34
$\begingroup$

exact duplicate of Lebesgue measurable but not Borel measurable

BUT! can you please translate Miguel's answer and expand it with a formal proof? I'm totally stuck...

In short: Is there a Lebesgue measurable set that is not Borel measurable?

They are an order of magnitude apart so there should be plenty examples, but all I can find is "add a Lebesgue-zero measure set to a Borel measurable set such that it becomes non-Borel-measurable". But what kind of zero measure set fulfills such a property?

  • 0
    Why not just ask for clarification in a comment there?2012-05-04
  • 1
    See also [this thread](http://math.stackexchange.com/q/137277/5363)2012-05-04
  • 1
    @TheChaz because nobody would ever see that comment (except for people randomly looking for this...)2012-05-04
  • 1
    It would notify Miguel and Jonas...2012-05-04
  • 11
    Prove: there are $\mathfrak{c} = \#\mathbb{R}$ Borel measurable sets. Every subset of the Cantor set is Lebesgue measurable. There are $2^\mathfrak{c} = \#P(\mathbb{R})$ subsets of the Cantor set.2012-05-04
  • 0
    @t.b. why are there only $\#\mathbb{R}$ Borel measurable sets? Depending on the construction, isn't the Cantor set either open or closed and thus Borel measurable?2012-05-04
  • 0
    @example But the Borel measure is not complete, so you don't get the subsets of a null set for free.2012-05-04
  • 1
    Also relevant: http://math.stackexchange.com/questions/70880/cardinality-of-borel-sigma-algebra2012-05-04
  • 0
    I have a doubt on choosing a non-Borel set S,too. The proof seems to be done by the cantor function, though.2018-09-29

1 Answers 1

33

Let $\phi(x)$ be the Cantor function, which is non-decreasing continuous function on the unit interval $\mathcal{U}_{(0,1)}$. Define $\psi(x) = x + \phi(x)$, which is an increasing continuous function $\psi: [0,1] \to [0,2]$, and hence for every $y \in [0,2]$, there exists a unique $x \in [0,1]$, such that $y = \psi(x)$. Thus $\psi$ and $\psi^{-1}$ maps Borel sets into Borel sets.

Now choose a non Borel subset $S \subseteq \psi(C)$. Its preimage $\psi^{-1}(S)$ must be Lebesgue measurable, as a subset of Cantor set, but it is not Borel measurable, as a topological mapping of a non-Borel subset.

  • 2
    This is. ${}{}{}{}$2012-05-05
  • 5
    Why is it possible to choose a non Borel subset $S\subset \psi(C)$ ? $C$ is the Cantor set, right ?2017-03-13