Highest occupied orbitals or bands I am not a student of physics, but a student of nano-technology, hence I might sound extremely stupid, but I just want to clarify my doubt even if sounds very trivial. 
The no. of free charge carriers, $n$, around an energy $E_{0}$ could be given as
$$n=\int_{E_{0}-\Delta/2}^{E_{0}+\Delta/2}\rho\left(E\right)f(E)\,\mathrm{d}E$$
where $\rho(E)$ is the density of states and $f(E)$ is the Fermi distribution. Now, because of the shape of the Fermi distribution, I would assume that using the integral below, $n$ would be zero only at $E_0=\infty$, and hence at finite temperatures, no state $E_0$ would be completely filled or completely empty and have a non-zero probability distribution value $f(E)\rho(E)$ of being occupied. But at $T=0K$, the story is different, since the fermi distribution becomes a kind of step. Does that mean that at finite temperatures, we cannot logically assign a highest filled orbital and that the concept of Highest or lowest occupied states exist only in context of the zero kelvin story??  
 A: As you already said the definition of the highest occupied level does indeed only really exist for T = 0K. For finite temperatures the term of "highest occupation" is still used but what is actually referred to is the position of the Fermi energy, which is mostly defined as the energy where the Fermi-distribution has the value 1/2.
Think for example about a large band-gap isolator. In a sloppy language physicists often say that the valence band is completely filled and the conduction band is completely empty. But we should add (at T = 0K), since the Fermi-distribution is non-zero also in the conduction band. A nice and simple experiment where you can see this is to measure the conductivity against temperature. For an insulator (or semiconductor) you will see that the conductivity increases with temperature which is due to the shift of the Fermi level ("occupation") towards higher energy, or put differently the probability of electron occupation in the conduction band is increasing.
Side note: Do not mix up the term of orbital and state. Orbital is used for the atomic language and if you think about a band diagram you do not fill up orbitals but rather k-states.
Hope that answered your question :)
