0

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.

Andrés E. Caicedo
  • 81,687
openspace
  • 6,766
  • 3
  • 19
  • 38

1 Answers1

2

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

GEdgar
  • 117,296