Outer Measure and the Caratheodory Extension Theorem — Examples

Exhausting the four subsets gives normalization and monotonicity. For countable subadditivity, an empty union has value $0$, while a nonempty union has value $1$ and contains a point lying in a nonempty cover member, so the covering sum is at least $1$; hence [F1] holds.

The sets $\mathrm{\varnothing }$ and $X$ satisfy [F2] identically. For either singleton $E$, the test $A=X$ gives $1={\mu }^{\ast }(X)$ but ${\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\setminus E)=1+1$, so neither singleton is measurable.

The measure axioms in [L1] give normalization and countable additivity on the full power set, hence monotonicity and countable subadditivity; thus counting measure is an outer measure.

For arbitrary $A,E\subseteq X$, the decomposition $A=(A\cap E)\bigsqcup (A\setminus E)$ and countable additivity in [L1] give ${\mathrm{\#}}_X(A)={\mathrm{\#}}_X(A\cap E)+{\mathrm{\#}}_X(A\setminus E)$, so [F1] holds for every $E$, including finite, infinite, and empty cases.

The function is the restriction of counting measure, so a disjoint sequence whose union lies in the finite-cofinite algebra has cardinality equal to the nonnegative sum of the member cardinalities, both for a finite union and for an infinite union; hence [F1] holds.

If $E$ is finite, its self-cover gives induced cost at most $\mathrm{\mid }E\mathrm{\mid }$, while [L1] applied to any cover of $E$ yields disjoint subordinate pieces whose total cardinality is $\mathrm{\mid }E\mathrm{\mid }$, so every cover costs at least $\mathrm{\mid }E\mathrm{\mid }$. If $E$ is infinite, a cover containing a cofinite member has infinite cost; a cover by finite members with finite total cardinality has finite union and cannot cover $E$, so every cover has infinite cost.

Step 1.2 gives the value $\mathrm{\mid }E\mathrm{\mid }$ for finite $E$ and $+\mathrm{\infty }$ for infinite $E$, exactly the counting outer measure of [L2].

By [L2], $d_{\mathbb{R}}(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is the usual metric on $\mathbb{R}$.

Positively separated nonempty sets are disjoint, so additivity of counting measure on disjoint sets in [L1] gives the equality required by [F1]; if either set is empty, both sides agree by the zero value. Thus counting outer measure is metric.

Applying [L3] gives Borel measurability, while [L1] gives the stronger conclusion that every subset of $\mathbb{R}$ is Carathéodory measurable.

Normalization and monotonicity follow by exhausting subset sizes. For subadditivity, a nonempty proper union has value $1$ and some member has value at least $1$; a union equal to $X$ either has an $X$ member of cost $2$ or requires at least two nonempty proper members of total cost at least $2$. Thus [F1] holds.

Every $E$ passes the whole-space equation: for nonempty proper $E$, the two pieces have value $1$ and sum to $2={\mu }^{\ast }(X)$. But choose $x\in E$ and $y\mathrm{\notin }E$ and test [F2] with $A=\{x,y\}$; then ${\mu }^{\ast }(A)=1$ while the two singleton pieces sum to $2$. Thus every nonempty proper set fails, and only $\mathrm{\varnothing },X$ are measurable.

The finite-cofinite family is an algebra. For a disjoint pair, two cofinite members cannot occur; two finite members and their union all have value $0$, while if one member is cofinite then both it and the union have value $+\mathrm{\infty }$, so ${\mu }_0$ is finitely additive.

The disjoint singleton sequence has union $\mathbb{N}\in {\mathcal{A}}_0$, but ${\mu }_0(\mathbb{N})=+\mathrm{\infty }$ and ${\sum }_{k=0}^{\mathrm{\infty }}{\mu }_0(\{k\})=0$, violating the countable-additivity clause of [F1].

The interval family is an algebra containing all finite half-open intervals, so [L2] gives $\mathcal{B}(\mathbb{R})\subseteq \sigma ({\mathcal{A}}_0)$. Every extended-endpoint half-open interval is Borel, and finite unions of Borel sets are Borel, so ${\mathcal{A}}_0\subseteq \mathcal{B}(\mathbb{R})$ and the reverse inclusion follows. The zero-or-infinity function satisfies [F1] because any nonempty disjoint union has a nonempty term. Every nonempty algebra member contains an interval component and an infinite midpoint-bisection sequence, so only $\mathrm{\varnothing }$ has finite premeasure and no sigma-finite cover exists.

Counting measure from [L1] restricts to a Borel extension because every nonempty algebra member is infinite. The Borel function $\nu (B)=0$ for $B=\mathrm{\varnothing }$ and $+\mathrm{\infty }$ otherwise is also countably additive by the same empty-versus-nonempty argument, and it too extends ${\mu }_0$.

For every $x\in \mathbb{R}$, counting measure gives $\mathrm{\#}(\{x\})=1$ while $\nu (\{x\})=+\mathrm{\infty }$, so the extensions are distinct.

Exhausting the four subsets proves normalization and monotonicity. For subadditivity, a union equal to a singleton has a singleton member of total cost at least $1$, while a union equal to $X$ either has an $X$ member of cost $3\mathrm{/}2$ or contains both singleton contributions of total cost $2\ge 3\mathrm{/}2$; hence [F1] holds.

For either singleton $E$, the test $A=X$ in [F2] would require $3\mathrm{/}2=1+1$, so neither singleton is measurable; $\mathrm{\varnothing }$ and $X$ are measurable directly.

The only measurable superset of a singleton is $X$, whose outer measure $3\mathrm{/}2$ is larger than the singleton value $1$; therefore [F3] gives no measurable hull for either singleton, and the outer measure is not regular.