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From my lecture notes:

In most cases a cross section is represented in a differential way with respect to one or more observables. While it is possible to make a theoretical prediction for a distribution dσ/dO with respect to a continuous observable O it is not possible in general to measure the differential cross section at an exact value of that observable. This is because the probability of the observable being exactly that value is 0, it is only non-vanishing if we consider the probability for the observable to be in an interval containing the value. For this reason differential cross sections are represented as histograms.

I know that the differential cross section $\frac{d\sigma}{d\Omega}$ represents the proportion of particles scattered in an infinitesimally small angle $d\Omega$, but I really don't understand the significance of the statement: the probability of the observable being exactly that value is 0, it is only non-vanishing if we consider the probability for the observable to be in an interval containing the value.

What is reasoning behind that?

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it is not possible in general to measure the differential cross section at an exact value of that observable.

This sounds clear, $\frac{d\sigma}{d\Omega}$ at exactly $\Omega_0$ would imply the need of an infinitely small detector.

This is because the probability of the observable being exactly that value is 0, it is only non-vanishing ...

yes, $\sigma$ measured with such a detector would be vanishing...

if we consider the probability for the observable to be in an interval containing the value.

"the observable" stands here for $\Omega$. I read the message as - we have a detector of the finite size $\Delta \Omega$ and $\Omega_0$ is inside.

For this reason differential cross sections are represented as histograms.

The final meaning of the sentence is for me a bit like - once we measure $\frac{d\sigma}{d\Omega}$ at some point, it is rather somewhat averaged quantity $\frac{\Delta\sigma}{\Delta\Omega}$ at that point.

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