# Riesz-Markov-Kakutani Representation Theorem

## Theorem

Let $\struct {X, \tau}$ be a locally compact 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:

$\ds \forall f \in \map {C_c} X: \Lambda f = \int_X f \rd \mu$

### Notation

For an open set $V \in \tau$ and a mapping $f \in \map {C_c} X$:

$f \prec V \iff \supp f \subset V$

where $\supp f$ denotes the support of $f$.

For a compact set $K \subset X$ and a mapping $f \in \map {C_c} X$:

$K \prec f \iff \forall x \in K: \map f x = 1$

### Construction of $\mu$ and $\MM$

For every $V \in \tau$, define:

$\map {\mu_1} V = \sup \set {\Lambda f: f \prec V}$

Note that $\mu_1$ is monotonically increasing.

That is, for all $V, W \in \tau$ such that $V \subset W$, we have:

 $\ds \map {\mu_1} V$ $=$ $\ds \sup \set {\Lambda f: \supp f \subset V}$ $\ds$ $\le$ $\ds \sup \set {\Lambda f: \supp f \subset W}$ $\ds = \map {\mu_1} W$

$\Box$

For every other subset $E \subset X$, define:

$\map \mu E = \inf \set {\map {\mu_1} V: V \supset E \land V \in \tau}$

Since $\mu_1$ is monotonically increasing:

$\mu_1 = \mu {\restriction_\tau}$

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:

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

## Proof

### Lemma $1$

$\mu$ is countably subadditive.

$\Box$

### Lemma $2$

Let $K \subset X$ be compact.

Then:

$K \in \MM_F$

and:

$\map \mu K = \inf \set {\Lambda f : K \prec f}$

$\Box$

### Lemma $3$

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

$\Box$

### Lemma $4$

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

$\Box$

### Lemma $5$

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

$\Box$

### Lemma $6$

For all $E \in \MM_F$ and $\epsilon \in \R_{>0}$, there exist some compact $K \in X$ and some $V \in \tau$ such that:

$K \subset E \subset V$

and:

$\map \mu {V \setminus K} < \epsilon$

$\Box$

### Lemma $7$

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

$\Box$

### Lemma $8$

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

$\Box$

### Lemma $9$

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

$\Box$

Let $K\subset X$ be compact.

Suppose $A \in \MM$.

Then, by Lemma $8$:

$A^C \cap K = K \setminus \paren {A \cap K} \in \MM_F$

So, $\MM$ is closed under complement.

Let $\sequence {A_n} \in \MM^\N$ and $A = \ds \bigcup_{i \mathop = 1}^\infty$.

Let $B_1 = A_1 \cap K$ and:

for $n \ge 2: B_n = \paren {A_n \cap K} \setminus \ds \bigcup_{i \mathop = 1}^{n - 1} B_i$

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

Then, by Lemma $7$:

$A \cap K = \ds \bigcup_{i \mathop = 1}^\infty \in \MM_F$

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

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

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

Let $K = \supp f$ and $\epsilon \in \R_{>0}$.

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

Choose $\sequence {y_i} \in \R^n$ such that for all $i$, $y_i - y_{i - 1} < \epsilon$ 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:

$\map \mu {W_i} < \map \mu {E_i} + \dfrac \epsilon n$

Define:

$V_i = W_i \cap \map {f^{-1} } {\openint \gets {y_i + \epsilon} }$

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

Then $\ds f = \sum h_i f$ on $K$ and, by Lemma 2:

$\ds \map \mu K \le \Lambda \sum_{i \mathop = 1}^n h_i = \sum_{i \mathop = 1}^n \Lambda h_i$

Since, for all $i$, for all $x \in E_i$:

$h_i f \le \paren {y_i + \epsilon} h_i$

and:

$y_i - \epsilon < \map f x$

we have:

 $\ds \Lambda f$ $=$ $\ds \sum_{i \mathop = 1}^n \map \Lambda {h_i f}$ $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \paren {y_i + \epsilon} \Lambda h_i$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {\size a + y_i + \epsilon} \Lambda h_i - \size a \sum_{i \mathop = 1}^n \Lambda h_i$ $\ds$ $\le$ $\ds \sum_{i \mathop = 1}^n \paren {\size a + y_i + \epsilon} \paren {\map \mu {E_i} + \dfrac \epsilon n} - \size a \map \mu K$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \paren {y_i - \epsilon} + 2 \epsilon \map \mu K + \frac \epsilon n \sum_{i \mathop = 1}^n \paren {\size a + y_i + \epsilon}$ $\ds$ $\le$ $\ds \int_X f \rd \mu + \epsilon \paren {2 \map \mu K + \size a + b + \epsilon}$

Since $\epsilon$ was arbitrary, for all real $f \in \map{C_c} X$:

$\Lambda f \le \int_X f \rd \mu$

Now, by linearity of $\Lambda$:

$-\Lambda f = \map \Lambda {-f} \le \int_X \paren {-f} \rd \mu = -\int_X f \rd \mu$

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 $\Lambda$:

$\map \Lambda {u + i v} = \Lambda u + i \Lambda v = \int_X u \rd \mu + i \int_X v \rd \mu = \int_X \paren {u + i v} \rd \mu$
 $\ds \map \Lambda {u + i v}$ $=$ $\ds \Lambda u + i \Lambda v$ $\ds$ $=$ $\ds \int_X u \rd \mu + i \int_X v \rd \mu$ $\ds$ $=$ $\ds \int_X \paren {u + i v} \rd \mu$

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 $\epsilon>0$, there exists an open $V \supset K$ such that:

$\map {\mu_2} V < \map {\mu_2} K + \epsilon$

By Urysohn's Lemma, there exists:

$f \in \map{C_c} X: K \prec f \prec V$

Then:

 $\ds \map {\mu_1} K$ $=$ $\ds \int_X \chi_K \rd \mu_1$ $\ds$ $\le$ $\ds \int_X f \rd \mu_1$ $\ds$ $=$ $\ds \Lambda f$ $\ds$ $=$ $\ds \int_X f \rd \mu_2$ $\ds$ $\le$ $\ds \int_X \chi_V \rd \map {\mu_2} V$ $\ds$ $=$ $\ds \map {\mu_2} V$ $\ds$ $<$ $\ds \map {\mu_2} K + \epsilon$

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.

$\blacksquare$

## Source of Name

This entry was named for Frigyes RieszAndrey Andreyevich Markov Jr. and Shizuo Kakutani.