ (respectively the expectation value $\langle \psi | D | \psi \rangle = -1$)? Intuitively I would say that the product of the results should give the eigenvalue of $\psi$ but I can't see how this should follow from any quantum mechanical postulate or mathematical reasoning like in the case of the common eigenvector.
 A: Taking $C=A_1A_2....A_n$, the problem arises because the eigenvector subspace corresponding to a eigenvalue of $C$ does not correspond to the eigenvector subspaces corresponding to a eigenvalue of the $A_i$
To see that, we will take an example with $C=A_1A_2$, with $A_1 = \sigma_x \otimes Id, A_2 = Id \otimes\sigma_x $. So $C= \sigma_x \otimes\sigma_x$ We have the following array : 
$$\begin{pmatrix} &\sigma_x \otimes Id&Id \otimes\sigma_x&\sigma_x \otimes\sigma_x
\\(|00\rangle+|10\rangle)+(|01\rangle+11\rangle)&+&+&+
\\(|00\rangle+|10\rangle)-(|01\rangle+11\rangle)&+&-&-
\\(|00\rangle+|01\rangle)-(|10\rangle+11\rangle)&-&+&-
\\(|00\rangle+|11\rangle)-(|01\rangle+10\rangle)&-&-&+\end{pmatrix}$$
The first column is made of the common eingenvectors, and the other columns correspond to the eigenvalues ($\pm$ means $\pm1$).
The subspace corresponding to the eigenvalue $+1$ of $\sigma_x \otimes\sigma_x$ is $2-$dimensional and corresponds to the first and last eigenvectors.
Now, if we add the first and the last eigenvector, we get the state $|00\rangle+|11\rangle$, and because the first and the last eigenvectors have the same eigenvalue $+1$ for $\sigma_x \otimes\sigma_x$, then $|00\rangle+|11\rangle$ is also a eigenvector with eigenvalue $+1$ for $\sigma_x \otimes\sigma_x$.
But the problem is that the first and last eigenvector have not the same eigenvalue for $\sigma_x \otimes Id$ and $Id\otimes\sigma_x$, so any combination of these $2$ eigenvectors cannot be an eigenvector for $\sigma_x \otimes Id$ and $Id\otimes\sigma_x$. And this is indeed the case for $|00\rangle+|11\rangle$
A: Firstly I think I did not state my question clearly enough: I know the mathematics behind commuting observables and their eigenvalues but I wanted to know what happens in a real physical experiment. I wanted to know if there is a connection between the eigenvalue $\lambda$ of a prepared eigenstate $\psi$ of $C=A_1 \ldots A_n$ and the individual results $\lambda_1, \ldots, \lambda_n$ of the measurements of $A_1, \ldots, A_n$. The situation is clear if $\psi$ is a common eigenvector of $A_1, \ldots, A_n$ but it was not clear to me what happens if $\psi$ is NOT a common eigenvector.
Trimok's answer did not answer my question directly but his/her example gave me an important insight which helped me to figure this out.
One of my misconceptions was the idea that an eigenstate $|\psi\rangle$ of $C=A_1 \ldots A_n$ does not change during a measurement of $C$ because of the mathematical relation $C|\psi\rangle = \lambda |\psi\rangle$. But this is not necessarily true. For example, the state $|00\rangle + |11\rangle$ changes during the measurement of $\sigma_x \otimes \sigma_x$: each observer will either get $|0\rangle$ or $|1\rangle$ so the state after the measurement will be $|00\rangle$ or $|11\rangle$ although $|00\rangle + |11\rangle$ is an eigenstate of $\sigma_x \otimes \sigma_x$ to the eigenvalue $+1$.
I think the connection between the different eigenvalues is as follows. Let $|\psi\rangle$ be an eigenstate of $C = A_1 \ldots A_n$ to the eigenvalue $\lambda$, but it does not necessarily have to be a common eigenvector of the observables $A_1, \ldots, A_n$. In general, $|\psi\rangle$ can be expressed as a linear combination of common eigenvectors:
$$
|\psi\rangle =\sum_{\substack{\alpha_1, \ldots, \alpha_n \\ \alpha_1 \cdots \alpha_n=\lambda}} c_{\alpha_1, \ldots, \alpha_n} |\alpha_1, \ldots, \alpha_n\rangle,
$$
where $|\alpha_1, \ldots, \alpha_n\rangle$ is an eigenvector of $A_1$ to the eigenvalue $\alpha_1$, etc. The product of the $\alpha_i$ has to be $\lambda$, because it is an abstract mathematical result that the set of eigenvalues of $C$ has this form.
Now, what happens if the observables $A_1, \ldots, A_n$ are measured separately (w.r.t. $|\psi\rangle)$? Each measurement will make the subspace where the measured state lives smaller: The measurement of $A_1$ will force the state $|\psi\rangle$ to change into a state $|\beta_1\rangle$ where $\beta_1$ is an eigenvalue of $A_1$. But it can still be expressed as a linear combination of common eigenvectors of $A_2, \ldots, A_n$. Then the measurement of $A_2$ will change the state into $|\beta_1, \beta_2\rangle$ and so on. After the measurement of $A_n$ we are left with a common eigenstate $|\beta_1, \ldots, \beta_n\rangle$ and $\beta_1 \cdots \beta_n = \lambda$ because $|\psi\rangle$ was originally a linear combination of common eigenvectors which fulfill this equation.
So the statement "The product of the individual results is equal to the eigenvalue $\lambda$ of the eigenstate $|\psi\rangle$ of $C$" still holds in the case, when $|\psi\rangle$ is not a common eigenvector. This intuitive result is now backed up by a correct argument.
Just as an interesting side note: When an observable $C$ is a composition of other observables, e.g. $C=A\circ B$ then this does not mean that one has to measure $B$ and $A$ consecutively in order to measure $C$. $C$ can represent a single measurement. For example, let's take $A=\sigma_x \otimes \sigma_y$ and $B=\sigma_y \otimes \sigma_x$, then $C = (\sigma_x \otimes \sigma_y) \circ (\sigma_y \otimes \sigma_x) = \sigma_z \otimes \sigma_z$. So $C$ can represent different experimental setups: either a consecutive measurement of $B$ and $A$ or a single measurement of two qubits along the z-axis. (I am new here and I don't know if such remarks are welcomed here or regarded as annoying. I remove them if somebody wishes.)
