# Basic question about the measure spaces behind examples of Lebesgue measurable sets that are not Borel

In trying to learn a few basic ideas on measure theory, I got to the mistaken idea that there Lebesgue measure was naturally associated with the Borel sigma algebra in the triple $$(\mathbb R, \mathcal B(\mathbb R), \lambda).$$ The opposite would manifestly not be true strictly, since there is the notorious $$(\Omega, \mathcal F, \mathbb P)$$ in probability.

But then I read a question that landed me on this online post on Lebesgue Measurable But Not Borel, and my question is whether the measure space would these examples belong to: what would be the sigma algebra and the measure?

In the completion mention in the comments, what other sets of measure zero are included besides the singletons (which are already Borel)?

In the quoted post, for instance one can read:

The Basic Idea: Our goal for today is to construct a Lebesgue measurable set which is not a Borel set. Such a set exists because the Lebesgue measure is the completion of the Borel measure. (The collection $$\mathcal B$$ of Borel sets is generated by the open sets, whereas the set of Lebesgue measurable sets $$\mathcal L$$ is generated by both the open sets and zero sets.) In short, $$B ⊂ L ,$$ where the containment is a proper one.

So some sets are Borel and of measure zero (singletons), but the completion would include other more bizarre zero-measure subsets?

• The sigma algebra would the Lebesgue sigma algebra - this is the Borel sigma algebra + all sets that have outer measure (from countable rectangle coverings) equal to zero. This known as the completion of the Borel sigma algebra. Apr 19, 2021 at 16:06
• math.stackexchange.com/questions/1168953/… Apr 19, 2021 at 16:07
• That is a deep question. Many things like this belong to the realm of descriptive set theory. In between Borel sets and Lebesgue sets there are also other slew of sets. Apr 19, 2021 at 21:14