Taken from the Motivation section of sigma-algebra article:
A measure on $X$ is a function that assigns a non-negative real number to subsets of $X$; this can be thought of as making precise a notion of "size" or "volume" for sets. We want the size of the union of disjoint sets to be the sum of their individual sizes, even for an infinite sequence of disjoint sets.
One would like to assign a size to every subset of $X$, but in many natural settings, this is not possible. For example the axiom of choice implies that when the size under consideration is the ordinary notion of length for subsets of the real line, then there exist sets for which no size exists, for example, the Vitali sets. For this reason, one considers instead a smaller collection of privileged subsets of $X$. These subsets will be called the measurable sets. They are closed under operations that one would expect for measurable sets, that is, the complement of a measurable set is a measurable set and the countable union of measurable sets is a measurable set. Non-empty collections of sets with these properties are called $\sigma$-algebras.
Does it explain the need for sigma-algebra in mathematics or measure theory? I just don't see how this description explains the motivation for sigma-algebra.
First, the article states that there are sets that are not Lebesgue-measurable, i.e. it's impossible to assign a Lebesgue measure to them (one satisfying the property that measure of a set is its length, which is very natural). Ok, fine. However, next it says:
For this reason, one considers instead a smaller collection of privileged subsets of $X$. These subsets will be called the measurable sets. They are closed under operations that one would expect for measurable sets, that is, the complement of a measurable set is a measurable set and the countable union of measurable sets is a measurable set.
What are the 'privileged subsets of X'? Borel sets for instance? Is it that because all Borel sets form the smallest sigma algebra (smallest meaning it has the least elements among all sigma algebras containing all Borel sets), and Borel sets have the nice property that they are made of open intervals, which means we can assign size to those sets that is equal to the length of that interval. That's obviously one of many possible measures we can use.
AFAIK, I can have a sigma-algebra that contains Vitali set. Everything can be a sigma-algebra, as long as it satisfies its 3 simple axioms. So if there is something you could add to clarify the quoted explanation, I'd be very grateful.
I've seen some amazing answers here om Math SE to related questions, like this one.