# Non-measurable sets and sigma-algebra definition

I´m starting to study about measure theory, but I have problems regarding the definition of measure space.

In my class we saw that there exists sets that are not measurable(Vitali sets in $\mathbb R$) and in order to avoid this problem we define a sigma-algebra on a set $X$ and we call the elements of $\Sigma$ "measurable sets"

But the problem I have is that how can we guarantee that with this "definition" we cannot construct a non-measurable set. Is there a theorem that says that there can´t be non-measurable sets on a sigma-algebra over $X$ with that definition?

Or I just need to assume that we cannot find such sets over these circumstances?

It's not that we define a $\sigma$-algebra "in order to avoid [the] problem" that non-(Lebesgue-)measurable sets exist. Rather, we define exactly what we mean by "measurable set", form the collection $\cal S$ of all such sets, and then prove that this collection is a \sigma$-algebra. Having done so, it's a triviality that every set in$\cal S$is measurable. Of course, there are still non-measurable sets! They're just not members of$\cal S$. • Maybe this is a silly question but can we define the concepts of measurable set and sigma-algebra independently? I have seen in plenty of books that they first define a sigma-algebra and then they say that a set is measurable iff is an element of the sigma algebra – user128422 Nov 15 '15 at 3:02 • The notion of$\sigma$-algebra is defined independently. For example, the power set of any set is a$\sigma$-algebra. It's true that the notion is involved in the definition of the measurable sets: they're the least$\sigma$-algebra such that blah-blah. This is not at odds with the existence of non-measurable sets! What the Vitali set shows is that the definition of measurable set isn't just trivial: the$\sigma\$-algebra of measurable sets is not the entire powerset of the reals. – BrianO Nov 15 '15 at 3:31