Consider a $\mathfrak{B}(\mathbb{R})$ - Borel $\sigma$-algebra on $\mathbb{R}$. It's easy to see that $\#\mathfrak{B}(\mathbb{R}) = 2^{\aleph_0}$. And we now that $\mathfrak{B}(\mathbb{R})\subset \mathfrak{L}(\mathbb{R})$ - Lebesgue $\sigma$-algebra. I want to show that there is exist such subset of real line , which are Borel non measurable and Lebesgue measurable. But I don't know how?

Actually , I know about some examples. But maybe it possible to show easier than constructing a counterexample? For example I thought about finding some Lebesgue measurable sets , which are bigger than $\mathfrak{B}(\mathbb{R})$. And I thought about Cantor set.


1 Answer 1


The Cantor set is a Lebesgue null set, so all of its subsets are Lebesgue measurable sets. It has $2^{\mathfrak c}$ subsets. But there are only $\mathfrak c$ Borel sets. ( $\mathfrak c = 2^{\aleph_0}$)

Another method. An analytic set that is not Borel

  • $\begingroup$ Thanks , I thought about that! $\endgroup$
    – openspace
    Commented Mar 11, 2018 at 11:09
  • $\begingroup$ Bad link....... $\endgroup$ Commented Mar 11, 2018 at 14:20
  • $\begingroup$ Link fixed again $\endgroup$
    – GEdgar
    Commented Mar 11, 2018 at 16:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .