A compact-metric probability representation using countable choice

For an open set $U\subseteq K$ define $t_U(x)={\inf }_{y\in K\setminus U}d(x,y)$ when $K\setminus U\mathrm{\ne }\mathrm{\varnothing }$, and $t_K=1$. The triangle inequality, followed by the infimum, shows that $t_U$ is 1-Lipschitz in the first case; it is positive at every point of $U$ because that point has a ball contained in $U$, and it vanishes outside $U$. If a nonempty closed $F$ lies in $U$, the sets $\{t_U>1\mathrm{/}m\}$ for positive integers $m$ cover $F$. Adjoining $K\setminus F$ and using [F1] gives a $\delta >0$ with $t_U>\delta$ on $F$. Consequently $h=\min (1,\max (0,2t_U\mathrm{/}\delta -1))$ equals 1 on $F$, takes values in $[0,1]$, and has support contained in $\{t_U\ge \delta \mathrm{/}2\}\subseteq U$. Here support means the closure in $K$ of the nonzero set, which is compact by [F2]. For $F=\mathrm{\varnothing }$ use $h=0$. Any compact subset of a metric space is closed: the empty subset is closed, and for a nonempty compact subset, if $x$ is outside it, its balls centered at $y$ of radii $d(x,y)\mathrm{/}3$ have a finite subcover; the minimum of these finitely many positive radii gives a ball at $x$ missing the subset. Thus these cutoffs apply also to every compact $F$.

Write $f\prec U$ for continuous $f$ with $0\le f\le 1$ and support contained in $U$. Positivity and linearity give monotonicity of $\mathrm{\Lambda }$ by applying positivity to differences, and $\mathrm{\mid }\mathrm{\Lambda }(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ by comparison with constants. The norm is finite: the open sets $\{\mathrm{\mid }f\mathrm{\mid }<m\}$ cover $K$ and a finite subcover bounds $f$. Define $\rho (U)={\sup }_{f\prec U}\mathrm{\Lambda }(f)$ and ${\mu }^{\ast }(E)={\inf }_{E\subseteq U\text{ open}}\rho (U)$. The zero function and the open set $K$ make both families nonempty; $0\le \rho (U)\le 1$, $\rho (\mathrm{\varnothing })=0$, and $\rho (K)=1$. Monotonicity immediately implies ${\mu }^{\ast }(U)=\rho (U)$ for open $U$, and monotonicity and zero empty-set value for ${\mu }^{\ast }$.

For a closed nonempty $F\subseteq {\bigcup }_{i=1}^s{U}_i$ with finitely many open $U_i$, put $t_i=t_{U_i}$. Compactness, applied to the sets $\{{\max }_i{t}_i>1\mathrm{/}m\}$ and $K\setminus F$, gives $\delta >0$ with ${\max }_i{t}_i>\delta$ on $F$. Set $w_i=(t_i-\delta \mathrm{/}2{)}_{+}$ and $w={\sum }_i{w}_i$. Define ${\phi }_i=w_i\min (2\mathrm{/}\delta ,1\mathrm{/}w)$ where $w>0$, and zero where $w=0$. Near a zero of $w$ the formula is $(2\mathrm{/}\delta )w_i$, proving continuity there. Each ${\phi }_i$ has support in $\{t_i\ge \delta \mathrm{/}2\}\subseteq U_i$, is nonnegative and at most 1, and ${\sum }_i{\phi }_i=\min (2w\mathrm{/}\delta ,1)=1$ on the open neighborhood $\{w>\delta \mathrm{/}2\}$ of $F$. For empty $F$ use all zero functions. This constructs a finite subordinate partition without any selection principle.

For open $U={\bigcup }_{n\ge 1}{U}_n$ and $f\prec U$, its closed support $F$ is covered by finitely many distinct $U_{n_i}$ by [F1] after adjoining $K\setminus F$. If $F$ is empty, $\mathrm{\Lambda }(f)=0$. Otherwise the partition in step 2.1 gives $f={\sum }_{i}f{\phi }_i$, with $f{\phi }_i\prec U_{n_i}$. Hence $\mathrm{\Lambda }(f)\le {\sum }_i\rho (U_{n_i})\le {\sum }_n\rho (U_n)$; taking the supremum proves open-set subadditivity. Given arbitrary $E_n$ and $\epsilon >0$, countable choice now selects simultaneously open $U_n\supseteq E_n$ with $\rho (U_n)<{\mu }^{\ast }(E_n)+\epsilon 2^{-n}$. These are a specified countable family of nonempty sets of admissible opens. Thus ${\mu }^{\ast }({\bigcup }_n{E}_n)\le \rho ({\bigcup }_n{U}_n)\le {\sum }_n{\mu }^{\ast }(E_n)+\epsilon$. Letting $\epsilon$ decrease to zero proves that ${\mu }^{\ast }$ is an outer measure; if the sum on the right is infinite the inequality is immediate and no selection is needed.

