Riesz-Markov-Kakutani Representation Theorem

Theorem
Let $\struct {X, \tau}$ be a Hausdorff space.

Let $\map {C_c} X$ be the space of continuous complex functions with compact support on $X$.

Let $\Lambda$ be a positive linear functional on $\map {C_c} X$.

There exists a $\sigma$-algebra $\MM$ over $X$ which contains the Borel $\sigma$-algebra of $\struct {X, \tau}$.

There exists a unique complete Radon measure $\mu$ on $\MM$ such that:
 * $\ds \forall f \in \map {C_c} X: \Lambda f = \int_X f \rd \mu$

Construction of $\mu$ and $\MM$
For every $V\in\tau$, define:
 * $\map{\mu_1}{v)}=\sup\set{\Lambda f: f\prec V}$

For every other set $E\subset X$, define:
 * $\map{\mu}{E}=\inf\set{\map{\mu}{V}: V\supset E \land V\in\tau}$

By monotonicity of $\mu_1$:
 * $\mu_1 = \mu {\restriction_\tau}$

Define:
 * $\MM_F=\set{E\subset X : \map{\mu}{E}<\infty \land \map{\mu}{E}=\sup\set{\map{\mu}{K}:K\subset E\land K\text{ compact}}}$

Define:
 * $\MM=\set{E\subset X : \forall K\subset X \text{ compact}, E\cap K\in\MM_F}$

Lemma $1$
$\mu$ is Countably Subadditive.

Proof
Let $V_1$ and $V_2$ be open subsets of $X$.

Let $g\in\map{C_c}{X}:g\prec V_1\cup V_2$.

Let $\set{h_1,h_2}$ be a partition of unity subordinate to $\set{V_1,V_2}$.

By linearity of $\Lambda$ and the definition of $\mu$,
 * $\Lambda g=\Lambda(h_1g+h_2g)=\Lambda h_1g+\Lambda h_2g\le\map{\mu}{V_1}+\map{\mu}{V_2}$.

Since $g$ was arbitrary, by the definition of $\mu$, $\mu$ is subadditive over $\tau$.

Applying induction yields the countable subadditivity of $\mu$ over $\tau$.

Let $\sequence{E_i}\in X^\N$.

By definition of $\mu$, for all $E_i$, for all $\varepsilon>0$, there exists an open $V_i\supset E_i$ such that
 * $\map{\mu}{V_i}\le\map{\mu}{E_i}+2^{-i}\varepsilon$.

Therefore, by $\ds\bigcup_{i=1}^\infty V_i\supset\bigcup_{i=1}^\infty E_i$, monotonicity of $\mu$, and countable subadditivity of $\mu$ over $\tau$,
 * $\ds\map{\mu}{\bigcup_{i=1}^\infty E_i}\le\map{\mu}{\bigcup_{i=1}^\infty V_i} \le \varepsilon +\sum_{i=1}^\infty \map{\mu}{E_i}$.

Since $\varepsilon$ was arbitrary, $\mu$ is countably subadditive over $X$.

Lemma $2$
$K\subset X$ compact $\implies$ $K\in\MM_F$ and $\map{\mu}{K}=\inf\set{\Lambda f : K \prec f}$.

Proof
Let $f\in\map{C_c}{X}: K \prec f$ and $\alpha\in\R\cap\openint{0}{1}$.

Define $V_\alpha = \map{f^{-1}}{\openint{\alpha}{\infty}}$.

Then, $K\in V_\alpha$, and for all $g:\map{C_c}{X}:g\prec V_\alpha$, $\alpha g\le f$.

Thus,
 * $\map{\mu}{K}\le\map{\mu}{V_\alpha}=\sup\set{\Lambda g : g\prec V_\alpha}\le \alpha^{-1}\Lambda f$.

Since $\alpha$ can be arbitrarily close to 1,
 * $\map{\mu}{K}\le\inf\set{\Lambda f:K\prec f}$.

By definition of $\mu$, for all $\epsilon>0$, there exists a $V\in\tau:V\supset K$ such that
 * $\map{\mu}{V}<\map{\mu}{K}+\epsilon$.

By Urysohn's Lemma, there exists an Urysohn function $f$ for $K$ and $V$.

