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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?

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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 $\cal S$.

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  • $\begingroup$ 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 $\endgroup$ – user128422 Nov 15 '15 at 3:02
  • $\begingroup$ 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. $\endgroup$ – BrianO Nov 15 '15 at 3:31

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