Radon Measures and the Riesz Markov Kakutani Theorem — Examples

Linearity of the Riemann integral makes $L$ linear, and $f\ge 0$ implies $L(f)\ge 0$. Since $[a,b]$ is compact, $C_c([a,b])=C([a,b])$.

Under the stated choice hypothesis, Lebesgue measure is regular on [given] $\mathbb{R}$; its restriction to the closed subspace $[a,b]$ is finite and Radon. For every $f\in C([a,b])$, equality of the Riemann and Lebesgue integrals gives $L(f)={\int }_{[a,b]}fd(\lambda {\mathrm{\mid }}_{[a,b]})$. Thus $\lambda {\mathrm{\mid }}_{[a,b]}$ is a Radon representing measure.

To prove uniqueness without a general LCH cutoff, let $\rho$ be any other Radon representing measure on $[a,b]$. If $K\subseteq O\subseteq [a,b]$ with $K$ compact and $O$ relatively open, A compact set and a disjoint closed set have a positive norm-distance gap gives a positive distance between nonempty $K$ and the closed set $[a,b]\setminus O$; when the latter is empty use the constant function $1$, and when $K$ is empty the comparison is trivial. Choose a smaller $\delta >0$. The function $u(x)=\max \{0,1-d(x,K)\mathrm{/}\delta \}$ is continuous by $\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y)$, so the distance to a fixed nonempty set is $1$-Lipschitz, equals $1$ on $K$, and vanishes outside $O$. Equality of the two representing integrals therefore gives $\lambda (K)\le \int ud\lambda =\int ud\rho \le \rho (O)$; symmetry gives $\rho (K)\le \lambda (O)$. Outer regularity yields equality on compact sets, inner regularity then equality on opens, and outer regularity yields equality on every Borel set. Thus $\rho =\lambda {\mathrm{\mid }}_{[a,b]}$.

Evaluation is linear, and $f\ge 0$ implies $f(x)\ge 0$, so $L_x$ is positive.

For a nonnegative simple function $s$, the definition of its integral [given] and the set formula for ${\delta }_x$ give $\int sd{\delta }_x=s(x)$. Increasing simple approximation and monotone convergence extend this identity to every nonnegative measurable function, and positive/negative parts extend it to every integrable real function. In particular, ${\int }_{X}fd{\delta }_x=f(x)=L_x(f)$ for every $f\in C_c(X)$.

The measure ${\delta }_x$ is finite on compact sets. If a Borel set $E$ [step 1.2] contains $x$, every open superset has ${\delta }_x$-measure one; if it does not, the open set $X\setminus \{x\}$ contains $E$ and has measure zero. Thus ${\delta }_x$ is outer regular. Likewise an open set containing $x$ contains the compact set $\{x\}$, while the empty compact set suffices otherwise, so ${\delta }_x$ is inner regular on opens. Hence ${\delta }_x$ is a Radon measure representing $L_x$ by step 1.2.

If $f\in C_c$ has support $K$, then $\int \mathrm{\mid }fg\mathrm{\mid }d\lambda \le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{K}gd\lambda <\mathrm{\infty }$. Hence $L_g$ is well defined and linear; it is positive because $g\ge 0$.

The density construction makes ${\mu }_g(E)={\int }_{E}gd\lambda$ a Borel measure. Every compact set is contained in a ball, so local integrability makes ${\mu }_g$ finite on compact sets. Moreover, ${\mathbb{R}}^n$ is LCH, and its rational open boxes form a countable basis. The second-countable regularity theorem therefore makes ${\mu }_g$ regular, hence Radon. Its defining integral gives $L_g(f)=\int fd{\mu }_g$, so ${\mu }_g$ is a Radon measure representing $L_g$.

The measure ${\mu }_F$ is finite on compact intervals, so the displayed integral is finite for compactly supported continuous $f$. It is linear and positive.

Lebesgue--Stieltjes regularity makes ${\mu }_F$ Radon without changing its half-open interval convention. Thus the definition already gives a Radon representation, and RMK uniqueness says it is the representing measure.

Compact subsets of a discrete space are finite, so $C_c(X)$ consists exactly of finite-support functions. The sum defining $L$ is therefore finite; it is linear and positive.

For counting measure $\mathrm{\#}$, integration of a finite-support function is its finite sum, so $L(f)=\int fd\mathrm{\#}$. Counting measure is Radon on a discrete space: compact sets are finite and every set is open and is the union of its finite subsets. Its total mass may be infinite when $X$ is infinite.

The space $X$ is compact Hausdorff and $\bar{m}(X)=1$. The open Borel subset $Y=[0,{\omega }_1)$ contains a club, so $\bar{m}(Y)=1$.

Every compact $K\subseteq Y$ is bounded below ${\omega }_1$, hence contains no club and has $\bar{m}(K)=0$. Therefore $Y$ cannot be approximated from within by compact sets, and $\bar{m}$ is not regular.

Since $X$ is compact, $C_c(X)=C(X)$. Eventual constancy gives $\int fd\bar{m}=f({\omega }_1)=\int fd\delta$ for every $f\in C_c(X)$.

Yet for $Y=[0,{\omega }_1)$, $\bar{m}(Y)=1$ and $\delta (Y)=0$. Thus the two Borel measures are distinct representations of the same functional; the missing condition is regularity.