Radon Measures and the Riesz Markov Kakutani Theorem

Since supports of $f$ and $g$ are compact, $g-f\in C_c(X)$; moreover $g-f\ge 0$.

Positivity gives $0\le \mathrm{\Lambda }(g-f)=\mathrm{\Lambda }(g)-\mathrm{\Lambda }(f)$, which is the claim.

Choose $V$ as in [L1]. The compact Hausdorff space $\bar{V}$ is normal; apply [L2] there to $K$ and $\bar{V}\setminus V$, obtaining $h=1$ on $K$ and $h=0$ on $\bar{V}\setminus V$.

Extend $h$ by $0$ off $\bar{V}$. Continuity of $h$ on $\bar{V}$ and its vanishing on the boundary $\bar{V}\setminus V$ make the extension continuous; it is supported in $\bar{V}\subseteq U$, is $1$ on $K$, and belongs to $C_c(X)$.

Consider all triples $(V,W,i)$ with $V,W$ open, $1\le i\le n$, [given, L1] and $\bar{V}\subseteq W\subseteq \bar{W}\subseteq U_i,$ where the displayed closures are compact. Local compactness and the Hausdorff property show that the sets $V$ occurring in these triples cover $K$. Compactness therefore gives finitely many triples $(V_j,W_j,i_j)$ whose $V_j$ cover $K$. Apply [L1] to $\overline{V_j}\subseteq W_j$ to obtain $h_j\in C_c(X)$ with $0\le h_j\le 1$, $h_j=1$ on $\overline{V_j}$, and $h_j=0$ off $W_j$. Consequently $\mathrm{supp}{h}_j\subseteq \overline{W_j}\subseteq U_{i_j}$.

Put $h={\sum }_j{h}_j$. Then $h\ge 1$ on $K$, so $\{h>1\mathrm{/}2\}$ is an [step 1.1, L1, choose] open neighbourhood of $K$. Choose open sets $O,W$ with $K\subseteq O\subseteq \bar{O}\subseteq W\subseteq \bar{W}\subseteq \{h>1\mathrm{/}2\},$ where the displayed closures are compact. Apply [L1] to $\bar{O}\subseteq W$ to obtain $g\in C_c(X)$ with $0\le g\le 1$, $g=1$ on $\bar{O}$, and $g=0$ off $W$. Thus $\mathrm{supp}g\subseteq \bar{W}\subseteq \{h>1\mathrm{/}2\}$, and $g=1$ on the open neighbourhood $O$ of $K$.

For $1\le i\le n$, set ${\phi }_i(x)=\{\begin{array}{ll}g(x){\displaystyle \frac{\sum _{j:i_j=i}{h}_j(x)}{h(x)},}& h(x)>0,\\ 0,& h(x)=0.\end{array}$ The quotient is only used on $\mathrm{supp}g\subseteq \{h>1\mathrm{/}2\}$, so extension by zero is continuous. Each ${\phi }_i$ is nonnegative, compactly supported in $U_i$, and ${\sum }_i{\phi }_i=g$, hence the sum is $1$ on $O$.

Every pointwise-admissible $f$ is nonnegative, so $\mathrm{\Lambda }(f)\ge 0$. The [given, L1] zero function is admissible for every open set, including the empty set; for $U=\mathrm{\varnothing }$ it is the only admissible function. Thus $\rho (U)\in [0,\mathrm{\infty }]$ and $\rho (\mathrm{\varnothing })=0$.

The cutoff supremum is at most the pointwise supremum. Conversely, let [L1, L2] $0\le f\le {\mathbf{1}}_U$ have compact support $K$, and choose $h\in C_c(X)$ with $0\le h\le 1$ and $h=1$ on $K$ by applying [L2] to $K\subseteq X$. For $t>0$ put $f_t=(f-th{)}_{+}$. Then $f_t\prec U$: its support lies in the compact set $\{f\ge t\}\subseteq U$. Moreover $0\le f-f_t\le th$, so positivity gives $\mathrm{\Lambda }(f)\le \mathrm{\Lambda }(f_t)+t\mathrm{\Lambda }(h)$. Letting $t\downarrow 0$ proves that the two displayed suprema defining $\rho (U)$ are equal.

