I'm confused by the derivation of the canonical ensemble, namely the origin of the probability density, that is the Boltzmann factor. Here's what I have:
We have a system of particles with $(N_{tot},V_{tot},E_{tot})$ in thermodynamical equilibrium that we divide into two subsystems, $A$ and $B$. Let $(N,V,E)$ describe $A$, and let $(N',V',E')$ describe $B$. We have the relations $$ N+N' = N_{tot} $$ $$V+V' = V_{tot}$$ $$E+E' \approx E_{tot}$$
where in the last one we've neglected any potential interaction between the particles, arguing that the number of degrees of freedom in any interaction is orders of magnitude smaller than the total amount of particles, because interactions are short ranged.
Now, consider the probability that subsystem $A$ has energy $E$. This is the ratio of the number of microstates of the entire system wherein $A$ has energy $E$ to the number of microstates such that the entire system has $E_{tot}$:
$$P(A \mbox{ has }E)= \frac{\Omega_{tot}(A \mbox{ has }E)}{\Omega_{tot}(E_{tot})}$$
but since we've neglected the interactions, we can factor the numerator:
$$P(A \mbox{ has }E)= \frac{\Omega(E)\Omega ' (E')}{\Omega_{tot}(E_{tot})}$$
where the omegas follow the same convention with primes and subscripts as the other variables. But now I don't understand how to get the Boltzmann factor... this relies on having
$$P(A \mbox{ has }E) = C \cdot \Omega ' (E')$$
But I don't see why that's true... $\Omega(E)$ isn't constant, is it?