What happens when measuring a subsystem of a product bipartite state? I have an horrible doubt, as I found that in my books the question is never directly addressed. How is collapse defined to work for measures of composite system? 
I always assumed that when we measure on a subsystem $I$ of a quantum system $I\otimes J$ we collapse the whole system unless the state is separable. 
For example if we have a state $\frac{1}{\sqrt{3}}\left|00\right\rangle +\frac{1}{\sqrt{3}}\left|01\right\rangle +\frac{1}{\sqrt{3}}\left|11\right\rangle$ and we measure the first q-bit finding $0$ we can say the state is collapsed in $\left|00\right\rangle$ or $\left|01\right\rangle$ with 50% probability each. The state can't be $\left|0\right\rangle(\frac{1}{\sqrt{2}}\left|0\right\rangle+\frac{1}{\sqrt{2}}\left|1\right\rangle)$ or something. It's either $\left|00\right\rangle$ or $\left|01\right\rangle$ am I right?   
 A: No -- the other option you propose is right: The result is the state obtained by applying $|0\rangle\langle0|\otimes1\!\!1$. In you example, this is $|0\rangle\otimes(|0\rangle+|1\rangle)/\sqrt{2}$.
Note that this is the only thing which makes sense if you don't want to have a preferred basis on the unmeasured system -- the scheme you propose is clearly basis-dependent.
A: If the system is in state 
$$
|\psi\rangle = \frac{1}{\sqrt{3}}|00\rangle + \frac{1}{\sqrt{3}}|01\rangle + \frac{1}{\sqrt{3}}|11\rangle
$$
and the first qubit only is measured to be in 0, then the combined state $|\psi\rangle$ gets projected onto the subspace of possible states that are consistent with this measurement outcome. In this case, we would $|\psi\rangle$ onto $|0\rangle_1\langle 0|$, where this is a projection operator for the left atom to be in 0.
Acting this projector onto $|\psi\rangle$, one obtain:
$$
|0\rangle_1\langle0| \psi \rangle = \frac{1}{\sqrt{3}}|00\rangle + \frac{1}{\sqrt{3}}|01\rangle
$$
The state should then be normalized after the projection, so the final resulting state would indeed be a superposition:
$$
|\psi\rangle \to \frac{1}{\sqrt{2}} |00\rangle + \frac{1}{\sqrt{2}}|01\rangle = |0\rangle \otimes (\frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle)
$$
Note that I have rewritten the state to emphasize that the left qubit is in the state $|0\rangle$ and the right qubit remains in a superposition of two states.
