Proof of The Radon Measure of a Positive Linear Functional on $C_c(X)$: the Outer Measure, Compact Sets, Borel Measurability and Regularity

Throughout, sums, scalar multiples and products of elements of $C_{c}(X)$ lie in $C_{c}(X)$, with the support inclusions of Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (ii), and every $f\in C_{c}(X)$ vanishes off $\operatorname{supp}f$ by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (iii). For an open set $V\subseteq X$ we write $f\prec V$ if $f\in C_{c}(X)$, $0\le f\le1$ and $\operatorname{supp}f\subseteq V$; thus $\mu(V)=\sup\{\Lambda f:f\prec V\}$.

Step 1 (monotonicity of $\Lambda$). If $f,g\in C_{c}(X)$ and $f\le g$, then $g-f\in C_{c}(X)$ and $g-f\ge0$, so by linearity and positivity

$$\Lambda g-\Lambda f=\Lambda(g-f)\ge0 .$$

Also $\Lambda0=\Lambda(0\cdot0)=0\cdot\Lambda0=0$.

Step 2 (elementary properties of $\mu$ and $\mu^{*}$). The constant function $0$ satisfies $0\prec V$ for every open $V$, its support being empty by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (i); so $\Lambda0=0$ lies in the set defining $\mu(V)$, and $\mu(V)\in[0,\infty]$. If $V\subseteq W$ are open, every $f\prec V$ satisfies $f\prec W$, so $\mu(V)\le\mu(W)$. If $f\prec\emptyset$ then $\operatorname{supp}f=\emptyset$ and $f=0$; hence $\mu(\emptyset)=0$. Since $X$ is open, the set defining $\mu^{*}(E)$ is a nonempty subset of $[0,\infty]$, so $\mu^{*}(E)\in[0,\infty]$ for every $E\subseteq X$, and $\mu^{*}(\emptyset)\le\mu(\emptyset)=0$. If $E\subseteq F\subseteq X$, every open set containing $F$ contains $E$, so $\mu^{*}(E)\le\mu^{*}(F)$. Finally, for open $V$ the set $V$ itself is admitted in the infimum defining $\mu^{*}(V)$, and $\mu(V)\le\mu(W)$ for every open $W\supseteq V$; hence $\mu^{*}(V)=\mu(V)$, which is the second assertion of (i).

Step 3 (finite subadditivity on open sets). Let $n\in\mathbb{N}$, let $V_{1},\dots,V_{n}\subseteq X$ be open and $V=V_{1}\cup\dots\cup V_{n}$; we show $\mu(V)\le\sum_{i=1}^{n}\mu(V_{i})$. Let $g\prec V$. Its support $K=\operatorname{supp}g$ is compact and contained in $V$, so Thickenings, Neighbourhoods with Compact Closure, Urysohn Functions and Partitions of Unity in a Locally Compact Metric Space (iv) gives $h_{1},\dots,h_{n}\in C_{c}(X)$ with $0\le h_{i}\le1$, $\operatorname{supp}h_{i}\subseteq V_{i}$, $\sum_{i}h_{i}\le1$ on $X$ and $\sum_{i}h_{i}=1$ on $K$. Each product $h_{i}g$ satisfies $0\le h_{i}g\le1$ and $\operatorname{supp}(h_{i}g)\subseteq\operatorname{supp}h_{i}\subseteq V_{i}$, so $h_{i}g\prec V_{i}$; and $g=\sum_{i=1}^{n}h_{i}g$, because $\sum_{i}h_{i}(x)=1$ for $x\in K$ while $g(x)=0$ for $x\notin K$. By linearity,

$$\Lambda g=\sum_{i=1}^{n}\Lambda(h_{i}g)\le\sum_{i=1}^{n}\mu(V_{i}),$$

and taking the supremum over $g\prec V$ gives the claim.