If $U\subseteq V$, every test function admissible for $U$ is admissible for $V$, hence $\rho (U)\le \rho (V)$. The family of open supersets of any $E$ is nonempty because it contains $X$, so ${\mu }^{\ast }(E)$ is well defined in $[0,\mathrm{\infty }]$.

For open $U$, using $U$ itself in the infimum gives [step 2.1] ${\mu }^{\ast }(U)\le \rho (U)$. Conversely, if $U\subseteq V$ with $V$ open, monotonicity gives $\rho (U)\le \rho (V)$; taking the infimum over such $V$ gives the reverse inequality. This also covers $U=\mathrm{\varnothing }$ and $\rho (U)=\mathrm{\infty }$.

The definition gives ${\mu }^{\ast }(\mathrm{\varnothing })=0$ and monotonicity: an open superset of $F$ is also one of $E$ when $E\subseteq F$.

Let $E\subseteq {\bigcup }_n{E}_n$ and choose open $U_n\supseteq E_n$. If $U={\bigcup }_n{U}_n$ and $0\le f\le {\mathbf{1}}_U$ has compact support $K$, finitely many $U_{n_j}$ cover $K$. By [L1] there are ${\phi }_j\in C_c(X)$ subordinate to those sets with sum $1$ near $K$. Then $f={\sum }_{j}f{\phi }_j$, each summand is admissible for $U_{n_j}$, and positivity and linearity give $\mathrm{\Lambda }(f)\le {\sum }_j\rho (U_{n_j})\le {\sum }_n\rho (U_n)\mathrm{.}$ Taking the supremum over $f$ yields $\rho (U)\le {\sum }_n\rho (U_n)$.

If ${\sum }_n{\mu }^{\ast }(E_n)=\mathrm{\infty }$, countable subadditivity is automatic. Otherwise, for $\epsilon >0$ choose $U_n\supseteq E_n$ with $\rho (U_n)\le {\mu }^{\ast }(E_n)+\epsilon 2^{-n-1}$. Step 1.2 and $E\subseteq U$ give ${\mu }^{\ast }(E)\le \rho (U)\le {\sum }_n{\mu }^{\ast }(E_n)+\epsilon \mathrm{.}$ Letting $\epsilon \downarrow 0$ proves countable subadditivity.

Fix open $G,V$ and $f\prec V\cap G$. Put [given] $K=\mathrm{supp}f$. For any $g\prec V\setminus K$, the supports of $f$ and $g$ are disjoint, so $f+g\prec V$. Hence $\rho (V)\ge \mathrm{\Lambda }(f)+\rho (V\setminus K)$. Since $V\setminus G\subseteq V\setminus K$, $\rho (V\setminus K)\ge {\mu }^{\ast }(V\setminus G)$. Taking the supremum over $f\prec V\cap G$ therefore gives $\rho (V)\ge \rho (V\cap G)+{\mu }^{\ast }(V\setminus G)={\mu }^{\ast }(V\cap G)+{\mu }^{\ast }(V\setminus G)\mathrm{.}$

For arbitrary $E$ and open $V\supseteq E$, monotonicity and step 1.1 [step 1.1] give $\rho (V)\ge {\mu }^{\ast }(E\cap G)+{\mu }^{\ast }(E\setminus G)\mathrm{.}$ Infimizing over $V$ gives the hard Caratheodory inequality; outer subadditivity gives the reverse inequality.

Therefore every open $G$ is Caratheodory measurable. By [L1], the measurable sets form a complete sigma-algebra; since they contain all opens, they contain $\mathcal{B}(X)$, and the restriction is a Borel measure.

Suppose ${\mathbf{1}}_K\le f\in C_c(X)$. For $0<\epsilon <1$, the open set $U_{\epsilon }=\{f>1-\epsilon \}$ contains $K$. Every $g\prec U_{\epsilon }$ satisfies $g\le f\mathrm{/}(1-\epsilon )$, so positivity gives $\rho (U_{\epsilon })\le \mathrm{\Lambda }(f)\mathrm{/}(1-\epsilon )$. Outer regularity therefore yields $\mu (K)\le \mathrm{\Lambda }(f)\mathrm{/}(1-\epsilon )$, and then $\mu (K)\le \mathrm{\Lambda }(f)$.

