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: $\mu^{*}(A) = inf(\sum_{i=1}^{+\infty}(b_{i}-a_{i}); A \subset \cup_{i=1}^{+\infty} [a_{i};b_{i}])$

  • $\begingroup$ Which of these is not true (or not clear) for the classical example you looked at? $\endgroup$ – Dap Oct 1 '17 at 15:12
  • $\begingroup$ 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. $\endgroup$ – Julien Oct 1 '17 at 15:23
  • $\begingroup$ Any set with zero exterior measure is a null set and hence Lebesgue measurable because Lebesgue measure is complete (=all null sets are measurable). $\endgroup$ – Dap Oct 1 '17 at 17:38
  • $\begingroup$ Indeed, I wrongly extrapolated elements of a proof by contradiction... I rephrased the question accordingly. Thanks $\endgroup$ – Julien Oct 1 '17 at 17:59

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.


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