Euclidean real Radon representation under Countable Choice

First let $P$ be bounded and positive. For relatively open $G\subseteq U$ put $p(G)=\sup \{P(f):f\in C_c(U),0\le f\le 1,\mathrm{supp}f\subseteq G\}$. Then $0\le p(G)\le \mathrm{\parallel }P\mathrm{\parallel }$, $p(\mathrm{\varnothing })=0$, and $p$ is monotone. If a compact $K$ is covered by open sets $G_j$, finitely many suffice, and there are continuous nonnegative compactly supported functions $b_1,\dots ,b_N$, each supported in one of these sets, whose sum is positive on $K$. Indeed cover $K$ by finitely many balls with doubled closures compactly contained in the relevant $G_j$ and use distance cutoffs on these balls. On a neighborhood of $K$ normalize the $b_i$ by their sum. Thus any $f$ supported on $K$ with $0\le f\le 1$ decomposes into finitely many $f_i\ge 0$ with $\sum f_i=f$ and each $f_i\le 1$ compactly supported in its assigned $G_j$; combine terms assigned to the same $j$. Consequently $p({\bigcup }_j{G}_j)\le {\sum }_{j}p(G_j)$. For pairwise disjoint opens, adding finitely many admissible functions gives the reverse inequality for every finite partial sum, so $p$ is countably additive on disjoint open unions. These compact partitions use only finite choices.

Define ${\mu }^{\ast }(A)=\inf \{p(G):A\subseteq G\subseteq U,G\text{ open}\}$. For a sequence $(A_j)$ and $\epsilon >0$, [A1] selects open $G_j\supseteq A_j$ with $p(G_j)<{\mu }^{\ast }(A_j)+\epsilon 2^{-j}$; step 1.1 proves countable subadditivity. Zero and monotonicity are immediate, so ${\mu }^{\ast }$ is an outer measure, and ${\mu }^{\ast }(G)=p(G)$ for open $G$. If $\mathrm{dist}(A,B)>0$, they have disjoint open neighborhoods $V,W$ in $U$. Every open $G\supseteq A\cup B$ satisfies $p(G)\ge p(G\cap V)+p(G\cap W)\ge {\mu }^{\ast }(A)+{\mu }^{\ast }(B)$, by step 1.1. Taking the infimum and using subadditivity proves the metric condition. By [F2], restriction to Borel sets is a finite measure $\mu$ with $\mu (U)\le \mathrm{\parallel }P\mathrm{\parallel }$.

For an admissible $f$ in the definition of $p(G)$, let $K=\mathrm{supp}f$. Every open $V\supseteq K$ admits $f$, hence $\mu (K)={\inf }_{V\supseteq K}p(V)\ge P(f)$. Taking the supremum over $f$ proves $\mu (G)={\sup }_{K\subseteq G\text{ compact}}\mu (K)$. Outer regularity on every Borel set is the definition of ${\mu }^{\ast }$. Finiteness gives compact inner regularity on every Borel set as well: given Borel $A$ and $\epsilon >0$, choose compact $H\subseteq U$ with $\mu (U\setminus H)<\epsilon$ and open $V\supseteq U\setminus A$ with $\mu (V\setminus (U\setminus A))<\epsilon$. Then $H\setminus V\subseteq A$ is compact and misses at most $2\epsilon$ of $\mu (A)$. Thus $\mu$ is finite Radon, including regularity on all Borel sets.

If $0\le g\in C_c(U)$ and $g\le {\mathbf{1}}_K$ for compact $K$, the argument in step 3.1 gives $P(g)\le \mu (K)$. If instead $g\ge {\mathbf{1}}_K$, then for $0<\eta <1$, the open set $G_{\eta }=\{g>1-\eta \}$ contains $K$ and every admissible $f$ for $p(G_{\eta })$ satisfies $f\le g\mathrm{/}(1-\eta )$ on all of $U$. Positivity yields $\mu (K)\le p(G_{\eta })\le P(g)\mathrm{/}(1-\eta )$. Letting $\eta \downarrow 0$ gives $\mu (K)\le P(g)$.

Fix $0\le f\in C_c(U)$ and $\epsilon >0$, take $N$ with $N\epsilon \ge \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and put $f_j=\min (\epsilon ,(f-(j-1)\epsilon {)}_{+})$, $K_0=\mathrm{supp}f$, $K_j=\{f\ge j\epsilon \}$ for $1\le j\le N$. The sum of the $f_j$ is $f$, and $\epsilon {\mathbf{1}}_{K_j}\le f_j\le \epsilon {\mathbf{1}}_{K_{j-1}}$. Step 4.1 gives the same lower and upper bounds $\epsilon \mu (K_j)$ and $\epsilon \mu (K_{j-1})$ for $P(f_j)$ as for $\int f_{j}d\mu$. Summing and telescoping gives $\mathrm{\mid }P(f)-\int fd\mu \mathrm{\mid }\le \epsilon \mu (K_0)$. Hence equality holds for $f\ge 0$, then for every real $f\in C_c(U)$ by positive and negative parts. Uniform density [F3], boundedness of $P$, and finiteness of $\mu$ extend the identity to $C_0(U)$.

Apply [F1] to $L=P-Q$ and the preceding construction to $P,Q$, obtaining finite positive Radon measures $\alpha ,\beta$. Their difference $\sigma =\alpha -\beta$ represents $L$. Its total variation is dominated by $\alpha +\beta$, hence is regular by compact and open approximation with that finite positive Radon measure. Uniqueness follows from [F3]. All countable selections were the open near-minimizers in step 2.1; all other constructions were explicit or involved finitely many choices.