For open $G,V$ and $f\prec V\cap G$, with support $F$. For any $g\prec V\setminus F$, the supports are disjoint, so $f+g\prec V$. Therefore $\rho (V)\ge \mathrm{\Lambda }(f)+\rho (V\setminus F)\ge \mathrm{\Lambda }(f)+{\mu }^{\ast }(V\setminus G)$. Taking the supremum over $f$ gives $\rho (V)\ge {\mu }^{\ast }(V\cap G)+{\mu }^{\ast }(V\setminus G)$. For arbitrary $E\subseteq V$, monotonicity replaces $V$ in the two terms by $E$; taking the infimum over open $V\supseteq E$ proves ${\mu }^{\ast }(E)\ge {\mu }^{\ast }(E\cap G)+{\mu }^{\ast }(E\setminus G)$. The opposite inequality is outer subadditivity. Thus every open $G$ is Carathéodory measurable, and [F3] gives a Borel measure $\mu$. It satisfies $\mu (K)=\rho (K)=1$ and is outer regular by its defining infimum and $\mu (U)=\rho (U)$.

For compact $F\subseteq K$, one has $\mu (F)=\inf \{\mathrm{\Lambda }(h):h\in C(K,\mathbb{R}),h\ge {\mathbf{1}}_F\}$. Indeed such $h$ is nonnegative everywhere, and for $0<\epsilon <1$ the open set $U=\{h>1-\epsilon \}$ contains $F$. Every $g\prec U$ satisfies $g\le h\mathrm{/}(1-\epsilon )$, whence $\mu (F)\le \rho (U)\le \mathrm{\Lambda }(h)\mathrm{/}(1-\epsilon )$, and then $\mu (F)\le \mathrm{\Lambda }(h)$. Conversely for each open $U\supseteq F$, step 1.1 supplies $h\prec U$ with $h=1$ on $F$; hence the displayed infimum is at most $\rho (U)$. Infimizing over $U$ gives the reverse inequality. Empty $F$ has both sides zero using $h=0$. If $f\prec U$ and $F=\mathrm{supp}f$, then $f\le h$ for every $h\ge {\mathbf{1}}_F$, so the compact formula gives $\mathrm{\Lambda }(f)\le \mu (F)$. Taking the supremum over $f$ proves $\mu (U)\le {\sup }_{F\subseteq U\text{ compact}}\mu (F)$; monotonicity proves equality. This is the required inner regularity on opens.

Let $0\le f\in C(K,\mathbb{R})$ and $\epsilon >0$. Choose a positive integer $N$ with $f\le N\epsilon$, put $F_0=K$, $F_n=\{f\ge n\epsilon \}$ for $1\le n\le N$, and $f_n=\min (\epsilon ,(f-(n-1)\epsilon {)}_{+})$. These are continuous, the $F_n$ are compact by [F2], $f={\sum }_{n=1}^N{f}_n$, and $\epsilon {\mathbf{1}}_{F_n}\le f_n\le \epsilon {\mathbf{1}}_{F_{n-1}}$. The compact formula gives the lower bound $\epsilon \mu (F_n)\le \mathrm{\Lambda }(f_n)$; comparison with every continuous majorant of ${\mathbf{1}}_{F_{n-1}}$ gives the upper bound $\mathrm{\Lambda }(f_n)\le \epsilon \mu (F_{n-1})$. By [F4] the same bounds hold for $\int f_{n}d\mu$. All these integrals are finite since $f_n\le \epsilon$ and $\mu (K)=1$. Summing and applying [F5] locates both $\mathrm{\Lambda }(f)$ and $\int fd\mu$ in the same interval of length $\epsilon (\mu (F_0)-\mu (F_N))\le \epsilon$. Since $\epsilon$ is arbitrary they are equal. Finally $f=f^{+}-f^{-}$ proves equality for every real continuous $f$, using positivity, linearity and [F5]. Countable choice was spent only on the admissible open supersets in step 3.1; the cutoffs and partitions are explicit metric formulas.