I am studying entanglement entropy.

It's fullfilled for any local quantum system that the entanglement entropy of a region $A$ in a mixed state is extensvie,

$$ S_A \sim \frac{\text{Vol}(A)}{\epsilon^d} $$

where $\epsilon$ is the length between sites (or the UV cutoff for a regularized QFT). This is because 

$$S_A=\log[\text{dim}(\mathcal{H}_A)]$$

where $\mathcal{H}_A$ is the Hilbert space of the degrees of freedom in $A$, and the dimension scales as the number of degrees of freedom per site by the number of the sites, $\sim \exp(\text{Vol}(A)/\epsilon^d)$.

On the other hand, for a pure state the entanglement entropy is not extensive, since $S_A=S_B$, where $B$ is the complement of $A$. In fact it is proven that it follows an 'area law'.

An usual intuituve argument for this area law is that to compute the entanglement entropy we count the pairs entangled in both sides of the boundary of $A$. Since the theory is local, the most entangled pairs will be those who lay at $\sim\epsilon$ of the boundary, while the further sites won't count for the entropy. So this way we have that the EE must scale as $\text{Vol}(\partial A)$.

My problem is that I don't know if I am understanding why are we counting all the states un $A$ in one case and only the states near to the boundary in the other.

The reason I found is that in the case of the pure state we are talking about the vaccum, the system is in the ground state of energy and a site can only see what is near around it, since the interactions are local.

But if we excite the system (for example, if we consider a thermal state $\rho = e^{-\beta H}/\text{tr }e^{-\beta H}$, what is a mixed state), the system has enough energy to go beyond the local behavior and then we need to considerate all the sites for the EE.

Am I right with that?