3

While looking at the "classical" example of a non-measurable set, I am wondering: given $\lambda >0$, can we find a non Lebesgue-measurable set $E \subset \mathbb{R}$ verifying:

  • $\mu^{*}(E) = \lambda$ (i.e. $E$ has some predetermined "thickness", so to speak)

  • There is no Lebesgue measurable set $F$ contained within $E$ such that $\mu^{*}(F)>0$ (i.e. $F$ is "essentially" non-measurable, so to speak)

N.B.: $\mu^{*}$ denotes the usual Lebesgue outer measure: $\mu^{*}(A) = inf(\sum_{i=1}^{+\infty}(b_{i}-a_{i}); A \subset \cup_{i=1}^{+\infty} [a_{i};b_{i}])$

Sourav Ghosh
  • 12,997
Julien
  • 363
  • Which of these is not true (or not clear) for the classical example you looked at? – Dap Oct 01 '17 at 15:12
  • The classical example consists in picking one element per set $E_{i}={i+q; q \in \mathbb{Q}}$ where $i \in \mathbb{R-Q}$ such that the sets $E_{i}$ are mutually disjoint. One can prove that the non-measurable subset thus obtained has exterior measure = 0, i.e. is "meager" to some extent. – Julien Oct 01 '17 at 15:23
  • Any set with zero exterior measure is a null set and hence Lebesgue measurable because Lebesgue measure is complete (=all null sets are measurable). – Dap Oct 01 '17 at 17:38
  • Indeed, I wrongly extrapolated elements of a proof by contradiction... I rephrased the question accordingly. Thanks – Julien Oct 01 '17 at 17:59

2 Answers2

3

The second property is that the inner measure of $E$ $$\mu _{*}(E)=\sup\{\mu (S):S {\text{ measurable and }}S\subseteq E\}$$ is zero. "Meagre" is already a standard term meaning something quite different, see meagre set. (A Vitali set happens to be non-meagre in this sense, since a countable union of Vitali sets contains $[0,1]$ which has non-empty interior.)

The inner measure of a Vitali set is in fact zero, by essentially the same argument as the usual argument that it is non-measurable: suppose $\mu(S)>0$ where $S$ is a measurable subset of a Vitali set. By construction of the Vitali set, the sets $S+q$ are disjoint for distinct $q\in \mathbb Q$, so by countable additivity of Lebesgue measure (for measurable sets), $\mu(\bigcup_{q\in [0,1]\cap\mathbb Q} (S+q))=\infty$, which contradicts monotonicity because $\bigcup_{q\in [0,1]\cap\mathbb Q} (S+q)\subseteq [0,2]$.

More generally any non-measurable set $E\subseteq[0,1]$ can be used to give a non-measurable set with inner measure zero. Take a sequence of measurable sets $S_n\subseteq E$ with $\mu(S)\geq \mu_*(E)-1/n$, which exist by definition of the supremum. Defining $S=\bigcup_{n\geq 1} S_n$ we find that $\mu(S)=\mu_*(E)$. And $\mu_*(E\setminus S)=0$ - otherwise there would exist a measurable $S'\subseteq E\setminus S$ of positive measure, giving $\mu(S\cup S')=\mu(S)+\mu(S')>\mu_*(E)$, contradicting the definition of $\mu_*(E)$. So $E\setminus S$ is a non-measurable set with inner measure zero.

Regarding the first point, given any set $E\subseteq [0,1]$ with $\mu^*(E)>0$, we can scale it to have any desired outer measure $\lambda>0$; the set $$\tfrac{\lambda}{\mu^*(E)}E = \left\{\tfrac{\lambda}{\mu^*(E)} x \middle\vert x\in E\right\}\subseteq [0,\tfrac{\lambda}{\mu^*(E)}]$$ satisfies $$\mu^*\left(\tfrac{\lambda}{\mu^*(E)}E\right)=\tfrac{\lambda}{\mu^*(E)}\mu^*(E)=\lambda.$$

This follows directly from your definition of the outer measure by noting that $E$ is covered by $\bigcup_{i\geq 1}[a_i,b_i]$ if and only if $\tfrac{\lambda}{\mu^*(E)}E$ is covered by $\bigcup_{i\geq 1}[\tfrac{\lambda}{\mu^*(E)}a_i,\tfrac{\lambda}{\mu^*(E)}b_i]$. By a similar scaling argument the inner measure transforms in the same way: $\mu_*\left(\tfrac{\lambda}{\mu^*(E)}E\right)=\tfrac{\lambda}{\mu^*(E)}\mu_*\left(E\right)$. So scaling does not destroy the property of having inner measure zero.

Dap
  • 25,286
0

Given any $\lambda>0$ , let us consider an interval $I=(0, \lambda) $

Let $E=\mathcal{B}\cap I$

where $\mathcal{B}$ is a Bernstein set.

  1. $\mathcal{B}$ is a non measurable set.

  2. Any measurable subset of a Bernstein set must be a null set.

  3. $\mu^*(\mathcal{B}\cap I) =m^*(I) =\ell(I) =\lambda$

  1. $E$ is non measurable set.

  2. $\mu^*(E) =\lambda$

  3. $F\subset E\subset \mathcal{B}$ measurable implies $\mu^*(F) =0$ . Hence $E$ doesn't contain any measurable set of positive measure.

Sourav Ghosh
  • 12,997