Excursus: “Extremely Nonmeasurable Sets”

$\Delta_{F, \mathcal{B}^{k}}^{*}\left(\left\{ \boldsymbol {x}\right\} \right) = 0$ for all $\boldsymbol {x} \in \mathbb {R}^{k}$ iff $F \in \mathfrak {S}_{k}$ is continuous.

Notation and framework (Section 4.9): Let $\mathcal{A}_{k}$ denote the semiring consisting of $\varnothing$ and the bounded $k$-dimensional rsc rectangles $(\boldsymbol {a}, \boldsymbol {b}]$, and let $F \in \mathfrak {S}_{k}$. Then $\Delta_{F}: \mathcal{A}_{k} \rightarrow \mathbb {R}$ is a measure. Let $\Delta_{F}^{*}: 2^{\mathbb {R}^{k}} \rightarrow \overline{\mathbb {R}}$ denote the outer measure induced by $\Delta_{F}$. Then the restriction $\Delta_{F, \mathcal{M}\left(\Delta_{F}^{*}\right)}^{*}$ of $\Delta_{F}^{*}$ to $\mathcal{M}\left(\Delta_{F}^{*}\right)$ is a measure, and the restriction $\Delta_{F, \mathcal{B}^{k}}^{*}$ of $\Delta_{F}^{*}$ to $\mathscr {B}^{k}$ is a measure as well.

Construction of the set $B$: Let $\left\{ F_{\eta }: \eta \prec \omega \right\}$ denote a well ordering of the $\mathfrak {c}$ uncountable closed subsets of $\mathbb {R}^{k}$, where $\omega$ denotes the smallest uncountable ordinal (this well ordering exists by (AC) together with the fact that the collection of uncountable closed subsets of $\mathbb {R}^{k}$ has cardinal number $\mathfrak {c}$). By transfinite induction, for each $\eta \prec \omega$ choose distinct points $\boldsymbol {x}_{\eta }, \boldsymbol {y}_{\eta } \in F_{\eta } - A_{\eta }$, where $A_{\eta } = \left\{ \boldsymbol {x}_{\gamma }: \gamma \prec \eta \right\} \cup \left\{ \boldsymbol {y}_{\gamma }: \gamma \prec \eta \right\}$ (this is always possible since $\operatorname {card}\left(A_{\eta }\right) < \mathfrak {c} = \operatorname {card}\left(F_{\eta }\right)$). Let $B = \left\{ \boldsymbol {x}_{\eta }: \eta \prec \omega \right\}$.

Claim 1: For any $A \in \mathcal{M}\left(\Delta_{F}^{*}\right)$ (with $F \in \mathfrak {S}_{k}$), $\Delta_{F, \mathcal{M}\left(\Delta_{F}^{*}\right)}^{*}(A) = \sup \left\{ \Delta_{F, \mathcal{B}^{k}}^{*}(K): K \subseteq A, K \text{ compact}\right\}$.

Claim 2: The collection of open subsets of $\mathbb {R}^{k}$ has cardinal number $\mathfrak {c}$.

Claim 3: The collection of uncountable closed subsets of $\mathbb {R}^{k}$ has cardinal number $\mathfrak {c}$.

Claim 4: For any $\eta \prec \omega$, $B \cap F_{\eta } \neq \varnothing$ and $B^{c} \cap F_{\eta } \neq \varnothing$.

Claim 5: If $F \in \mathfrak {S}_{k}$ is continuous, then $\Delta_{F, \mathcal{B}^{k}}^{*}\left(\left\{ \boldsymbol {x}\right\} \right) = 0$ for every $\boldsymbol {x} \in \mathbb {R}^{k}$.

Let $F \in \mathfrak {S}$ be continuous. For each $\varepsilon >0$ there exists a nowhere dense perfect set $A \subseteq [0,1]$ such that $A \in \mathcal{M}\left(\Delta_{F}^{*}\right)$ and $\Delta_{F, \mathcal{M}\left(\Delta_{F}^{*}\right)}^{*}(A) > \Delta_{F, \mathcal{B}}^{*}([0,1]) - \varepsilon$. In the case of Lebesgue measure, this says [roughly] that there is a subset of $[0,1]$ with Lebesgue measure ["length"] arbitrarily close to 1, yet this set contains no open subinterval of $[0,1]$.

Let $F \in \mathfrak {S}$ be continuous. There exists a set $B \subseteq [0,1]$ with (i) $B \in \mathcal{M}\left(\Delta_{F}^{*}\right)$, (ii) $B$ is a countable union of nowhere dense sets, and (iii) $\Delta_{F, \mathcal{M}\left(\Delta_{F}^{*}\right)}^{*}(B) = \Delta_{F, \mathcal{B}}^{*}([0,1])$.

Let $F \in \mathfrak {S}$ be continuous. There exists a set $C \subseteq [0,1]$ such that (i) $C$ is the countable intersection of open subsets of $[0,1]$, (ii) $C$ is not a countable union of nowhere dense sets, and (iii) the $\Delta_{F, \mathcal{M}\left(\Delta_{F}^{*}\right)}^{*}$-measure of $C$ is 0.

The Continuum Hypothesis (CH) and the Axiom of Choice (AC) imply that there exists a set $S \subseteq \mathbb {R}$ with $\operatorname {card}(S) = \mathfrak {c}$, $S \in \mathcal{M}\left(\Delta_{F}^{*}\right)$, and $\Delta_{F, \mathcal{M}\left(\Delta_{F}^{*}\right)}^{*}(S) = 0$ for every continuous $F \in \mathfrak {S}$. Follow this outline.