Conversely, for every open $U\supseteq K$, choose an open $V$ with [L1, choose] $K\subseteq V\subseteq \bar{V}\subseteq U$ and compact closure. [L1] applied to $K\subseteq V$ supplies $f\in C_c(X)$ with ${\mathbf{1}}_K\le f\le {\mathbf{1}}_V$. Since $f=0$ off $V$, $\mathrm{supp}f\subseteq \bar{V}\subseteq U$, and hence $f\prec U$. Thus $\mathrm{\Lambda }(f)\le \rho (U)$. Taking first the infimum over $f$, then over $U$, gives the reverse inequality. For $K=\mathrm{\varnothing }$, $f=0$ gives both sides zero.

Choosing one relatively compact open neighbourhood $U$ of $K$ and the [step 1.1, step 1.2, L1] cutoff produced in step 1.2 gives $\mu (K)\le \mathrm{\Lambda }(f)<\mathrm{\infty }$.

If $f\prec U$ and $K=\mathrm{supp}f$, then [L1] $0\le f\le {\mathbf{1}}_K$. For every $h\in C_c(X)$ with ${\mathbf{1}}_K\le h$, monotonicity gives $\mathrm{\Lambda }(f)\le \mathrm{\Lambda }(h)$; taking the infimum in [L1] yields $\mathrm{\Lambda }(f)\le \mu (K)$. Hence $\rho (U)\le {\sup }_{K⋐U}\mu (K)$.

The reverse inequality is monotonicity of the measure. Since $\mu (U)=\rho (U)$, the displayed equality follows. Outer regularity is built into ${\mu }^{\ast }$, and [L1] gives compact finiteness, so all Radon clauses hold.

Let $0\le f\in C_c(X)$ and choose $N$ with $f\le N\epsilon$. For $1\le n\le N$ set $K_n=\{f\ge n\epsilon \}$ and $f_n=\min \{\epsilon ,(f-(n-1)\epsilon {)}_{+}\}\mathrm{.}$ Then $f={\sum }_{n=1}^N{f}_n$, $\epsilon {\mathbf{1}}_{K_n}\le f_n\le \epsilon {\mathbf{1}}_{K_{n-1}}$, where $K_0=\mathrm{supp}f$.

The upper compact estimate $0\le h\le {\mathbf{1}}_K\Rightarrow \mathrm{\Lambda }(h)\le \mu (K)$ follows directly from [L1] by comparing $h$ with functions majorizing ${\mathbf{1}}_K$; the lower estimate ${\mathbf{1}}_K\le h\Rightarrow \mu (K)\le \mathrm{\Lambda }(h)$ is [L1]. Applying these to $f_n\mathrm{/}\epsilon$ gives $\epsilon \mu (K_n)\le \mathrm{\Lambda }(f_n)\le \epsilon \mu (K_{n-1})\mathrm{.}$ The same inequalities hold for $\int f_{n}d\mu$.

Summing step 1.2 and using $\mu (K_0)<\mathrm{\infty }$ shows $\mid \mathrm{\Lambda }(f)-\int fd\mu \mid \le \epsilon \mu (K_0)\mathrm{.}$ Letting $\epsilon \downarrow 0$ proves equality for nonnegative $f$. Applying it to $f^{+}$ and $f^{-}$ and using linearity proves the result for every real $f\in C_c(X)$.

If $K\subseteq U$ with $K$ compact and $U$ open, choose $f$ by [L1]. Then $\mu (K)\le \int fd\mu =\int fd\nu \le \nu (U)\mathrm{.}$ Taking the infimum over $U\supseteq K$ gives $\mu (K)\le \nu (K)$ by outer regularity; symmetry gives equality on compact sets.

Inner regularity on open sets now gives $\mu (U)=\nu (U)$ for every open $U$. Outer regularity on Borel sets then gives $\mu (E)=\nu (E)$ for every Borel $E$, including infinite values.

