Probability densities at the LHC Let $\sigma$ be a cross-section for the collision of two protons as given by
$$
\sigma = \intop_0^1 \mathrm{d}x_1 \intop_0^1 \mathrm{d}x_2 \, \sum_{a,b}f_a(x_1,Q^2) f_b(x_2,Q^2) \frac{1}{2\hat{s}} \intop \mathrm{d}\Phi ~ \overline{\left|M_{ab}\right|^2},
$$
where $f_i$ are the parton density functions, $\Phi$ is the available phase-space and $M_{ab}$ ist the (let's say tree-level) Matrix element of the partonic process.
What I fail to understand is which of the integrands can be understood as probability densities in the sense that the integral over the whole domain  equals $1$.
I have for example always thought of $M_{ab}$ as a probability amplitude, so $\overline{\left|M_{ab}\right|^2}$ would be a probability density. But in which variables? It can't be the momenta of the particles or else the integral over the full phase-space would always be $1$.
Also $f_i(x_1,Q^2)$ is said to give the probabilty of finding parton $i$ with momentum fraction $x_1$ inside the proton at scale $Q$, but does that mean that
$$
\intop_0^1 \mathrm{d}x_1 f_i(x_1,Q^2) = 1?
$$
Finally people usually say that $\sigma$ represents the probabilty of a process taking place, but since it has a dimension ($\text{GeV}^{-2}$), this can't be true in the mathematical sense. 
Can anyone help me connect the dots here?
 A: None of it is a probability density in that sense. As you've noticed, the end result is dimensionful, so it is not a probability either.
The cross-section is proportional to probability for any given process. To get something more like a probability, you have to multiply by a quantity called the "luminosity." The luminosity effectively lumps together all the "collider" details into one quantity, as opposed to the "physics" details that are taken care of in the cross section. 
Even this doesn't quite give you a probability- it gives you an expected event rate or number of events in a given interval. Of course, if the time interval is sufficiently short that the expected number of events is much less than one, you can treat this effectively as a probability.
As for $\intop_0^1 \mathrm{d}x_1 f_i(x_1,Q^2) \stackrel{?}{=} 1$, almost. The parton distribution functions are weighted by the expected number of each parton in the nucleon. So, for instance, the function for an up quark in a proton would integrate to more than the function for the down quark.
