Proof of The Riesz–Markov Representation Theorem on a Locally Compact Metric Space

Throughout, sums, scalar multiples, products and absolute values 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). Finite sums over $\{1,\dots,n\}$ exist by Iterating a Binary Operation along a Finite List (i) and are handled with Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (v), Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (ii), Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (i), Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (ii), Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (iii) and Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (v). For an open set $V\subseteq X$ we write $f\prec V$ if $f\in C_{c}(X)$, $0\le f\le1$ and $\operatorname{supp}f\subseteq V$.

Step 1 (the measure). Let $\mu(V)$, for open $V\subseteq X$, and $\mu^{*}$ be as in The Radon Measure of a Positive Linear Functional on C_c(X): the Outer Measure, Compact Sets, Borel Measurability and Regularity, for $\Lambda$, so that $\mu(V)=\sup\{\Lambda g:g\prec V\}$; and let $\bar{\mu}=\mu^{*}|_{\mathcal{B}(X)}$, a Radon measure on $X$ by The Radon Measure of a Positive Linear Functional on C_c(X): the Outer Measure, Compact Sets, Borel Measurability and Regularity (iv). We use three of its properties. First, for open $V$ and $f\prec V$ we have $\Lambda f\le\mu(V)=\bar{\mu}(V)$, the equality by The Radon Measure of a Positive Linear Functional on C_c(X): the Outer Measure, Compact Sets, Borel Measurability and Regularity (i). Second, for compact $K$ and $g\in C_{c}(X)$ with $g\ge\mathbf{1}_{K}$ we have $\bar{\mu}(K)\le\Lambda g$, and $\bar{\mu}(K)<\infty$, by The Radon Measure of a Positive Linear Functional on C_c(X): the Outer Measure, Compact Sets, Borel Measurability and Regularity (ii). Third, for $B\in\mathcal{B}(X)$ with $\bar{\mu}(B)<\infty$ and real $\eta>0$ there is an open $W\supseteq B$ with $\bar{\mu}(W)<\bar{\mu}(B)+\eta$, since $\bar{\mu}(B)=\mu^{*}(B)$ is the infimum of the values $\mu(W)=\bar{\mu}(W)$ over open $W\supseteq B$. Moreover $\Lambda$ is monotone: if $f,g\in C_{c}(X)$ and $f\le g$, then $\Lambda g-\Lambda f=\Lambda(g-f)\ge0$ by linearity and positivity.

Step 2 (reduction to functions with values in $[0,1]$). We show that $\bar{\mu}$ satisfies the identity of (i); every $f\in C_{c}(X)$ is $\bar{\mu}$-integrable by Radon Measures and Finite Signed Borel Measures Are Determined by Their Integrals of Compactly Supported Continuous Functions (i). It suffices to show

$$\Lambda f=\int_{X}f\,d\bar{\mu}\qquad(f\in C_{c}(X),\ 0\le f\le1).$$

Indeed, let $f\in C_{c}(X)$ and $c=\|f\|_{u}$, so that $|f|\le c$ by The Space of Continuous Functions with Compact Support and the Uniform Norm (iii). If $c=0$, then $f=0$ and both sides of the identity vanish. If $c>0$, the functions $f^{\pm}=\frac12(|f|\pm f)$ lie in $C_{c}(X)$ and satisfy $f=f^{+}-f^{-}$ and $0\le f^{\pm}\le|f|\le c$, so $g^{\pm}=c^{-1}f^{\pm}$ lie in $C_{c}(X)$ and satisfy $0\le g^{\pm}\le1$. By linearity of $\Lambda$ and of the integral (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)),

$$\Lambda f=c\,\Lambda g^{+}-c\,\Lambda g^{-}=c\int_{X}g^{+}\,d\bar{\mu}-c\int_{X}g^{-}\,d\bar{\mu}=\int_{X}f\,d\bar{\mu}.$$

Step 3 (slicing $f$). Fix $f\in C_{c}(X)$ with $0\le f\le1$, and fix $N\in\mathbb{N}$. Let $K_{0}=\operatorname{supp}f$, a compact set, and for $j=1,\dots,N$ let

$$K_{j}=\{x\in X:f(x)\ge j/N\}.$$

For $j\ge1$ the set $K_{j}$ is closed, being the complement of the preimage of the open set $(-\infty,j/N)$ under the continuous function $f$ (Continuity of a Map Between Metric Spaces via Preimages of Open Sets (i)), and it lies in $K_{0}$, since $f(x)\ge j/N>0$ gives $f(x)\neq0$ and hence $x\in\operatorname{supp}f$ by Support of a Real Function on a Metric Space (i). Thus every $K_{j}$ is closed and compact (for $j\ge1$ as a closed subset of $K_{0}$, by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (i)), hence Borel (Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (ii)) and of finite measure $\bar{\mu}(K_{j})<\infty$ (Step 1). For $j=1,\dots,N$ let