If $U$ is open, write $U={\bigcup }_n{L}_n$ with $L_n$ compact. The compact sets $K_n={\bigcup }_{j\le n}{L}_j$ increase to $U$, so continuity from below gives $\mu (U)={\sup }_n\mu (K_n)$. Thus $\mu$ is inner regular on opens.

Since $X$ itself is sigma-compact, local compactness and finite [step 1.1, choose] subcovers give increasing relatively compact open sets $V_n$ with $X={\bigcup }_n{V}_n$. Fix one such $V=V_n$. Its total measure is finite because $\mu (V)\le \mu (\bar{V})<\mathrm{\infty }$. Every relatively open subset of $V$ is open in $X$, hence is compact-inner-regular by step 1.1. Every relatively closed subset of $V$ is sigma-compact: intersect it with a compact exhaustion of the sigma-compact open space $V$. It is therefore also compact-inner-regular by continuity from below.

Let ${\mathcal{R}}_V$ be the Borel subsets of $V$ that are both outer [step 2.1] regular in $V$ and inner regular by compact sets. The relatively open sets belong to ${\mathcal{R}}_V$: inner regularity is step 2.1 and outer regularity is immediate. A relatively closed set $F$ is inner regular by step 2.1; if $W=V\setminus F$, choose compact $K\subseteq W$ with $\mu (W\setminus K)<\epsilon$, and then the relatively open set $V\setminus K$ contains $F$ with excess below $\epsilon$. Thus closed sets also belong to ${\mathcal{R}}_V$. Because $\mu (V)<\mathrm{\infty }$, taking complements interchanges the two approximation properties. Countable unions preserve outer regularity by summable open errors and preserve inner regularity by first taking a finite partial union and then a finite union of compact cores. Hence ${\mathcal{R}}_V$ is a sigma-algebra containing the relatively open sets, so every Borel subset of $V$ is regular.

Let $E$ be Borel. Since $E\cap V_n\uparrow E$, the regularity just proved [step 3.1] on each $V_n$ gives $\mu (E)=\sup \{\mu (K):K\subseteq E,K\text{ compact}\}\mathrm{.}$ Indeed, when $\mu (E)<\mathrm{\infty }$, first choose $n$ so that $\mu (E\setminus V_n)$ is small and then take a compact core of $E\cap V_n$; when $\mu (E)=\mathrm{\infty }$, choose $n$ and then a compact core with arbitrarily large finite measure.

If $\mu (E)=\mathrm{\infty }$, outer regularity is automatic. Otherwise partition [step 3.1, step 4.1] $E$ into the Borel slices $E_n=E\cap (V_n\setminus V_{n-1})$, with $V_{-1}=\mathrm{\varnothing }$. Relative outer regularity in $V_n$ gives a relatively open, hence open-in-$X$, set $O_n\subseteq V_n$ containing $E_n$ with $\mu (O_n\setminus E_n)<\epsilon 2^{-n-1}$. Then $O={\bigcup }_n{O}_n$ is open, contains $E$, and $\mu (O\setminus E)<\epsilon$. Thus every Borel set is outer regular as well as compact-inner-regular; in particular $\mu$ is Radon and is regular in the stronger convention.

Refining a countable base by [L1] gives a countable base $(V_n)$ with compact closures. Every open $U$ is the union of those $V_n$ whose closures lie in $U$, hence is a countable union of compact sets $\overline{V_n}$.

Thus every open set is sigma-compact, and [L2] applies to $\mu$.

Compact subsets of ${\mathbb{R}}^n$ are bounded, so they have finite measure by [L1]. Outer regularity on Borel sets and compact inner regularity on open sets are also direct instances of [L1]. These are precisely the Radon clauses.

The last part of [L1] applies to every Borel set, so the stronger regularity assertion also holds.

By [L1], ${\mu }_F$ is finite on compact sets, outer regular on Borel sets, and inner regular by compact sets on Borel sets, hence in particular on open sets. These imply every clause of the LCH Radon definition.

No measure is replaced in this comparison, so the defining half-open interval formula and the Borel sigma-algebra from the construction remain unchanged.