Step 4 (countable subadditivity). Let $(A_{m})_{m\in\mathbb{N}}$ be a sequence of subsets of $X$ and $A=\bigcup_{m}A_{m}$. If $\sum_{m}\mu^{*}(A_{m})=\infty$ there is nothing to prove, so let every $\mu^{*}(A_{m})$ be real and the sum finite, and let $\varepsilon>0$. By the definition of $\mu^{*}(A_{m})$ as an infimum there is, for each $m$, an open set $V_{m}\supseteq A_{m}$ with $\mu(V_{m})<\mu^{*}(A_{m})+2^{-m}\varepsilon$; we fix such a sequence $(V_{m})$ (by countable choice, which follows from the setting's standing hypothesis of dependent choice). The set $V=\bigcup_{m}V_{m}$ is open and contains $A$. Let $g\prec V$. The open sets $U_{N}=V_{1}\cup\dots\cup V_{N}$ ($N\in\mathbb{N}$) increase and their union $V$ contains the compact set $\operatorname{supp}g$, so $\operatorname{supp}g\subseteq U_{N}$ for some $N$ by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (iii). Then $g\prec U_{N}$, and by Step 3

$$\Lambda g\le\mu(U_{N})\le\sum_{m=1}^{N}\mu(V_{m})\le\sum_{m=1}^{N}\bigl(\mu^{*}(A_{m})+2^{-m}\varepsilon\bigr)\le\sum_{m\in\mathbb{N}}\mu^{*}(A_{m})+\varepsilon .$$

Taking the supremum over $g\prec V$ and using Step 2, $\mu^{*}(A)\le\mu(V)\le\sum_{m}\mu^{*}(A_{m})+\varepsilon$; as $\varepsilon>0$ was arbitrary, $\mu^{*}(A)\le\sum_{m}\mu^{*}(A_{m})$. Together with Step 2 this shows that $\mu^{*}$ is an outer measure on $X$, completing the proof of (i).

Step 5 (compact sets: the upper bound). Let $K\subseteq X$ be compact and put $I(K)=\inf\{\Lambda f:f\in C_{c}(X),\ f\ge\mathbf{1}_{K}\}$. Let $f\in C_{c}(X)$ with $f\ge\mathbf{1}_{K}$, so in particular $f\ge0$, and let $0<\alpha<1$. The set $V_{\alpha}=\{x\in X:f(x)>\alpha\}$ is open, as $f$ is continuous, and contains $K$. If $g\prec V_{\alpha}$ then $\alpha g\le f$: for $x\in V_{\alpha}$ we have $\alpha g(x)\le\alpha<f(x)$, and for $x\notin V_{\alpha}$ we have $x\notin\operatorname{supp}g$, so $\alpha g(x)=0\le f(x)$. By Step 1, $\alpha\Lambda g\le\Lambda f$ for every $g\prec V_{\alpha}$, hence

$$\mu^{*}(K)\le\mu(V_{\alpha})\le\alpha^{-1}\Lambda f .$$

As this holds for every $\alpha\in(0,1)$, $\mu^{*}(K)\le\Lambda f$, and taking the infimum over $f$ gives $\mu^{*}(K)\le I(K)$.

Step 6 (compact sets: the lower bound). Let $V\supseteq K$ be open. By Thickenings, Neighbourhoods with Compact Closure, Urysohn Functions and Partitions of Unity in a Locally Compact Metric Space (iii) there is $f\in C_{c}(X)$ with $0\le f\le1$, $f=1$ on $K$ and $\operatorname{supp}f\subseteq V$. Then $f\ge\mathbf{1}_{K}$ and $f\prec V$, so $I(K)\le\Lambda f\le\mu(V)$; taking the infimum over $V$ gives $I(K)\le\mu^{*}(K)$. Thus $\mu^{*}(K)=I(K)$, and taking $V=X$ shows $\mu^{*}(K)\le\Lambda f<\infty$ for the function $f$ so obtained. This proves (ii).

