Non Measurable Sets and the Cost of Choice: Examples and Counterexamples

Two points $x,y\in [0,1]$ lie in the same class exactly when they differ by a rational, and [F1] says that $V$ contributes one and only one representative to each such class. Thus the pieces $(x+\mathbb{Q})\cap [0,1]$ are pairwise disjoint and each is hit once by $V$.

If $t\in [0,1]$, let $v\in V$ be the unique representative of its class. Then $t-v\in \mathbb{Q}$ and, because $0\le t,v\le 1$, also $-1\le t-v\le 1$; so $t\in V+q$ for some $q\in \mathbb{Q}\cap [-1,1]$. Conversely every $v+q$ with $v\in V$ and $q\in [-1,1]$ lies in $[-1,2]$.

Step [L3] gives $\psi (x)=x+1\mathrm{/}2$ for $x\in (1\mathrm{/}3,2\mathrm{/}3)$, so $\psi (1\mathrm{/}3)=5\mathrm{/}6$, $\psi (2\mathrm{/}3)=7\mathrm{/}6$, and $\psi [(1\mathrm{/}3,2\mathrm{/}3)]=(5\mathrm{/}6,7\mathrm{/}6)$. Also $\psi (0)=0$ and $\psi (1)=2$.

These computations sit inside the global picture from [L1] and [L2]: the map is a homeomorphism of the whole interval, and the image of the Cantor set itself is the compact measure-one set obtained by removing the translated gaps.

By [L1] choose a Vitali set $V\subseteq [0,1]$.

The set $V$ is a subset of $\mathbb{R}$ and is not Lebesgue measurable by [L2], so it refutes the claim.

By [L1] choose a Bernstein set $B$. If some nonempty open interval were disjoint from $B$, it would contain a nondegenerate closed subinterval, hence a nonempty perfect subset of $\mathbb{R}$, contradicting the Bernstein property. So $B$ meets every nonempty open interval and is dense in $\mathbb{R}$.

The given Axiom of Countable Choice supplies the hypothesis of [L2] and [L3]. Hence ${\lambda }_{\ast }(B)=0$ by [L2], and $B$ is not Lebesgue measurable by [L3].

Therefore $B$ is a dense subset of $\mathbb{R}$ of inner measure $0$ that is not Lebesgue measurable, so it refutes the claim.

Choose a Vitali set $V\subseteq [0,1]$ by [L1], and put $W:=[0,1]\setminus V$. Then $V$ and $W$ are disjoint and $V\cup W=[0,1]$, which is measurable by [L3].

The set $V$ is not measurable by [L2]. If $W$ were measurable, then $V=[0,1]\setminus W$ would also be measurable because $[0,1]$ is measurable, contradiction. So $W$ is not measurable either, and the pair $(V,W)$ refutes the claim.

By [L1] choose a Hamel basis $B$, a basis vector $b_{\star }\in B$, and the corresponding coefficient map ${\mathrm{\Lambda }}_{b_{\star }}:\mathbb{R}\to \mathbb{Q}$. Its kernel $W:=\{x\in \mathbb{R}:{\mathrm{\Lambda }}_{b_{\star }}(x)=0\}$ is a subgroup of $(\mathbb{R},+)$, and it is proper because ${\mathrm{\Lambda }}_{b_{\star }}(b_{\star })=1$. Every real is of the form $q{b}_{\star }+w$ with $q\in \mathbb{Q}$ and $w\in W$.

If $W$ were measurable with positive measure, [L2] would force $W=\mathbb{R}$, contradicting step 1.1.

If $W$ were measurable with measure $0$, then every translate $q{b}_{\star }+W$ would also be measurable with measure $0$ by [L5], and step 1.1 says these countably many translates cover $\mathbb{R}$. Their union would be null by [L4] and [L6], yet it contains the measurable interval $[0,1]$ of measure $1$ by [L3], a contradiction.

So the proper subgroup $W$ is not Lebesgue measurable, and it refutes the claim.

By [L1] choose a Vitali set $V\subseteq [0,1]$.

The set $V$ is a subset of $\mathbb{R}$ and is not Lebesgue measurable by [L2], so the universal claim is false.

By [L1] choose a measurable subset $E\subseteq \mathbb{R}$ and a continuous map $f:E\to \mathbb{R}$ whose image is not measurable.

This single witness $(E,f)$ refutes the universal claim.

By [L1] choose a measurable subset $E\subseteq \mathbb{R}$ and a continuous map $g$ whose preimage of $E$ is not measurable.

This witness refutes the universal claim.

The set ${\mathbb{Q}}_{\mathbb{R}}$ is dense by [L1], and it has outer measure $0$ because it is countably infinite and therefore Lebesgue null by [L2].

Let $I\subseteq \mathbb{R}$ be a bounded nondegenerate interval. Then $I\setminus {\mathbb{Q}}_{\mathbb{R}}$ is measurable, because $I$ is measurable by [L5] and ${\mathbb{Q}}_{\mathbb{R}}$ has measure $0$ by step 1.1. Also $\lambda (I\setminus {\mathbb{Q}}_{\mathbb{R}})=\lambda (I)$, so [L3] gives ${\lambda }_{\ast }(I\setminus {\mathbb{Q}}_{\mathbb{R}})=\lambda (I)$.

Every nonempty open interval contains both a rational and an irrational by [L1]. So ${\mathbb{Q}}_{\mathbb{R}}$ is dense, and $\mathbb{R}\setminus {\mathbb{Q}}_{\mathbb{R}}$ is also dense. Step 2.1 shows that inside every bounded nondegenerate interval the latter has full inner measure. These two dense sets therefore coexist, and the statement is false.

By [L1], the stated consistency hypothesis supplies a model of ZF + DC in which every set of reals is Lebesgue measurable.

If ZF + DC proved that a nonmeasurable subset of $\mathbb{R}$ exists, every model of ZF + DC would contain one. The model from step 1.1 contains none, so the asserted theorem of ZF + DC is false relative to the stated consistency hypothesis. This is precisely the consistency-strength obstruction recorded in [L2].