I'll try to give an intuitive explanation:
So the probability is defined as a measure, i.e. a map that maps subsets of $\Omega$ into values in $[0,1]$ that satisfies certain axiom we desire.
Naturally people would want to ask "can we find a good map that properly assigns a value to every subset of $\Omega$ that follows the axiom we desire?"
The answer is yes for the finite $\Omega$, but is complicated, and is usually a NO for infinite $\Omega$. (EDIT: to be more precise, yes for finite or countable space, and no for uncountable space).
Since we cannot properly assign values for all subsets of $\Omega$ in lots of cases, we start to ask "what is the collection of subsets that could be properly assigned values to, by our map/measure?". Preferably, we hope such collection to be as big as possible so that we could properly measure as much subsets as possible - usually we end up expanding such collection into a sigma-algebra.
EDIT - an example
Let $\Omega$ be the set of all points on the unit circle, and the action on $\Omega$ by a group $G$ consisting of all rational rotations (rotations by angles which are rational multiples of $\pi$). Thus $G$ is countable while $\Omega$ is uncountable. Hence $\Omega$ breaks up into uncountably many orbits under $G$, with each orbit that corresponds to rational number $q$ consisting of points that are $q\pi$ angle away from each other. Using the axiom of choice, we could pick a single point from each orbit, obtaining an uncountable subset $ A\subset \Omega$ with the property that all of its translates by $G$ are disjoint from $A$ and from each other - to see this: if otherwise, then $\exists a \in (A+q_1\pi) \cap (A+q_2\pi)$. But since there are no two elements in $A$ from the same orbits, we have $\arg((a-q_1\pi),(a-q_2\pi)) \notin \mathbb Q\pi$, contradiction. (here I use +/- to denote rotation of a point for certain angle)
The set of those translates partitions the circle into a countable collection of disjoint sets, which are all pairwise congruent (by rational rotations). Then the set $A$ will be non-measurable for any rotation-invariant countably additive probability measure on $\Omega$, because: if $A$ has zero measure, countable additivity would imply that the whole circle has zero measure. If $A$ has positive measure, countable additivity would show that the circle has infinite measure.