$$f_{j}=\min\Bigl(\max\Bigl(f-\frac{j-1}{N},\,0\Bigr),\,\frac1N\Bigr).$$

Each $f_{j}$ is continuous: the function $t\mapsto\min(\max(t-(j-1)/N,0),1/N)$ on $\mathbb{R}$ changes by at most $|s-t|$ between $s$ and $t$, so $|f_{j}(x)-f_{j}(y)|\le|f(x)-f(y)|$ for all $x,y\in X$, and the continuity of $f$ at a point gives that of $f_{j}$ with the same $\delta$. Moreover $0\le f_{j}\le1/N$, and $f_{j}(x)>0$ only if $f(x)>(j-1)/N$, which gives $x\in K_{j-1}$: for $j\ge2$ by the definition of $K_{j-1}$, and for $j=1$ because then $f(x)\neq0$. Thus the set $\{x\in X:f_{j}(x)\neq0\}$ lies in the closed set $K_{j-1}$, and so does its closure:

$$\operatorname{supp}f_{j}\subseteq K_{j-1}\subseteq K_{0}.$$

Hence $\operatorname{supp}f_{j}$, a closed subset of the compact set $K_{0}$, is compact by Closed Subsets of Compact Sets Are Compact, and Compact Subsets of a Metric Space Are Closed and Bounded (i), and $f_{j}\in C_{c}(X)$. Finally, checking the three cases $f(x)\le(j-1)/N$, $(j-1)/N<f(x)\le j/N$ and $f(x)>j/N$ shows

$$f_{j}(x)=\min\Bigl\{f(x),\frac{j}{N}\Bigr\}-\min\Bigl\{f(x),\frac{j-1}{N}\Bigr\}\qquad(x\in X),$$

so summing over $j$ the right-hand sides telescope to $\min\{f(x),1\}-\min\{f(x),0\}=f(x)$, as $0\le f\le1$:

$$f=\sum_{j=1}^{N}f_{j}.$$

Step 4 (bounds for $f_{j}$). Let $j\in\{1,\dots,N\}$. On $K_{j}$ we have $f-(j-1)/N\ge1/N$, so $f_{j}=1/N$ there; together with $0\le f_{j}\le1/N$ and $f_{j}=0$ off $K_{j-1}$ (Step 3) this gives

$$\frac1N\mathbf{1}_{K_{j}}\le f_{j}\le\frac1N\mathbf{1}_{K_{j-1}}.$$

Moreover $Nf_{j}\in C_{c}(X)$, $0\le Nf_{j}\le1$, and $\operatorname{supp}(Nf_{j})\subseteq\operatorname{supp}f_{j}\subseteq K_{j-1}$ by the support inclusion for scalar multiples and Step 3.

Step 5 (the functional on $f_{j}$). Let $j\in\{1,\dots,N\}$. Since $K_{j}$ is compact and $Nf_{j}\ge\mathbf{1}_{K_{j}}$ by Step 4, the second property in Step 1 gives $\bar{\mu}(K_{j})\le\Lambda(Nf_{j})$. Let $\eta>0$ be real. As $K_{j-1}$ is Borel with $\bar{\mu}(K_{j-1})<\infty$ (Step 3), the third property in Step 1 gives an open $W\supseteq K_{j-1}$ with $\bar{\mu}(W)<\bar{\mu}(K_{j-1})+\eta$; by Step 4, $Nf_{j}\prec W$, so the first property gives $\Lambda(Nf_{j})\le\bar{\mu}(W)<\bar{\mu}(K_{j-1})+\eta$. As $\eta$ is arbitrary, $\Lambda(Nf_{j})\le\bar{\mu}(K_{j-1})$, and since $\Lambda(Nf_{j})=N\Lambda f_{j}$,

$$\frac1N\bar{\mu}(K_{j})\le\Lambda f_{j}\le\frac1N\bar{\mu}(K_{j-1})\qquad(j=1,\dots,N).$$

Step 6 (the integral of $f_{j}$). For $j=0,\dots,N$ the indicator $\mathbf{1}_{K_{j}}$ is a nonnegative real-valued function, measurable into $[0,\infty]$ with $\int_{X}\mathbf{1}_{K_{j}}\,d\bar{\mu}=\bar{\mu}(K_{j})<\infty$ by Basic Properties of the Integral of Nonnegative Measurable Maps (ii), and so measurable as a real-valued function by Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii); hence it is $\bar{\mu}$-integrable with this integral by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi); and each $f_{j}$ is $\bar{\mu}$-integrable by Radon Measures and Finite Signed Borel Measures Are Determined by Their Integrals of Compactly Supported Continuous Functions (i). Linearity (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii)) and monotonicity (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (ix)) of the integral, applied to the bounds of Step 4, give

$$\frac1N\bar{\mu}(K_{j})\le\int_{X}f_{j}\,d\bar{\mu}\le\frac1N\bar{\mu}(K_{j-1})\qquad(j=1,\dots,N).$$

