What is an ensemble in the classical statistics context? In classical statistics an ensemble is a collection of copies of a system, each copy being in a microstate which is compatible with a given macrostate, right?
My doubt would actually be about this classical concept. If you have 1 single particle moving three-dimensionally in your system, how many systems (let me call every copy a system too) do you have in your ensemble? I'm having trouble understanding how this works for continuous variables like position and momentum. There are infinite positions this particle could be in so it seems that the ensemble would have an infinity number of systems constituing it. Is this right?
 A: When dealing with ensembles in statistical physics, there is no "number of states" the system can be in, because, as you stated correctly, there would be infinitely many for continuous variables (classically speaking). Instead, this concept is used to calculate properties of the system, where we take many copies of it, each with some probably in some microstate (depending on the ensemble and the set-up), and then averaging the property over these systems. How to deal with "many" depends on the problem at hand.
When doing this computationally, we sample the phase space with suitably distributed random manifestations of the system until the result converges. Often times, however, we can do this analytically by integrating (notice: not summing) over all microstates, when the probabilies are known.
A: An ensemble is essentially a collection of all possible energy states a system can be in. That is why in its most general form, the canonical ensemble is given as
\begin{equation}
Q= \sum_i e^{-\beta E_i}
\end{equation}
Which shows that the ensemble is a sum over all possible thermodynamical configuarations of the system. Statistical physics gives us the possibility of using probability to understand extremely large, complex systems so you can obtain the probability of the system being in some discrete energy state, $E_s$.
\begin{equation}
\mathcal{P_s} = \frac{e^{-\beta E_s}}{\displaystyle \sum_i e^{-\beta E_i}}
\end{equation}
Now you are right that when dealing with position and momentum space, that we have an infinite number of possible "energy states". This is why we utilize the classical picture of the ensemble. Think back to the general form where we sum over discrete energy states of the system. Since there are theoretically many states in position and momentum space, we can approximate the states to be so close together that they are continuous as opposed to discrete. This is why we can instead utilize an integration over all the "energy states" as opposed to a sum. 
\begin{equation}
Q= \frac{1}{h^3} \int e^{-\beta H(q, p)} d\Gamma
\end{equation}
Where $d\Gamma$ is the integration over all position and momentum space, $d^3qd^3p$ and $H(q, p)$ is the Hamiltonian that describes the system. So for example, a single particle with translational kinetic energy and no potential with the Hamiltonian of $H = \displaystyle \frac{p^2}{2m}$, would give you the translational ensemble
\begin{equation}
Q = V \left( \frac{2 \pi m k_B T}{h^2} \right)^{3/2}
\end{equation}
