When talking about means (which are highly non-constructive objects) it is very useful to use a much more constructive approach based on the Day-Reiter characterization of amenability. Namely, a sequence of probability measures $\lambda_n$ on a countable group $G$ (in your case it is just the group of integers $\mathbb Z$) is called (left) **approximately invariant** if $\|g\lambda_n-\lambda_n\|\to 0$ for any group element $g\in G$ (here $\|\cdot\|$ denotes the total variation). Any weak$^*$ limit point of an approximately invariant sequence is an invariant mean, and, conversely, any invariant mean can be obtained in this way (one can read more about it in the classical book of Greenleaf, which, at the end of Section 2.4, also contains a discussion pretty close to your question). The easiest example is provided by any sequence of uniform measures on intervals $I_n\subset\mathbb Z$ whose lengths go to infinity (in particular, any sequence of Cesaro averages, unilateral or bilateral, will do).

Now, back to your question. By the definition of the lower and upper densities, there are sequences of bilateral Cesaro averages whose limits realize these values. Therefore, by the above, there are invariant means which also realize the same values. By taking their convex combinations one can realize any intermediate value as well (one can also argue in a somewhat different way: for any given intermediate value there is a sequence of Cesaro averages whose limit realizes this value).

On the other hand, **the converse is not true**. Namely, it is well possible that an invariant mean gives to a set $A$ a value which is not sandwiched between its lower and upper densities. In particular, even if the lower and upper densities coincide, there may exist means with other values. The reason is very simple: one can modify any set $A\subset\mathbb Z$ in such a way that the new set $A'$ has the same upper and lower densities as $A$, but both $A'$ and its complement contain arbitrarily long intervals. Then (as it follows from the above example with intervals) there exist invariant means which give the set $A'$ values 0 and 1 (and therefore any intermediate value as well).

meanis a linear functional $m:l^\infty(\mathbb Z)\rightarrow\mathbb R$ such that $m(\chi_{\mathbb Z})=1$.Invariantmean is a mean such that $m(\chi_{n+A}=m(\chi_A)$ for all $A\subseteq\mathbb Z$ $\endgroup$ – Valerio Capraro Apr 7 '11 at 10:11