First let $B$ be Borel of finite measure and let $\delta >0$. By [L1], choose open $U\supseteq B$ of finite measure and compact $L\subseteq U$ with $\mu (U\setminus L)<\delta \mathrm{/}2$. By outer regularity choose open $O\supseteq U\setminus B$ with $\mu (O)<\mu (U\setminus B)+\delta \mathrm{/}2$. Then $H=L\setminus O$ is compact and contained in $B$. Since $B$ is disjoint from $U\setminus B$, $\mu (B\cap O)<\delta \mathrm{/}2$, and hence $\mu (B\setminus H)\le \mu (U\setminus L)+\mu (B\cap O)<\delta$. This proves compact approximation for each finite-measure Borel set using only [L1].

Since $f$ is complex-valued and $\mu (E)<\mathrm{\infty }$, continuity from above gives an $M>0$ for which $B=\{x\in E:\mathrm{\mid }f(x)\mathrm{\mid }\le M\}$ satisfies $\mu (E\setminus B)<\epsilon \mathrm{/}4$. For each $m\ge 0$, partition the disk $\mathrm{\mid }z\mathrm{\mid }\le M$ into finitely many nonempty disjoint Borel cells $C_{m,j}$, $1\le j\le J_m$, each of diameter less than $2^{-m}$. Such cells can be obtained by intersecting the disk with a finite half-open square grid. Their preimages $B_{m,j}$ partition $B$ into finite-measure Borel sets.

Apply step 1.1 to choose compact $H_{m,j}\subseteq B_{m,j}$ with $\mu (B_{m,j}\setminus H_{m,j})<\epsilon 2^{-m-3}\mathrm{/}{J}_m$. Put $H_m={\bigcup }_{j=1}^{J_m}{H}_{m,j}$ and $K={\bigcap }_{m\ge 0}{H}_m$. Each $H_m$ is compact and thus closed in $X$; their intersection is a closed subset of the compact $H_0$, hence compact. Moreover $K\subseteq B\subseteq E$ and $\mu (E\setminus K)\le \mu (E\setminus B)+{\sum }_{m\ge 0}\mu (B\setminus H_m)<\epsilon \mathrm{/}4+{\sum }_{m\ge 0}\epsilon 2^{-m-3}=\epsilon \mathrm{/}2<\epsilon$.

For each $m$, the finitely many disjoint compact sets $K\cap H_{m,j}$ cover $K$. Each is relatively clopen, because its complement is a finite union of closed sets. On it, the values of $f$ lie in $C_{m,j}$ and have oscillation below $2^{-m}$. Given $x\in K$ and a positive tolerance, choose $m$ so $2^{-m}$ is smaller; its clopen cell piece is a neighbourhood witnessing continuity at $x$. The empty $K$ case is vacuous. Thus $f{\mathrm{\mid }}_K$ is continuous.

It suffices by [L1] to approximate ${\mathbf{1}}_E$ when $\mu (E)<\mathrm{\infty }$. [L1, choose] Given $\eta >0$, outer regularity gives open $U\supseteq E$ with $\mu (U\setminus E)<{\eta }^p\mathrm{/}2$, and inner regularity of $U$ gives compact $K\subseteq U$ with $\mu (U\setminus K)<{\eta }^p\mathrm{/}2$.

Choose $f\in C_c(X)$ with ${\mathbf{1}}_K\le f\le {\mathbf{1}}_U$ by [L2]. [step 1.1, L2] Then $\mathrm{\mid }f-{\mathbf{1}}_E\mathrm{\mid }\le {\mathbf{1}}_{U\setminus K}+{\mathbf{1}}_{U\setminus E}\mathrm{.}$ Since both indicators take only the values zero and one and their union is $U\setminus (K\cap E)$, while $0\le f,{\mathbf{1}}_E\le 1$, the error is bounded by the indicator of that union. Hence $\mathrm{\parallel }f-{\mathbf{1}}_E{\mathrm{\parallel }}_p^p\le \mu (U\setminus K)+\mu (U\setminus E)<{\eta }^p\mathrm{.}$ Thus $\mathrm{\parallel }f-{\mathbf{1}}_E{\mathrm{\parallel }}_p<\eta$.