Step 7 (open sets are $\mu^{*}$-measurable). Let $U\subseteq X$ be open. By Carathéodory's Theorem: the Measurable Sets of an Outer Measure Form a σ-Algebra on Which It Is a Measure (i) it suffices to show $\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)\le\mu^{*}(A)$ for every $A\subseteq X$, and for this that $\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)\le\mu(V)$ for every open $V\supseteq A$. This is clear if $\mu(V)=\infty$, so let $\mu(V)<\infty$ and $\varepsilon>0$. Since $\mu(V\cap U)\le\mu(V)<\infty$, there is $f\prec V\cap U$ with $\Lambda f>\mu(V\cap U)-\varepsilon$. Its support $D=\operatorname{supp}f$ is compact, hence closed, so $V\setminus D$ is open with $\mu(V\setminus D)\le\mu(V)<\infty$, and there is $g\prec V\setminus D$ with $\Lambda g>\mu(V\setminus D)-\varepsilon$. Now $f+g\prec V$: its support lies in $\operatorname{supp}f\cup\operatorname{supp}g\subseteq V$, and $0\le f+g\le1$ because $f$ vanishes off $D$ while $g$ vanishes on $D$. Since $A\cap U\subseteq V\cap U$ and $A\setminus U\subseteq V\setminus D$ (as $D\subseteq U$), Step 2 gives

$$\mu(V)\ge\Lambda f+\Lambda g>\mu(V\cap U)+\mu(V\setminus D)-2\varepsilon\ge\mu^{*}(A\cap U)+\mu^{*}(A\setminus U)-2\varepsilon .$$

As $\varepsilon>0$ was arbitrary, the claim follows.

Step 8 (the Borel measure). By Carathéodory's Theorem: the Measurable Sets of an Outer Measure Form a σ-Algebra on Which It Is a Measure (ii) the set $\mathcal{M}(\mu^{*})$ of $\mu^{*}$-measurable sets is a $\sigma$-algebra on $X$, and by Step 7 it contains every open set; by The σ-Algebra Generated by a Set of Subsets (ii) it therefore contains the $\sigma$-algebra $\mathcal{B}(X)$ generated by the open sets. By Carathéodory's Theorem: the Measurable Sets of an Outer Measure Form a σ-Algebra on Which It Is a Measure (iv), $\mu^{*}|_{\mathcal{M}(\mu^{*})}$ is a measure on $(X,\mathcal{M}(\mu^{*}))$. Its restriction $\bar{\mu}$ to the sub-$\sigma$-algebra $\mathcal{B}(X)$ vanishes at $\emptyset$ and is countably additive, since a disjoint sequence in $\mathcal{B}(X)$ is one in $\mathcal{M}(\mu^{*})$; so $\bar{\mu}$ is a Borel measure on $X$. This proves (iii).

Step 9 (finiteness on compact sets and outer regularity). A compact $K\subseteq X$ is closed, hence Borel, and $\bar{\mu}(K)=\mu^{*}(K)<\infty$ by (ii). For $B\in\mathcal{B}(X)$, the definition of $\mu^{*}$ and (i) give

$$\bar{\mu}(B)=\mu^{*}(B)=\inf\{\mu(V):V\supseteq B\text{ open}\}=\inf\{\bar{\mu}(V):V\supseteq B\text{ open}\},$$

since $\bar{\mu}(V)=\mu^{*}(V)=\mu(V)$ for open $V$.

Step 10 (inner regularity of open sets). Let $V\subseteq X$ be open. Every compact $K\subseteq V$ has $\bar{\mu}(K)\le\bar{\mu}(V)$. Conversely let $t$ be a real number with $t<\bar{\mu}(V)=\mu(V)$, and choose $f\prec V$ with $\Lambda f>t$. Then $K=\operatorname{supp}f$ is compact and $K\subseteq V$, and $f\prec W$ for every open $W\supseteq K$, so $\Lambda f\le\mu(W)$ for all such $W$; hence $\bar{\mu}(K)=\mu^{*}(K)\ge\Lambda f>t$. Therefore

$$\bar{\mu}(V)=\sup\{\bar{\mu}(K):K\subseteq V\text{ compact}\},$$

and with Step 9 this shows that $\bar{\mu}$ is a Radon measure, proving (iv).