Step 7 (conclusion). The $\bar{\mu}(K_{j})$ are real numbers by Step 3; put

$$A=\frac1N\sum_{j=1}^{N}\bar{\mu}(K_{j}),\qquad B=\frac1N\sum_{j=1}^{N}\bar{\mu}(K_{j-1}).$$

Since the sum in $C_{c}(X)$ is the pointwise sum (The Space of Continuous Functions with Compact Support and the Uniform Norm (ii)), the pointwise sum $\sum_{j=1}^{N}f_{j}=f$ of Step 3 is also the sum in $C_{c}(X)$, so $\Lambda f=\sum_{j=1}^{N}\Lambda f_{j}$ by linearity of $\Lambda$, and $\int_{X}f\,d\bar{\mu}=\sum_{j=1}^{N}\int_{X}f_{j}\,d\bar{\mu}$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (viii). Summing the bounds of Steps 5 and 6 over $j$ therefore shows that $\Lambda f$ and $\int_{X}f\,d\bar{\mu}$ both lie between $A$ and $B$, and the difference $B-A$ telescopes:

$$\Bigl|\Lambda f-\int_{X}f\,d\bar{\mu}\Bigr|\le B-A=\frac1N\sum_{j=1}^{N}\bigl(\bar{\mu}(K_{j-1})-\bar{\mu}(K_{j})\bigr)=\frac{\bar{\mu}(K_{0})-\bar{\mu}(K_{N})}{N}\le\frac{\bar{\mu}(K_{0})}{N}.$$

Here $N\in\mathbb{N}$ was arbitrary, while $K_{0}=\operatorname{supp}f$ does not depend on $N$ and $\bar{\mu}(K_{0})<\infty$. If $\delta=\bigl|\Lambda f-\int_{X}f\,d\bar{\mu}\bigr|$ were positive, The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (ii) would give $N\in\mathbb{N}$ with $\bar{\mu}(K_{0})<N\delta$, contradicting the estimate for this $N$. Hence $\Lambda f=\int_{X}f\,d\bar{\mu}$, and by Step 2 the Radon measure $\bar{\mu}$ satisfies the identity of (i), which proves (i).

Step 8 (uniqueness). Let $\mu_{1}$ and $\mu_{2}$ be the measures of (ii): Radon measures on $X$ that both satisfy the identity of (i). Then $\int f\,d\mu_{1}=\Lambda f=\int f\,d\mu_{2}$ for every $f\in C_{c}(X)$, and Radon Measures and Finite Signed Borel Measures Are Determined by Their Integrals of Compactly Supported Continuous Functions (ii) gives $\mu_{1}(B)=\mu_{2}(B)$ for every $B\in\mathcal{B}(X)$. This proves (ii).

Step 9 (the total mass equals the supremum). Let now $\Lambda\in C_{c}(X)^{*}$. By (ii) the Radon measure of (i) is the measure $\bar{\mu}$ of Step 1, and since $X$ is open, The Radon Measure of a Positive Linear Functional on C_c(X): the Outer Measure, Compact Sets, Borel Measurability and Regularity (i) and the definition of $\mu(X)$ as in The Radon Measure of a Positive Linear Functional on C_c(X): the Outer Measure, Compact Sets, Borel Measurability and Regularity give

$$\bar{\mu}(X)=\mu^{*}(X)=\mu(X)=\sup\{\Lambda f:f\prec X\}=\sup\{\Lambda f:f\in C_{c}(X),\ 0\le f\le1\}=:S .$$

Step 10 (the supremum is at most the dual norm). If $f\in C_{c}(X)$ and $0\le f\le1$, then $\|f\|_{u}\le1$, and since the dual norm is a bound for $\Lambda$ (The Nonnegative Bounds of a Bounded Linear Functional Have an Infimum, Which Is a Bound (ii)),

$$\Lambda f\le|\Lambda f|\le\|\Lambda\|_{*}\|f\|_{u}\le\|\Lambda\|_{*}.$$

Hence $S\le\|\Lambda\|_{*}<\infty$, and in particular $\bar{\mu}$ is finite.

Step 11 (the dual norm is at most the supremum). Let $f\in C_{c}(X)$ and $c=\|f\|_{u}$. If $c=0$ then $f=0$ and $|\Lambda f|=0$. If $c>0$, the function $g=c^{-1}|f|$ lies in $C_{c}(X)$ and satisfies $0\le g\le1$, so $\Lambda|f|=c\,\Lambda g\le cS$; and since $-|f|\le f\le|f|$, monotonicity gives $|\Lambda f|\le\Lambda|f|$. Thus

$$|\Lambda f|\le S\,\|f\|_{u}\qquad(f\in C_{c}(X)),$$

so $S\ge0$ (as $\Lambda0=0$ lies in the set defining $S$) is a nonnegative bound for $\Lambda$, and $\|\Lambda\|_{*}\le S$ because $\|\Lambda\|_{*}$ is the infimum of these bounds by The Dual Norm (i). With Steps 9 and 10, $\bar{\mu}(X)=S=\|\Lambda\|_{*}$, which proves (iii).