Does every non null lebesgue measurable set contain a non-measurable subset?

Question

We know that the Lebesgue measure obtained via the usual Caratheodory extension is complete. As such, the subset of every null set is null.

Is it possible to prove that every non-null measurable subset $A\subseteq \mathbb R$ contains a non-measurable subset $B\subseteq A$?

I suppose it must be true, if it has non-empty interior then it must be true (just use an isometry to take a "copy" of a vitali set inside the interval).

See http://math.stackexchange.com/questions/1287212/non-measurable-subset-of-a-positive-measure-set. — Noah Schweber, Jan 01 '17 at 15:27
Does this answer your question? Non measurable subset of a positive measure set — Felix B., Sep 14 '22 at 14:24

score 3 · Accepted Answer · answered Jan 01 '17 at 15:24

3

You can just imitate the usual construction - let $ A $ be a measurable subset of $ \mathbb R $, and choose a bounded interval $ I $ such that $ A \cap I = X $ has nonzero measure. By translating and scaling, we may wlog assume $ I = [0, 1] $. Let $ V $ be a set of representatives in $ I $ for the cosets of $ \mathbb R / \mathbb Q $ intersecting $ X $ (obviously $ V $ may be chosen to be a subset of $ X $, and thus of $ A $), and note that choosing an enumeration $ q_k $ of the rationals in $ [-1, 1] $ and defining $ V_k = V + q_k $ gives that

$$ \mu(X) \leq \sum_{k=1}^{\infty} \mu(V_k) = \sum_{k=1}^{\infty} \mu(V) \leq 3 $$

which is a contradiction, since $ \mu(X) \neq 0 $.

answered Jan 01 '17 at 15:24

Ege Erdil

17,747

$3$?? ${}{}{}{}{}{}$ – Asinomás Jan 01 '17 at 15:32
$ 3 = \mu([-1, 2]) $. – Ege Erdil Jan 01 '17 at 15:36
So what is the contradiction? – Asinomás Jan 01 '17 at 15:41
Summing countably many copies of a real number can only yield $ 0 $ or $ \pm \infty $, it cannot yield a nonzero real number (as it does in this case). – Ege Erdil Jan 01 '17 at 15:43
thank you very much, I get it now. – Asinomás Jan 01 '17 at 15:49
It is not obvious - and in fact wrong - that $V$ may be chosen as a subset of $X$. If $A=\mathbb{R}\setminus\mathbb{Q}$, then you will not be able to find any representative of the $[0]$ equivalence class in $A$ and therefore also not in $X$ – Felix B. Sep 14 '22 at 14:17
@FelixB. I said "cosets of $ \mathbb R/\mathbb Q $ intersecting $ X $". In your example the coset $ \mathbb Q $ does not intersect $ A $. – Ege Erdil Sep 15 '22 at 15:09
fair enough - I think I overread the "intersecting X" bit. sorry – Felix B. Sep 16 '22 at 07:33

Does every non null lebesgue measurable set contain a non-measurable subset?

1 Answers1

Linked