By definition of $\mu$,
 * $\Lambda f\le\map{\mu}{V}<\map{\mu}{K}+\varepsilon$.

Since $\varepsilon$ was arbitrary,
 * $\map{\mu}{K}\ge\inf\set{\Lambda f: K\prec f}$.

Therefore,
 * $\map{\mu}{K}=\inf\set{\Lambda f:K\prec f}$.

By definition of positive linear functionals, $\Lambda$ is finite. Thus, $\map{\mu}{K}<\infty$.

So, $K\in\MM_F$.

Lemma $3$
$\mu$ is countably additive over pairwise disjoint collections of compact sets

Proof
Let $K_1$ and $K_2$ be disjoint compact subsets of $X$.

By Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods, there exists an open $V\supset K_1$ such that $V$ and $K_2$ are disjoint.

By Urysohn's Lemma, there exists an $f\in\map{C_c}{X}$ such that $f(K_1)=\set{1}$ and $f(K_2)=\set{0}$.

By Lemma 2 and union of compact sets is compact, for all $\varepsilon>0$, there exists a $g\in\map{C_c}{X}$ such that
 * $K_1\cup K_2\prec g \text{ and } \Lambda g<\map{\mu}{K_1\cup K_2}+\varepsilon$.

Now, $K_1\prec fg$ and $K_2=\prec (1-f)g$.

By linearity of $\Lambda$ and Lemma 2,
 * $\map{\mu}{K_2}+\map{\mu}{K_2}\le\Lambda(fg)+\Lambda(g-fg)=\Lambda g<\map{\mu}{K_1\cup K_2}+\varepsilon$.

Thus, by Lemma 1, $\mu$ is additive for disjoint compact set.

Applying induction yields countable additivity.

Lemma $4$
$\mu$ is countably additive over pairwise disjoint collections of subsets of $\MM_F$

Proof
Let $\sequence{E_i}\in\paren{\MM_F}^\N$ be pairwise disjoint with union $E$.

If $\map{\mu}{E}=\infty$, then, by countable subadditivity,
 * $\ds\infty=\map{\mu}{E}\le\sum_{i=1}^\infty \map{\mu}{E_i}$.

So,
 * $\ds\map{\mu}{E}=\sum_{i=1}^\infty\map{\mu}{E}$.

Now, suppose $\map{\mu}{E}<\infty$.

By definition of $\MM_F$, for all $\varepsilon>0$, for each $i$, there exists a compact $H_i\subset E_i$ such that $\map{\mu}{H_i}>\map{\mu}{E_i}-2^{-i}\varepsilon$.

So,

This holds for all $n\in\N$. So, $\mu$ is countably superadditive over pairwise disjoint collections of subsets of $\MM_F$.

Therefore, by Lemma 1, $\mu$ is countably additive over pairwise disjoint collections of subsets of $\MM_F$.

Lemma $5$
$\set{V\in\tau:\map{\mu}{V}<\infty}\subset\MM_F$

Proof
Since $\mu$ and $\mu_1$ coincide on $\tau$, by definition of $\mu_1$, for all $\alpha<\map{\mu}{V}$, there exists some $f\in\map{C_c}{X}:f\prec V$ such that $\Lambda f >\alpha$.

Let $K=\supp f$.

For all $W\in\tau$ such that $K\subset W$, $f\prec W$.

So, by definition of $\mu_1$, $\Lambda f\le\map{\mu}{W}$.

By definition of $\mu$, $\Lambda f \le \map{\mu}{K}$.

So, $\alpha<\map{\mu}{K}$.

That is, for all $\alpha<\map{\mu}{V}$, there exists some compact $K\subset V$ such that $\alpha<\map{\mu}{K}\le\map{\mu}{V}$.

since $V$ was arbitrary,
 * $\set{V\in\tau:\map{\mu}{V}<\infty}\subset\MM_F$.

Lemma $6$
For all $E\in\MM_F$ and $\varepsilon>0$, there exist some compact $K\in X$ and some $V\in\tau$ such that $K\subset E\subset V$ and $\map{\mu}{V-K}<\varepsilon$.

