I've seen two kinds of demonstrations of Vitali's sets being not measurable (for example, answer number 2 here: https://math.stackexchange.com/questions/137949/the-construction-of-a-vitali-set, I adress it specifically because proof number one I haven't seen very often).

In both cases, the fact that the Lebesgue measure on $\mathbb{R}$ is traslation invariant is used to demonstrate that a Vitali set is not measurable:

"Given $V$ a Vitali set, define, for $k \in \mathbb{Q} \cap [-1,1]$, $V_k = \{v + k | v \in V\}$. Then, if $\mathcal{L}(A)$ is the Lebesgue measure for $A$, it is true that $\mathcal{L}(V) = \mathcal{L}(V_k) , \forall k \in \mathbb{Q} \cap [-1,1]$" And then this fact is used to demonstrate $V$ immesurability w.r.t Lebesgue's measure.

Since this statement does not hold for an arbitrary measure, my question is if it is or is not true that a Vitali set is not measurable for every measure $\mathcal{M}$ absolutely continuous with $\mathcal{L}$. To be more precise, $\mathcal{M}$ must be absolutely continuous with $\mathcal{L}$ when restricted to the Borel sets, $\mathcal{M}$ may be defined on a broader $\sigma$-algebra.

This can be easily answered if you don't ask $\mathcal{M}$ to be absolutely continuous w.r.t Lebesgue's measure:

Taking as probability space $(\mathbb{R},\mathcal{P(\mathbb{R})},\mathcal{I_x})$, where $\mathcal{P(\mathbb{R})}$ is $2^\mathbb{R}$ and $\mathcal{I_x}$ is the Dirac measure concentrated on $x$ (a pre-set, arbitrary point), then $V$ would be measurable with $\mathcal{I_x}(V)$ being 1 or 0 depending on whether $x \in V$ or $x \notin V$.

But i cannot get a general result when I impose the absolute continuity requirement.


2 Answers 2


user's answer, stated another way

Say the Vitali set $V \subseteq [0,1]$ has inner measure zero, outer measure 1. We may extend Lebesgue measure to a larger $\sigma$-algebra. Let $\mathscr F$ be the Lebesgue-measurable sets in $[0,1]$, and let the extension be $$ \mathscr G = \{ (A\cap V) \cup (B \setminus V)\;|\; A,B \in \mathscr F\} $$ Let $\mathcal L$ be Lebesgue measure, and define measure $\mu$ on $\mathscr G$ by $$ \mu\big((A\cap V) \cup (B \setminus V)\big) = \mathcal L(A), \qquad A,B \in \mathscr F . $$ Then:
$\mathscr G$ is a $\sigma$-algebra on $[0,1]$,
$\mu$ is $\sigma$-additive on $\mathscr G$,
$\mathscr G \supset \mathscr F$,
$V \in \mathscr G$,
$\mu(E)=\mathcal L(E)$ for $E \in \mathscr F$.


Let $\mu$ be the restriction of $\mathcal{L}$ on $V$ (note $\mathcal{L}(V)>0 )$, then $\mu$ is automatically an outer measure and absolutely continuous w.r.t $\mathcal{L}$, and $V$ is $\mu$ -measurable though not $\mathcal{L}$- measurable, however $\mu$ may not be Borel.

  • 1
    $\begingroup$ I'm sorry but I don't understand this: $\mathcal{L}(V) > 0$. Since $V$ is a Vitali set, and $\mathcal{L}$ the Lebesgue measure, then $\not\exists\mathcal{L}(V)$ $\endgroup$ Commented Apr 7, 2015 at 13:52
  • $\begingroup$ Lebesgue measure is defined for all sets, every set has its Lebesgue measure no matter it is measurable or not. $\endgroup$
    – user201043
    Commented Apr 7, 2015 at 13:59
  • $\begingroup$ He used $\mathcal L$ for what you would call Lebesgue outer measure. $\endgroup$ Commented Apr 7, 2015 at 14:10
  • $\begingroup$ @GeraldEdgar thanks for the clarification. Let me go to the answer, tough, I still do not fully catch it. $\endgroup$ Commented Apr 7, 2015 at 14:18

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