Approximate the finitely many indicator terms of a simple function separately and sum the resulting $C_c$ functions. Then use [L1] and the triangle inequality.

Restrict $L$ to $C_c(X)$ and apply [L1], obtaining a Radon measure $\mu$. For every compact $K$, a cutoff $0\le h\le 1$ equal to $1$ on $K$ gives $\mu (K)\le L(h)\le \mathrm{\parallel }L\mathrm{\parallel }$. Inner regularity on $X$ therefore yields $\mu (X)\le \mathrm{\parallel }L\mathrm{\parallel }<\mathrm{\infty }$.

Conversely, $\mathrm{\mid }L(f)\mathrm{\mid }\le L(\mathrm{\mid }f\mathrm{\mid })\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mu (X)$ first for $f\in C_c(X)$. Since $C_c(X)$ is uniformly dense in $C_0(X)$ and $\mu$ is finite, both sides extend continuously to $C_0(X)$, giving the representation and $\mathrm{\parallel }L\mathrm{\parallel }\le \mu (X)$. Thus equality holds.

A finite Radon measure is compact-inner-regular on every Borel set: apply open inner regularity to an open superset of the complement and use finite complements. Hence $\mu$ is regular. Uniqueness follows from [L1].

For $f\ge 0$ define $L^{+}(f)=\sup \{L(g):0\le g\le f\}\mathrm{.}$ The supremum is finite because $\mathrm{\mid }L(g)\mathrm{\mid }\le \mathrm{\parallel }L\mathrm{\parallel }\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. It is positively homogeneous and monotone.

If $f,h\ge 0$, decompositions $0\le g\le f+h$ satisfy $g=g_1+g_2$ with $g_1=\min (g,f)$ and $g_2=g-g_1$, where $0\le g_1\le f$ and $0\le g_2\le h$. This gives $L^{+}(f+h)\le L^{+}(f)+L^{+}(h)$; the reverse inequality follows by adding independent approximants. Thus $L^{+}$ is additive on the positive cone.

Extend $L^{+}$ linearly by $L^{+}(u)=L^{+}(u^{+})-L^{+}(u^{-})$. Cone additivity makes this well defined and positive. Put $L^{-}=L^{+}-L$; for $f\ge 0$, the competitor $g=f$ in step 1.1 gives $L^{+}(f)\ge L(f)$, so $L^{-}$ is positive. The bounds in step 1.1 give $\mathrm{\parallel }{L}^{\pm }\mathrm{\parallel }\le \mathrm{\parallel }L\mathrm{\parallel }$, and $L=L^{+}-L^{-}$.

On the real vector space of real-valued functions put $A(u)=\mathrm{Re}L(u)$ and $B(u)=\mathrm{Im}L(u)$. Apply [L1], then [L2], to the positive decompositions of both $A$ and $B$. This gives finite regular signed measures $\alpha$ and $\beta$ representing $A$ and $B$. Put $\mu =\alpha +i\beta$. If $f=u+iv$, complex linearity gives $L(f)=L(u)+iL(v)$, whose real and imaginary parts agree exactly with those of $\int fd(\alpha +i\beta )$; hence $\mu$ represents $L$.

If two finite regular complex measures $\mu$ and $\nu$ represent $L$, [step 1.1, L2] then the real and imaginary signed parts of their difference $\mu -\nu$ integrate every real $C_c$ function to zero. For either signed part, move its negative Jordan component to the other side; the two resulting positive Radon measures have equal integrals on $C_c$. The positive-measure uniqueness in [L2] makes those positive measures equal, so both signed parts of $\mu -\nu$ vanish and $\mu =\nu$.

Conversely, $\mathrm{\mid }\int fd\mu \mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mu \mathrm{\mid }(X)$, so integration is bounded with norm at most $\mathrm{\mid }\mu \mathrm{\mid }(X)$. The definition of total variation and regular approximation by compactly supported phase functions gives functions with $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and integrals arbitrarily close to $\mathrm{\mid }\mu \mathrm{\mid }(X)$; hence equality of norms.