Proof
Let $E\in\MM_F$. Then, by definition of $\mu$ and $\MM_F$, there exist compact $K\subset E$ and open $V\supset E$ such that
 * $\map{\mu}{V}-\frac{\varepsilon}{2}<\map{\mu}{E}<\map{\mu}{K}+\frac{\varepsilon}{2}$.

By compact subspace of Hausdorff space is closed, $V-K\in\tau$.

So, by Lemma 5, $V-K\in\MM_F$.

By Lemma 4,
 * $\map{\mu}{V-K}=\mu{V}-\mu{K}<\epsilon$.

Lemma $7$
The union, if of finite measure, of countable pairwise disjoint subsets of $\MM_F$ is in $\MM_F$.

Proof
Let $\sequence{E_i}\in\paren{\MM_F}^\N$ be pairwise disjoint.

Suppose $\ds E=\bigcup_{i=1}^\infty E_i$ has finite measure.

By Lemma 4, for all $\varepsilon>0$, there exists some $N\in\N$ such that
 * $\ds\map{\mu}{E}\le\varepsilon\sum_{i=1}^\infty\map{\mu}{E_i}$.

By definition of $\MM_F$, for all $i$, there exists some compact $H_i\subset E_i$ such that
 * $\map{\mu}{H_i}>\map{\mu}{E_i}-2^{-i}\varepsilon.$

Then,
 * $\ds\map{\mu}{E}\le\map{\mu}{\bigcup_{i=1}^N H_i}+2\varepsilon.$

By Finite union of compact sets is compact, $\bigcup_{i=1}^N\subset E$.

Therefore, $E\in\MM_F$.

Lemma $8$
$\MM_F$ is closed under set difference, union, and intersection.

Proof
Suppose $\paren{A,B}\in\paren{\MM_F}^2$.

By Lemma 6, there exist compact sets $K_1,K_2$ and open sets $V_1,V_2$ such that
 * $K_1\subset A\subset V_1, K_2\subset B\subset V_2, \text{ and } \forall i\in\set{1,2},\map{\mu}{V_i-K_i}<\frac{\varepsilon}{2}$.

Now,
 * $A-B\subset V_1-K_2\subset (V_1-K_1)\cup(K_1-V_2)\cup(V_2-K_2)$.

So, by Lemma 1,
 * $\map{\mu}{A_B}\le\map{\mu}{K_1-V_2}+\varepsilon$.

By closed subset of compact set is compact, $K_1-V_2$ is compact.

Since $K_1-V_2\subset A-B$, there exist compact subsets of $A-B$ arbitrarily close in measure to $A-B$.

So, $A-B\in\MM_F$.

Now, by Lemma 7,
 * $A\cup B=\paren{A-B}\cup B\in\MM_F$

and
 * $A\cap B=A-\paren{A-B}\in\MM_F$.

Lemma $9$
$\MM_F=\set{E\in\MM:\map{\mu}{E}<\infty}$

Proof
By Lemma 2 and Lemma 8,
 * $E\in\MM_F\implies \forall K\text{ compact}, E\cap K\in\MM_F\implies E\in\M$.

That is, $\MM_F\subset\MM$.

Conversely, suppose $E\in\MM$, $\map{\mu}{E}<\infty$, and $\epsilon>0$.

By definition of $\mu$, there exists an open $V\supset E$ such that $\map{\mu}{V}<\infty$.

By Lemma 5 and Lemma 6, there exists a compact $K\subset V$ such that $\map{\mu}{V-K}<\varepsilon$.

By definition of $\MM$, $E\cap K\in\MM_F$.

So, by definition of $\MM_F$, there exists a compact $H\subset E\cap K$ such that
 * $\map{\mu}{E\cap K}<\map{\mu}{H}+\varepsilon$.

Since $E\subset\paren{E\cap K}\cup\paren(V-K)$, by Lemma 1,
 * $\map{\mu}{E}\le\map{\mu}{E\cap K}+\map{\mu}{V-K}+2\varepsilon$.

Thus, $E\in\MM_F$.

Therefore, $\MM_F=\set{E\in\MM:\map{\mu}{E}<\infty}$.

Let $K\subset X$ be compact.

Suppose $A\in\MM$.

