Construct directly a subset in $\mathbb{R}^{2}$ that is not Lebesgue measurable. (Don't use the corresponding result in $\mathbb{R}$).

Since I could not use the result in $\mathbb{R}$ where we can find a "choice" set as a nonmeasurable set. It would require some new idea to construct one in $\mathbb{R}^{2}$. And I can't come up with it immediately.

  • $\begingroup$ It's almost surely going to require some form of the axiom of choice $\endgroup$
    – Aweygan
    Jul 25, 2016 at 5:34
  • 2
    $\begingroup$ I assume what is meant is that you cannot use the choice set in $\mathbb{R}$ to construct your choice set in $\mathbb{R}^2$. You would have to use the idea behind it, however. $\endgroup$ Jul 25, 2016 at 5:41
  • $\begingroup$ I don't really like the work "construct" when Choice is involved ... $\endgroup$ Jul 25, 2016 at 6:34
  • $\begingroup$ @HagenvonEitzen I concede your point. One certainly cannot use the word 'construct' in the sense of constructive mathematics when the axiom of choice is involved. Perhaps 'contrive' would be a better word choice. $\endgroup$ Jul 25, 2016 at 17:14

3 Answers 3


Here is how to contrive the set in plain English.

Start with the set of all points in the plane with rational coordinates, call it $S$. No translation of $S$ intersects $S$ unless it is $S$.

Let $C$ denote the set of all translations of $S$, including $S$ itself.

For each element $S^\prime$ of $C$ pick one point of $S^\prime$ in a unit square $M$ and let the set of all those points be your choice set $E$.

If $E$ is translated left, right, up or down by a rational amount, the translated set cannot intersect $E$.

Let $G$ denote the set of all translations of $E$ left, right, up or down by an amount equal to a rational number in the interval $[-1,1]$

Then $G$ is a countable collection of mutually exclusive sets which covers the unit square $M$ but whose union lies within a $3\times3$ square centered at $M$.

I assume you already know how this contradicts any assumption of $E$ having positive measure or measure $0$.


In fact, the result is the same in the concruction of Vitali set in $\Bbb R$.

Consider $\Bbb R^2/\Bbb Q^2$, construct the Vitali set $V\subset [0,1]^2$ by choosing only one point in each coset of $\Bbb Q^2$.

Now, check this

$[0,1]^2\subset \cup _{q\in \Bbb Q^2\cap [-1,1]^2} (V+q)\subset [-10,10]^2$,

then you can proof $1\leq \sum _{i=1}^{\infty}m(V)\leq 100$, which is abuse.

Remark: you can never construct a non-measurable set without Axiom of choice! Can one construct a non-measurable set without Axiom of choice?

Hope this help!


Def'ns: (D1).The cardinal ordinal $c$ is the cardinal of $R$, also of $R^2.$

(D2). $F$ is the set of all uncountable closed subsets of $R^2$.

Preparations: (P1). The cardinal of $F$ is $c$.

(P2). The cardinal of every member of $F$ is $c$.

We construct $S\subset R^2$ such that $\forall f\in F \;(f\cap S\ne \emptyset \ne f\backslash S)$ as follows:

Let $F=\{f_a: a< c \}.$ For $a<c$ let $S(a)$ and $T(a)$ be unequal members of $f_a\backslash (\cup_{b<a}\{S(b), T(b)\}).$ This is possible by (P2).

Let $S=\{S(a):a<c\}.$ For any $a<c$ we have $S(a)\in f_a\cap S$ and $T(a)\in f_a\backslash S.$

Let $m^i$ be Lebesgue inner measure on $R^2.$ Now any closed subset of $S$ is countable, and any closed subset of $R^2\backslash S$ is countable . So $S$ is not measurable because $m^i(S)=m^i(R^2\backslash S)=0.$


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