Then, by Lemma 8,
 * $A^C\cap K=K-\paren(A\cap K)\in\MM_F$.

So, $\MM$ is closed under complement.

Let $\sequence{A_n}\in\MM^\N$ and $A=\ds\bigcup_{i=1}^\infty$. Let $B_1=A_1\cap K$ and
 * $B_n=\paren{A_n\cap K}-\ds\bigcup_{i=1}^{n-1}B_i, \qquad n\ge2$.

$\sequence{B_n}$ is disjoint and, by Lemma 8, a sequence of $\MM_F$.

Then, by Lemma 7,
 * $A\cap K=\ds\bigcup_{i=1}^\infty\in\MM_F$.

Therefore, $M$ is closed under countable union.

Now,
 * $C\text{ closed}\implies C\cap K\text{ compact} \implies C\cap K\in\MM_F\implies C\in\MM$.

Therefore, by Borel $\sigma$-algebra generated by closed sets, $\MM$ is $\sigma$-algebra which contains the Borel $\sigma$-algrebra of $\struct{X,\tau}$.

Now, by Lemma 2, the definition of $\mu$, Lemma 9, and Lemma 4, $\mu$ is Radon measure on $\MM$.

Let $f\in\map{C_c}{X}$ be real.

Let $K=\supp f$ and $\varepsilon>0$.

Let $\paren{a,b}\in\R^2$ be such that $\closedint{a}{b}\supset K$.

Choose $\sequence{y_i}\in\R^n$ such that for all $i$, $y_i-y_{i-1}<\varepsilon$ and
 * $y_0<a<y_1<\cdots<y_n=b$.

Define
 * $E_i=\map{f^{-1}}{\hointl{y_{i-1}}{y_1}}\cap K$.

By definition of $\mu$, there exist open sets $W_i\supset E_i$ such that $\map{\mu}{W_i}<\map{\mu}{E_i}+\varepsilon/n$.

Define
 * $V_i=W_i\cap\map{f^{-1}}{\openint{-\infty}{y_i+\varepsilon}}$.

Let $\set{h_i}$ be a partition of untiy on $K$ subordinate to $\set{V_i}$.

Then $f=\sum h_i f$ on $K$ and, by Lemma 2,
 * $\ds\map{\mu}{K}\le\Lambda\sum_{i=1}^n h_i=\sum_{i=1}^n \Lambda h_i$.

Since, for all $i$, for all $x\in E_i$, $h_i f\le\paren{y_i+\varepsilon}h_i$ and $y_i-\varepsilon<\map{f}{x}$,

Since $\varepsilon$ was arbitrary, for all real $f\in\map{C_c}{X}$, $\Lambda f \le \int_X f\dr\mu$.

Now, by linearity of $\Lambda$,
 * $-\Lambda f = \Lambda(-f)\le\int_X\paren{-f}\rd\mu=-\int_X f\rd\mu$.

Therefore, for all real $f\in\map{C_c}{X}$,
 * $\Lambda f = \int_X f\rd\mu$.

Suppose $u,v\in\map{C_c}{X}$ are real. Then by linearity of $\Lambda$,
 * $\map{\Lambda}{u+iv}=\Lambda u +i\Lambda v=\int_X u\rd\mu +i\int_X v\rd\mu=\int_X\paren{u+iv}\rd\mu$.

That is, equality holds for all functions in $\map{C_c}{X}$.

Suppose $\mu_1$ and $\mu_2$ are two Radon measures on $\MM$.

For all compact $K\subset X$ and $\varepsilon>0$, there exists an open $V\supset K$ such that
 * $\map{\mu_2}{V}<\map{\mu_2}{K}+\varepsilon$.

By Urysohn's lemma, there exists an $f\in\map{C_c}{X}:K\prec f\prec V$.

Then

Thus, $\map{\mu_1}{K}\le\map{\mu_2}{K}$. Interchanging $\mu_1$ and $\mu_2$ yields the opposite inequality. Therefore, $\mu_1$ and $\mu_2$ coincide on all compact sets.

By the definition of a Radon measure, the measure of all measurable sets is uniquely determined by the measures of the compact sets.

So, $\mu_1=\mu_2$ and the measure is unique, completing the proof.